Post by thread:[EE]: PIC overheating a 7805 !

Hi, I've got this system up and running where a series of modules containing 16F628 (running @ 20MHz) and individually fed by LM7805 (TO-220 package) are all run from the same 24VDC line (long run) coming from a switching supply. Each module also has an LM7806 for driving an R/C Servo motor. All my modules work fine except that the 7805s are running rather hot even though the R/C motors are pulling much more current from the 7806 than the little PIC is from the 7805. The 7806's temperature is ok though. At first I thought that it would only be necessary to heat-sink the motor's regulator but know I think I might have to do the same with the PIC's. I can almost burn my finger from the 7805 ... I understand that 24V isn't the optimal voltage to run the 7805 from if I intended to draw a lot of current from it but National's datasheet says that it can take up to 35V and considering the very low current consumption I don't understand why it is overheating. I suspected the length of my supply line but running a module even a few minutes on a short length gives the same result. Besides, the voltage drop is barely perceptible and I have pretty good wire gauge. Could my choice of caps for the regulators be causing the problem ? I have one 470uF 35V electrolytic before the 2 regulators and a 0.1uF ceramic cap after each of them. thanks, Tobie Horswill spam_OUTthorswilTakeThisOuTexmachina.qc.ca -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

At 10:07 PM 08/11/2001 -0500, you wrote: {Quote hidden}>Hi, I've got this system up and running where a series of modules >containing 16F628 (running @ 20MHz) and individually fed by LM7805 (TO-220 >package) are all run from the same 24VDC line (long run) coming from a >switching supply. Each module also has an LM7806 for driving an R/C Servo >motor. All my modules work fine except that the 7805s are running rather >hot even though the R/C motors are pulling much more current from the 7806 >than the little PIC is from the 7805. The 7806's temperature is ok though. >At first I thought that it would only be necessary to heat-sink the >motor's regulator but know I think I might have to do the same with the >PIC's. I can almost burn my finger from the 7805 ... > >I understand that 24V isn't the optimal voltage to run the 7805 from if I >intended to draw a lot of current from it but National's datasheet says >that it can take up to 35V and considering the very low current >consumption I don't understand why it is overheating. I suspected the >length of my supply line but running a module even a few minutes on a >short length gives the same result. Besides, the voltage drop is barely >perceptible and I have pretty good wire gauge. Could my choice of caps for >the regulators be causing the problem ? I have one 470uF 35V electrolytic >before the 2 regulators and a 0.1uF ceramic cap after each of them. Maybe the 7805 is oscillating? You should have a 10uF capacitor on the 7805 output as well. Regards... -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

I had a similar problem with a 7805 running from a 24V source, the load was a 555 timer, 4020, 2716 eprom and a 74ls373 driving 3 opto triacs IIRC. not much load, but the 7805 had to be mounted on a 1.0oC/W heatsink to keep it working, since then I have mad a design rule, all 7805's are fed from 8 Vdc. I know they have a wide input voltage (I have stack (not sure of PN) that were 7805 compatible that had 120VDC input rating). These days unless cost is an issue I either use a multi tap transformer, or I use switching regulators. Regards, Kat. ____________________________________________________________________________ /"\ ASCII Ribbon Campaign | K.A.Q. Electronics \ / - NO HTML/RTF in e-mail | Software and Electronic Engineering X - NO Word docs in e-mail | Perth Western Australia / \ | Ph +61 419 923 731 ____________________________________________________________________________ {Original Message removed}

Kathy Wrote: > I had a similar problem with a 7805 running from a 24V source, the load was > a 555 timer, 4020, 2716 eprom and a 74ls373 driving 3 opto triacs IIRC. not > much load, but the 7805 had to be mounted on a 1.0oC/W heatsink to keep it > working, since then I have mad a design rule, all 7805's are fed from 8 Vdc. > I know they have a wide input voltage (I have stack (not sure of PN) that > were 7805 compatible that had 120VDC input rating). I see problems like this all the time and it bears going back and understanding what is happening. As a rough guess, I would think the circuit was drawing 300 to 350 mA (assuming the 555 is bipolar and 2716 is NMOS) - according to the datasheets, well within the output range of the 7805. But, remembering that the formula for power dissipated is: Power = Voltage Across the device * Current In your application, the 7805 has 19 Volts across it (24 Volts in - 5 Volts out) so the total power dissipated in the 7805 is: Power = 19 Volts * 300 mA = 5.7 Watts This power level is going to put the 7805 into thermal shutdown (crowbar) pretty quickly. > These days unless cost is an issue I either use a multi tap transformer, or > I use switching regulators. Maxim has some nice regulator chips that are pretty inexpensive - on a production board, a power supply using these chips would cost you about as much as a 7805 and you don't have the heat issues (and if you are running from batteries, the circuit will run a *lot* longer). myke -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

