Post by thread:[OT]:Current Sensor

Hi everyone, I am building a robot and I want to monitor the current drawn by the motor. The motor can draw from 250mA up to 2.5Amp! What do I have to do to measure this current? I was thinking of puting a 1-ohm resistor probe in series with the motor, and feed the voltage drop across the resistor into a ADC, but at 2.5A, the voltage drops across the resistor is too high and it leaves nothing for the motor! Is there any current sensor circuit out there that have very low loading resistance? Is there any IC doing this job? Thanks Thomas _________________________________________________________________________ Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com. -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

Use a small resistor and a curent to voltage converter. example: 0.1 ohm , 25mV to 250mV drop-out, convert this potential with a good operational amplifier to 0-5V Vasile On Thu, 5 Apr 2001, Thomas N wrote: {Quote hidden}> Hi everyone, > > I am building a robot and I want to monitor the current drawn by the motor. > The motor can draw from 250mA up to 2.5Amp! What do I have to do to measure > this current? > > I was thinking of puting a 1-ohm resistor probe in series with the motor, > and feed the voltage drop across the resistor into a ADC, but at 2.5A, the > voltage drops across the resistor is too high and it leaves nothing for the > motor! > > Is there any current sensor circuit out there that have very low loading > resistance? Is there any IC doing this job? > > Thanks > > Thomas > _________________________________________________________________________ > Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com. > > -- > http://www.piclist.com hint: The PICList is archived three different > ways. See http://www.piclist.com/#archives for details. > > -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email spam_OUTlistservTakeThisOuTmitvma.mit.edu with SET PICList DIGEST in the body

part 1 2648 bytes content-type:text/plain; charset=us-asciiHall effect sensors can also be used in this application to detect the flux lines radiating from the motor supply leads. Don't know a part number off the top of my head, but a quick search on the web should turn up some hits. Scott F. Touchton 1550 Engineering Manager JDS Uniphase |--------+-----------------------> | | Vasile | | | Surducan | | | <vasile@L30.I| | | TIM-CJ.RO> | | | | | | 04/05/01 | | | 03:28 AM | | | Please | | | respond to | | | pic | | | microcontroll| | | er discussion| | | list | | | | |--------+-----------------------> >----------------------------------------------------------------------------| | | | To: .....PICLISTKILLspam@spam@MITVMA.MIT.EDU | | cc: (bcc: Scott Touchton/US/UNIPHASE) | | Subject: Re: [OT]:Current Sensor | >----------------------------------------------------------------------------| Use a small resistor and a curent to voltage converter. example: 0.1 ohm , 25mV to 250mV drop-out, convert this potential with a good operational amplifier to 0-5V Vasile On Thu, 5 Apr 2001, Thomas N wrote: {Quote hidden}> Hi everyone, > > I am building a robot and I want to monitor the current drawn by the motor. > The motor can draw from 250mA up to 2.5Amp! What do I have to do to measure > this current? > > I was thinking of puting a 1-ohm resistor probe in series with the motor, > and feed the voltage drop across the resistor into a ADC, but at 2.5A, the > voltage drops across the resistor is too high and it leaves nothing for the > motor! > > Is there any current sensor circuit out there that have very low loading > resistance? Is there any IC doing this job? > > Thanks > > Thomas > _________________________________________________________________________ > Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com. > > -- > http://www.piclist.com hint: The PICList is archived three different > ways. See http://www.piclist.com/#archives for details. > > -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listservKILLspammitvma.mit.edu with SET PICList DIGEST in the body part 2 3003 bytes content-type:application/octet-stream; (decode) part 3 136 bytes -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email .....listservKILLspam.....mitvma.mit.edu with SET PICList DIGEST in the body

Do you know the part number for the current to Voltage converter IC? THomas {Quote hidden}>From: Vasile Surducan <EraseMEvasilespam_OUTTakeThisOuTL30.ITIM-CJ.RO> >Reply-To: pic microcontroller discussion list <PICLISTspam_OUTMITVMA.MIT.EDU> >To: @spam@PICLISTKILLspamMITVMA.MIT.EDU >Subject: Re: [OT]:Current Sensor >Date: Thu, 5 Apr 2001 10:28:22 +0300 