> Subject: [EE]: PIC overheating a 7805 ! > From: TH (.....thorswilKILLspam@spam@VIDEOTRON.CA) > Date: Thu Nov 08 2001 - 19:07:20 PST > Hi, I've got this system up and running where a series of > modules containing 16F628 (running @ 20MHz) and ... [snip] Well, from 24Vdc the Vdrop will be 19V, even if your circuit is draining only 200mA, it means 3.8Watts. Without a proper heatsink this can be hot. Oscillation can be an issue, but not likely. The 7806 does not heat so much, probably because the servors could work in intermitent fashion so the thermal average would be low. The Microchip datasheet shows 7mA @ 20MHz, what would heats the 7805 @ less than 150mW. However, the maximum current into VDD could reach @ 250mA before it blows up the chip. Check your design, make sure no late night soldering got a short circuit at any unused port pin or by mistake in soldering or design. Also check the crystal oscillator for a proper signal shape and level. Just for testing, change the crystal for a 4MHz and see what happens, perhaps a not so good 20MHz units lot? Also check Microchip for 16F628 Errata. By any analogy, what is sinking current to heats the 7805 is also heating at least @ 5V, so it should heat 1/4 of the 7805 thermal reference. A good wet finger sometimes can sense what component is doing this. In case you have an ampermeter, try and measure everything. In case you have a thermal noncontact measuring gun, use it, or simply point your video camera to the unit in a dark room, you will be surprise what the camera actually can shows you as thermal image. Most B&W video cameras are better sensitive to IR than the color ones... have fun. Wagner Lipnharski http://www.ustr.net -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Sometime you gotta love this list. Sometime people go right out and let the smoke out of their bits long before your iron is hot. That and someone blurts out things that save time n money. Just so I'm on topic slightly. We never design anything with a 7805 that doesn't have a head voltage in the 3-5 range. Preferably 3 if its going to pull near rated current. I have run one with 18.5 head and it ran red hot even with a huge sink on it. It was one of the least reliable designs I let out. >Maxim has some nice regulator chips that are pretty inexpensive - on a >production board, a power supply using these chips would cost you about as >much as a 7805 and you don't have the heat issues (and if you are running >from batteries, the circuit will run a *lot* longer). > >myke Now this just made my day. I had a quick look at some of Maxims stuff and found one of these that eliminates me having to design two separate power supplies. I can heat sink it for worst case and laugh. You know I think I've learned two things from this list this month. That's almost scary ;-] As a direct result I didn't have to think nearly as hard and can now afford to kill off those extra brain cells with beer this weekend. Cheers Dave -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Hi, try LM2574 - fine, cheap, cool! Imre +-----------------------------------------------------------------------+ | The information transmitted is intended only for the person or entity | | to which it is addressed and may contain confidential and/or | | privileged material. Any review, retransmission, dissemination or | | other use of, or taking of any action in reliance upon, this | | information by persons or entities other than the intended recipient | | is prohibited. If you received this in error, please contact the | | sender and delete the material from any computer. | +-----------------------------------------------------------------------+ On Fri, 9 Nov 2001, Kathy Quinlan wrote: {Quote hidden}> I had a similar problem with a 7805 running from a 24V source, the load was > a 555 timer, 4020, 2716 eprom and a 74ls373 driving 3 opto triacs IIRC. not > much load, but the 7805 had to be mounted on a 1.0oC/W heatsink to keep it > working, since then I have mad a design rule, all 7805's are fed from 8 Vdc. > I know they have a wide input voltage (I have stack (not sure of PN) that > were 7805 compatible that had 120VDC input rating). > > These days unless cost is an issue I either use a multi tap transformer, or > I use switching regulators. > > Regards, > > Kat. > > > ____________________________________________________________________________ > /"\ ASCII Ribbon Campaign | K.A.Q. Electronics > \ / - NO HTML/RTF in e-mail | Software and Electronic Engineering > X - NO Word docs in e-mail | Perth Western Australia > / \ | Ph +61 419 923 731 > ____________________________________________________________________________ > > > {Original Message removed}

Am I the only person who has just received multiple copies of this email ?? ----- Original Message ----- From: "Dave King" <KingDWSKILLspamHOME.COM> To: <.....PICLISTKILLspam.....MITVMA.MIT.EDU> Sent: Friday, November 09, 2001 4:22 PM Subject: Re: [EE]: PIC overheating a 7805 ! > Sometime you gotta love this list. Sometime people go right out and let the > smoke > out of their bits long before your iron is hot. That and someone blurts out {Quote hidden}> things that > save time n money. > > Just so I'm on topic slightly. We never design anything with a 7805 that > doesn't > have a head voltage in the 3-5 range. Preferably 3 if its going to pull > near rated > current. I have run one with 18.5 head and it ran red hot even with a huge > sink on > it. It was one of the least reliable designs I let out. > > > >Maxim has some nice regulator chips that are pretty inexpensive - on a > >production board, a power supply using these chips would cost you about as > >much as a 7805 and you don't have the heat issues (and if you are running > >from batteries, the circuit will run a *lot* longer). > > > >myke > > Now this just made my day. I had a quick look at some of Maxims stuff and > found > one of these that eliminates me having to design two separate power supplies. > I can heat sink it for worst case and laugh. > > You know I think I've learned two things from this list this month. That's > almost scary ;-] > As a direct result I didn't have to think nearly as hard and can now afford {Quote hidden}> to kill off > those extra brain cells with beer this weekend. > > Cheers > > Dave > > -- > http://www.piclist.com hint: The list server can filter out subtopics > (like ads or off topics) for you. See http://www.piclist.com/#topics > > > -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

On Fri, 9 Nov 2001, Dave King wrote: > Sometime you gotta love this list. Sometime people go right out and let the > smoke > out of their bits long before your iron is hot. That and someone blurts out > things that > save time n money. > > Just so I'm on topic slightly. We never design anything with a 7805 that > doesn't > have a head voltage in the 3-5 range. Preferably 3 if its going to pull > near rated 3V it's not enough ! Do a simple computation for a 220V +10% -15% ( or 120V, whatever...) mains variation and see you need at least 4V in-out to be sure with output ripple. Better are: LM low dropout series ( 1.25...1.5 V in-out ) KA78R05 ( 0.5...1V in-out) and of course the standard 7805 with good design consideration up to +35V input. regards, Vasile -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Tobie said - Hi, I've got this system up and running where a series of modules containing 16F628 (running @ 20MHz) and individually fed by LM7805 (TO-220 package) are all run from the same 24VDC line (long run) coming from a switching supply. Each module also has an LM7806 for driving an R/C Servo motor. All my modules work fine except that the 7805s are running rather hot even though the R/C motors are pulling much more current from the 7806 than the little PIC is from the 7805. The 7806's temperature is ok though. At first I thought that it would only be necessary to heat-sink the motor's regulator but know I think I might have to do the same with the PIC's. I can almost burn my finger from the 7805 ... I understand that 24V isn't the optimal voltage to run the 7805 from if I intended to draw a lot of current from it but National's datasheet says that it can take up to 35V and considering the very low current consumption I don't understand why it is overheating. I suspected the length of my supply line but running a module even a few minutes on a short length gives the same result. Besides, the voltage drop is barely perceptible and I have pretty good wire gauge. Could my choice of caps for the regulators be causing the problem ? I have one 470uF 35V electrolytic before the 2 regulators and a 0.1uF ceramic cap after each of them. ____________ It doesn't SOUND like it makes sense. IF the 6v supply IS drawing more current from the same input then it DOESN'T make sense. - As others note, ensure the regulator is not oscillating. A scope helps a lot for this. Make sure caps are per the databook. - A 7805 / LM340 in a TO220 package is good for about 1 watt (preferably less) without a heatsink. At 24 volts in, 5v out this gives Current = 1 watt / (24-5) = ~50 mA. A PIC by itself will draw much less than that. - Check that the 5 v supply is not drawing more than you think. - A series resistor will take the load off the regulator. eg at 100 mA and 15v drop a 15/.1 = 150 ohm resistor will dissipate V^2/R = 225/150 = 1.5 watts and allow the regulator to dissipate (24-15-5) * 0.1 = 0.4 watt. If you place a 150 ohm resistor in series and the regulator can't support the 5v output then it is drawing rather more than 100 mA. Note that you need a larger decoupling capacitor at the regulator input if you use the series input resistor. Using a 5 watt wire wound here will move most of the dissipation out of the regulator. If the 24v supply varies to below 24v you will have to design the resistor for the lowest input voltage. Rseries = (Vin_minimum - 5v - Vheadroom) / (Imaximum) V headroom is the minimum voltage you want across the regulator. For a 7805 2.0v is safe. How's the Autonomous Boat coming? regards Russell McMahon -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

24V is a bit high! Calculate out the power dissipation (24V-5V)*LoadCurrent. Take this (in watts) and multiply by the rating of your heat sink in degrees C/ watt. This should give temperature rise from ambient. Another thing to watch for is oscillating regulators, which get hot. Check the output on a scope and make sure you've got DC without a bunch of RF. Finally, consider a switching regulator. The inefficiency of using a linear regulator to go from 24V down to 5V is pretty impressive. Harold FCC Rules Online at http://hallikainen.com/FccRules Lighting control for theatre and television at http://www.dovesystems.com ________________________________________________________________ GET INTERNET ACCESS FROM JUNO! Juno offers FREE or PREMIUM Internet access for less! Join Juno today! For your FREE software, visit: dl.http://www.juno.com/get/web/. -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

On Fri, 9 Nov 2001 08:26:27 -0600 Mark Peterson <EraseMEmarkpspam_OUTTakeThisOuTCANNONTECH.COM> writes: > An easy way to unoad the regulator is to put a series resistor in the > input. Let it dissipate as much of the heat as possible. Size it > such > that you always maintain the minimum input voltage required by the > regulator. If, for example, the 7805 needs a minimum of 7 volts at > its > input and you have a 24 volt source, use a resistor that will drop > 17 volts > at your load's maximum current draw. > Or a series zener diode so the input to the regulator does not vary as much with load... Harold FCC Rules Online at http://hallikainen.com/FccRules Lighting control for theatre and television at http://www.dovesystems.com ________________________________________________________________ GET INTERNET ACCESS FROM JUNO! Juno offers FREE or PREMIUM Internet access for less! Join Juno today! For your FREE software, visit: dl.http://www.juno.com/get/web/. -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Your problem is very simple. You are dropping 19V accross the regulator. Multiply that by the current and you will see that you need to dissipate considerable wattage in the regulator. I have several devices that need to run on 24 or 12V and use a 7805 in an enclosure with no way to use a heat sink. I put a 12V zeener in series with the input positive supply before the regulator. I put a jumper on the board in parallel with the zeener so customers can cut the jumper if they are using a 24V supply and leave it if they have a 12V supply. Use a 5W zeener and try to have it as far as possible from the regulator. I also try to have big fat traces going to the regulator and the zeener to conduct the hear away from it also. Power Trends(now owned by TI) mini switching regulators are also great if your project can justify the fourteen bucks a pop. TH wrote: {Quote hidden}>Hi, I've got this system up and running where a series of modules containing 16F628 (running @ 20MHz) and individually fed by LM7805 (TO-220 package) are all run from the same 24VDC line (long run) coming from a switching supply. Each module also has an LM7806 for driving an R/C Servo motor. All my modules work fine except that the 7805s are running rather hot even though the R/C motors are pulling much more current from the 7806 than the little PIC is from the 7805. The 7806's temperature is ok though. At first I thought that it would only be necessary to heat-sink the motor's regulator but know I think I might have to do the same with the PIC's. I can almost burn my finger from the 7805 ... > >I understand that 24V isn't the optimal voltage to run the 7805 from if I intended to draw a lot of current from it but National's datasheet says that it can take up to 35V and considering the very low current consumption I don't understand why it is overheating. I suspected the length of my supply line but running a module even a few minutes on a short length gives the same result. Besides, the voltage drop is barely perceptible and I have pretty good wire gauge. Could my choice of caps for the regulators be causing the problem ? I have one 470uF 35V electrolytic before the 2 regulators and a 0.1uF ceramic cap after each of them. > >thanks, > >Tobie Horswill >thorswilspam_OUTexmachina.qc.ca > >-- >http://www.piclist.com hint: PICList Posts must start with ONE topic: >[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads > > > -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

The power used and dissipated by the voltage regulator is NOT calculated that way. the power dissipated by the device is approx. = (Vin-Vout) * quiescent current and NOT load current. Check the data sheets! @spam@haroldhallikainenKILLspamJUNO.COM on 09/11/2001 18:44:56 Please respond to KILLspamPICLISTKILLspamMITVMA.MIT.EDUKILLspamINTERNETKILLspamwtgw To: RemoveMEPICLISTTakeThisOuTMITVMA.MIT.EDUTakeThisOuTINTERNETTakeThisOuTwtgw cc: Paper Mail: Subject: Re: [EE]: PIC overheating a 7805 ! *** WARNING *** This mail has originated outside your organization, either from an external partner or the Global Internet. Keep this in mind if you answer this message. 24V is a bit high! Calculate out the power dissipation (24V-5V)*LoadCurrent. Take this (in watts) and multiply by the rating of your heat sink in degrees C/ watt. This should give temperature rise from ambient. Another thing to watch for is oscillating regulators, which get hot. Check the output on a scope and make sure you've got DC without a bunch of RF. Finally, consider a switching regulator. The inefficiency of using a linear regulator to go from 24V down to 5V is pretty impressive. Harold FCC Rules Online at http://hallikainen.com/FccRules Lighting control for theatre and television at http://www.dovesystems.com ________________________________________________________________ GET INTERNET ACCESS FROM JUNO! Juno offers FREE or PREMIUM Internet access for less! Join Juno today! For your FREE software, visit: dl.http://www.juno.com/get/web/. -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics ******************************************************************** This email and any attachments are confidential to the intended recipient and may also be privileged. If you are not the intended recipient please delete it from your system and notify the sender. You should not copy it or use it for any purpose nor disclose or distribute its contents to any other person. ******************************************************************** -- http://www.piclist.com hint: To leave the PICList spamBeGonepiclist-unsubscribe-requestspamBeGonemitvma.mit.edu

----- Original Message ----- From: "Mike Blakey" <TakeThisOuTmike.blakeyEraseMEspam_OUTBAESYSTEMS.COM> To: <RemoveMEPICLISTTakeThisOuTMITVMA.MIT.EDU> Sent: Monday, November 12, 2001 8:58 AM Subject: Re: [EE]: PIC overheating a 7805 ! > The power used and dissipated by the voltage regulator is NOT calculated that way. > the power dissipated by the device is approx. = (Vin-Vout) * quiescent current and > NOT load current. Check the data sheets! > >I beg to differ. I think you mix up power consumption and power diddipation. Power consumption is what the regulator uses to do the regulation, so to speak. That is very little indeed. But the dissipation is (Vin-Vout)*load current (plus the power consumption of course, but that is normally insignificant). Look at it as a (variable) resistor with a voltage difference of Vin -Vout, and a current = load current. If it was us you seem to think, no regulator would ever become hot and Ohms law would be violated... Regards, jan Didden {Quote hidden}> > > > haroldhallikainenEraseME.....JUNO.COM on 09/11/2001 18:44:56 > Please respond to EraseMEPICLISTMITVMA.MIT.EDUINTERNETwtgw > > To: RemoveMEPICLISTEraseMEEraseMEMITVMA.MIT.EDUEraseMEEraseMEINTERNETEraseMEEraseMEwtgw > cc: > Paper Mail: > Subject: Re: [EE]: PIC overheating a 7805 ! > > *** WARNING *** > > This mail has originated outside your organization, > either from an external partner or the Global Internet. > Keep this in mind if you answer this message. > > > 24V is a bit high! Calculate out the power dissipation > (24V-5V)*LoadCurrent. Take this (in watts) and multiply by the rating of > your heat sink in degrees C/ watt. This should give temperature rise from > ambient. > Another thing to watch for is oscillating regulators, which get hot. {Quote hidden}> Check the output on a scope and make sure you've got DC without a bunch > of RF. > Finally, consider a switching regulator. The inefficiency of using a > linear regulator to go from 24V down to 5V is pretty impressive. > > Harold > > > FCC Rules Online at http://hallikainen.com/FccRules > Lighting control for theatre and television at http://www.dovesystems.com > > ________________________________________________________________ > GET INTERNET ACCESS FROM JUNO! > Juno offers FREE or PREMIUM Internet access for less! > Join Juno today! For your FREE software, visit: > dl.http://www.juno.com/get/web/. > > -- > http://www.piclist.com hint: The list server can filter out subtopics > (like ads or off topics) for you. See http://www.piclist.com/#topics > > > > > > > > > ******************************************************************** > This email and any attachments are confidential to the intended > recipient and may also be privileged. If you are not the intended > recipient please delete it from your system and notify the sender. > You should not copy it or use it for any purpose nor disclose or > distribute its contents to any other person. > ******************************************************************** > > -- > http://www.piclist.com hint: To leave the PICList > RemoveMEpiclist-unsubscribe-requestspam_OUTKILLspammitvma.mit.edu > > -- http://www.piclist.com hint: To leave the PICList RemoveMEpiclist-unsubscribe-requestTakeThisOuTspammitvma.mit.edu

Sorry Mike, but is this case your datasheet is wrong. Any device working as linear device ( not switching ) is dissipating delta_V * I where delta_v is in-out difference and I is the current flow through device. ( Ic and Uce at transistors, Uin_out and Iload at 7805 ) Regards, Vasile On Mon, 12 Nov 2001, Mike Blakey wrote: {Quote hidden}> The power used and dissipated by the voltage regulator is NOT calculated that way. > the power dissipated by the device is approx. = (Vin-Vout) * quiescent current and > NOT load current. Check the data sheets! > > > > > > EraseMEharoldhallikainenspamspamBeGoneJUNO.COM on 09/11/2001 18:44:56 > Please respond to RemoveMEPICLISTKILLspamMITVMA.MIT.EDUKILLspamINTERNETKILLspamwtgw > > To: PICLISTSTOPspamspam_OUTMITVMA.MIT.EDUSTOPspamspam_OUTINTERNETSTOPspamspam_OUTwtgw > cc: > Paper Mail: > Subject: Re: [EE]: PIC overheating a 7805 ! > > *** WARNING *** > > This mail has originated outside your organization, > either from an external partner or the Global Internet. > Keep this in mind if you answer this message. > > > 24V is a bit high! Calculate out the power dissipation > (24V-5V)*LoadCurrent. Take this (in watts) and multiply by the rating of > your heat sink in degrees C/ watt. This should give temperature rise from > ambient. > Another thing to watch for is oscillating regulators, which get hot. > Check the output on a scope and make sure you've got DC without a bunch > of RF. > Finally, consider a switching regulator. The inefficiency of using a > linear regulator to go from 24V down to 5V is pretty impressive. > > Harold > > > FCC Rules Online at http://hallikainen.com/FccRules > Lighting control for theatre and television at http://www.dovesystems.com > > ________________________________________________________________ > GET INTERNET ACCESS FROM JUNO! > Juno offers FREE or PREMIUM Internet access for less! > Join Juno today! For your FREE software, visit: > dl.http://www.juno.com/get/web/. > > -- > http://www.piclist.com hint: The list server can filter out subtopics > (like ads or off topics) for you. See http://www.piclist.com/#topics > > > > > > > > > ******************************************************************** > This email and any attachments are confidential to the intended > recipient and may also be privileged. If you are not the intended > recipient please delete it from your system and notify the sender. > You should not copy it or use it for any purpose nor disclose or > distribute its contents to any other person. > ******************************************************************** > > -- > http://www.piclist.com hint: To leave the PICList > spamBeGonepiclist-unsubscribe-requestSTOPspamEraseMEmitvma.mit.edu > > -- http://www.piclist.com hint: To leave the PICList KILLspampiclist-unsubscribe-requestspamBeGonemitvma.mit.edu

Hi Mike, I have searched data sheets from natsemi and TI, and cannot see how to calculate power disapation. Could you please provide a reference. I disagree with your comment, to me a linear device working in a linear region has to dissipate the power not consumed by the load ie If I have 24 volts in, 5volt out to the load, load consumes say 1 amp. Power into regulator is 24W, power out is 5W, so the regulator must consume the missing 19W. With your idea, the max power consumption would be 152mW ( vin - vout * Qc) = (24 -5) *0.008A (from the TI data sheet maximum value) Regards, Kat. ____________________________________________________________________________ /"\ ASCII Ribbon Campaign | K.A.Q. Electronics \ / - NO HTML/RTF in e-mail | Software and Electronic Engineering X - NO Word docs in e-mail | Perth Western Australia / \ | Ph +61 419 923 731 ____________________________________________________________________________ {Original Message removed}

At 20:32 11/12/2001 +0800, Kathy Quinlan wrote: >I disagree with your comment, to me a linear device working in a linear >region has to dissipate the power not consumed by the load ie > >If I have 24 volts in, 5volt out to the load, load consumes say 1 amp. > >Power into regulator is 24W, power out is 5W, so the regulator must consume >the missing 19W. Of course :) Just one comment to add: the device doesn't have to work in a linear region for this to be true. Ah, yes: the difference between current in and current out has to do with the power consumption of the device (which is probably what Mike confused with the dissipation under load). ge -- http://www.piclist.com hint: To leave the PICList EraseMEpiclist-unsubscribe-requestEraseMEmitvma.mit.edu

At 15:38 11/12/2001 +0200, Vasile Surducan wrote: >On Mon, 12 Nov 2001, Gerhard Fiedler wrote: > > > Of course :) Just one comment to add: the device doesn't have to work in a > > linear region for this to be true. > > > But also the relationship is different then. > In switching mode for example, Of course. The complete phrase I was referring to was "to me a linear device working in a linear region ..." A switching device is not considered a linear device, mostly, probably. OTOH, a DC current through a transistor may well be in a region where the transistor is not considered linear, but the simple calculation of dissipated power is still valid enough for most purposes. I guess "linear" is not defined well enough to be really useful here... :) There are few if any fully linear devices in real world circuits, and what you consider linear depends on your specific point of view. "Linear" power regulators have some quite unlinear regions, and we're dealing with them every now and then :) ge -- http://www.piclist.com hint: To leave the PICList spamBeGonepiclist-unsubscribe-requestKILLspammitvma.mit.edu

Hi Kat, I Think I misunderstood the original point of the thread and only interperated what I thought. According to National Semiconductors data sheets the TOTAL Power is Pd = ((Vin -Vout) * Load Current) + (Vin * Quesient Current) MAximum allowable temperature is Tr(max) = Tj(max) - Ta(max) Tj(max) is 125deg C and Ta(max) is the maximum ambiant for your application. Thermal Resistance(ja) = Tr(max) / Pd Use this to calculate the TR of the heatsink. I hope that helps? .....katinkaspam_OUTMAGESTOWER.COM on 12/11/2001 22:28:47 Please respond to TakeThisOuTPICLIST.....TakeThisOuTMITVMA.MIT.EDU.....TakeThisOuTINTERNET.....TakeThisOuTwtgw To: TakeThisOuTPICLISTKILLspamspamMITVMA.MIT.EDUKILLspamspamINTERNETKILLspamspamwtgw cc: Paper Mail: Subject: Re: [EE]: PIC overheating a 7805 ! *** WARNING *** This mail has originated outside your organization, either from an external partner or the Global Internet. Keep this in mind if you answer this message. Hi Mike, I have searched data sheets from natsemi and TI, and cannot see how to calculate power disapation. Could you please provide a reference. I disagree with your comment, to me a linear device working in a linear region has to dissipate the power not consumed by the load ie If I have 24 volts in, 5volt out to the load, load consumes say 1 amp. Power into regulator is 24W, power out is 5W, so the regulator must consume the missing 19W. With your idea, the max power consumption would be 152mW ( vin - vout * Qc) = (24 -5) *0.008A (from the TI data sheet maximum value) Regards, Kat. ____________________________________________________________________________ /"\ ASCII Ribbon Campaign | K.A.Q. Electronics \ / - NO HTML/RTF in e-mail | Software and Electronic Engineering X - NO Word docs in e-mail | Perth Western Australia / \ | Ph +61 419 923 731 ____________________________________________________________________________ {Original Message removed}

I'll respectfully disagree. Harold On Mon, 12 Nov 2001 07:58:13 +0000 Mike Blakey <.....mike.blakeyRemoveMEBAESYSTEMS.COM> writes: {Quote hidden}> The power used and dissipated by the voltage regulator is NOT > calculated that way. > the power dissipated by the device is approx. = (Vin-Vout) * > quiescent current and > NOT load current. Check the data sheets! > > > > > > RemoveMEharoldhallikainenspamBeGoneJUNO.COM on 09/11/2001 18:44:56 > Please respond to spamBeGonePICLIST@spam@spam_OUTMITVMA.MIT.EDU@spam@spam_OUTINTERNET@spam@spam_OUTwtgw > > To: TakeThisOuTPICLISTspamMITVMA.MIT.EDUspamINTERNETspamwtgw > cc: > Paper Mail: > Subject: Re: [EE]: PIC overheating a 7805 ! > > *** WARNING *** > > This mail has originated outside your organization, > either from an external partner or the Global Internet. > Keep this in mind if you answer this message. > > > 24V is a bit high! Calculate out the power dissipation > (24V-5V)*LoadCurrent. Take this (in watts) and multiply by the > rating of > your heat sink in degrees C/ watt. This should give temperature rise > from > ambient. > Another thing to watch for is oscillating regulators, which > get hot. > Check the output on a scope and make sure you've got DC without a > bunch > of RF. > Finally, consider a switching regulator. The inefficiency of > using a > linear regulator to go from 24V down to 5V is pretty impressive. > > Harold > > > FCC Rules Online at http://hallikainen.com/FccRules > Lighting control for theatre and television at > http://www.dovesystems.com > > ________________________________________________________________ > GET INTERNET ACCESS FROM JUNO! > Juno offers FREE or PREMIUM Internet access for less! > Join Juno today! For your FREE software, visit: > dl.http://www.juno.com/get/web/. > > -- > http://www.piclist.com hint: The list server can filter out > subtopics > (like ads or off topics) for you. See http://www.piclist.com/#topics > > > > > > > > > ******************************************************************** > This email and any attachments are confidential to the intended > recipient and may also be privileged. If you are not the intended > recipient please delete it from your system and notify the sender. > You should not copy it or use it for any purpose nor disclose or > distribute its contents to any other person. > ******************************************************************** > > -- > http://www.piclist.com hint: To leave the PICList > piclist-unsubscribe-requestEraseMEmitvma.mit.edu > > FCC Rules Online at http://hallikainen.com/FccRules Lighting control for theatre and television at http://www.dovesystems.com ________________________________________________________________ GET INTERNET ACCESS FROM JUNO! Juno offers FREE or PREMIUM Internet access for less! Join Juno today! For your FREE software, visit: dl.http://www.juno.com/get/web/. -- http://www.piclist.com hint: To leave the PICList RemoveMEpiclist-unsubscribe-requestEraseMEspam_OUTmitvma.mit.edu

For regulators with a large voltage drop across them the power drain of the regulator proper becomes a significant part of Pd at low loads. With devices used without a heatsink this can be the straw that breaks the camel's back and cause strange faults in cased devices (regulator enters thermal shutdown under hard to duplicate conditions - i.e. at the customer's location, over the bridge, down the lane, etc, on hot Sundays exclusively). Pdreg = U1 * Iddreg + (U1 - U0) * Iload 7805 series regulators use little extra Iddreg to drive the output transistors because they are darlingtons and the base currents also flow to the load. With other regulators this is not so. The self-heating problem is especially visible with TO92 regulators. The usual fix is to add a R in series with the voltage input such that at max. load Uin on the regulator is higher than the required min. input voltage. A decoupling capacitor is required after the resistor. R = (U1 - Uinmin) / Iloadmax It is possible to split an arbitrary part of the power to be dissipated into the resistor (i.e. 50%, 80% etc) and it need not be just one resistor, it can be a series combo. Me, when I do this, I arrange for the power dissipation in each device to be equal (including in the regulator), so I have no hot spots on the board (with similar sized devices - e.g. TO220 7805 and 2W resistors). The formula for that is: Rtot = (U1 - Uload) / (2 * Iload) This is inefficient but nobody cares whether a couple of resistors dissipate 2W in a unit that draws 100 or 3000W. You can put other things in series if the current is reasonably constant, like dial lights, pilot lights, anti-dew heaters, LCD backlights, even a small fan with decoupling. The fan trick is neat because as the circuit draws more current the fan accelerates ... (assuming DC brushless fan). A 7805 TO220 is not to dissipate more than about 1.5W without a heatsink (at 50C ambient) and it will run very hot to the touch at that. Up to about 1W into a device of that size will keep it 'touchably' warm. At 24Vin and 5 Vout that is less than 50mA total, leaving only 40mA for the load (acc. NS data book). With an 'equal dissipation' R the current can be doubled under the same conditions (R = 180 Ohms, 2W). Decouple the regulator input after the resistor with 10 uF 16V to GND. The resistor will run about as hot as the TO220 case (not as hot as the TO220 case without the R). This will allow loads up to 80mA and may save your design as is now (cut trace, connect resistor with wires - after packing it in heatshrink tubing and add decoupling capacitor on bottom of board - if you have room). Remember that heat does not go away, you will have to put the resistor in a place where it does not heat something else - like making ugly dents in the plastic case after a while). Peter -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

As fixed regulators tend to be cheaper than power resistors, I prefer to use an other regulator in front with even better results, let´s say you got 24V and need 5V with like 100mA, so take a 12V regulator first (24V-12V=12V*100mA=1.2W) and than the 5V regulator (12V-5V=7V*100mA=0.7W) and you get off without needing the extra cap, which results even cheaper. Friedel At 09:30 p.m. 13/11/01 +0200, you wrote: {Quote hidden}>For regulators with a large voltage drop across them the power drain of >the regulator proper becomes a significant part of Pd at low loads. With >devices used without a heatsink this can be the straw that breaks the >camel's back and cause strange faults in cased devices (regulator enters >thermal shutdown under hard to duplicate conditions - i.e. at the >customer's location, over the bridge, down the lane, etc, on hot Sundays >exclusively). > >Pdreg = U1 * Iddreg + (U1 - U0) * Iload > >7805 series regulators use little extra Iddreg to drive the output >transistors because they are darlingtons and the base currents also flow >to the load. With other regulators this is not so. > >The self-heating problem is especially visible with TO92 regulators. The >usual fix is to add a R in series with the voltage input such that at max. >load Uin on the regulator is higher than the required min. input voltage. >A decoupling capacitor is required after the resistor. > >R = (U1 - Uinmin) / Iloadmax > >It is possible to split an arbitrary part of the power to be dissipated >into the resistor (i.e. 50%, 80% etc) and it need not be just one >resistor, it can be a series combo. Me, when I do this, I arrange for the >power dissipation in each device to be equal (including in the regulator), >so I have no hot spots on the board (with similar sized devices - e.g. >TO220 7805 and 2W resistors). The formula for that is: > >Rtot = (U1 - Uload) / (2 * Iload) > >This is inefficient but nobody cares whether a couple of resistors >dissipate 2W in a unit that draws 100 or 3000W. You can put other things >in series if the current is reasonably constant, like dial lights, pilot >lights, anti-dew heaters, LCD backlights, even a small fan with >decoupling. The fan trick is neat because as the circuit draws more >current the fan accelerates ... (assuming DC brushless fan). > >A 7805 TO220 is not to dissipate more than about 1.5W without a heatsink >(at 50C ambient) and it will run very hot to the touch at that. Up to >about 1W into a device of that size will keep it 'touchably' warm. At >24Vin and 5 Vout that is less than 50mA total, leaving only 40mA for the >load (acc. NS data book). > >With an 'equal dissipation' R the current can be doubled under the same >conditions (R = 180 Ohms, 2W). Decouple the regulator input after the >resistor with 10 uF 16V to GND. The resistor will run about as hot as the >TO220 case (not as hot as the TO220 case without the R). This will allow >loads up to 80mA and may save your design as is now (cut trace, connect >resistor with wires - after packing it in heatshrink tubing and add >decoupling capacitor on bottom of board - if you have room). Remember that >heat does not go away, you will have to put the resistor in a place where >it does not heat something else - like making ugly dents in the plastic >case after a while). > >Peter > >-- >http://www.piclist.com hint: PICList Posts must start with ONE topic: >[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics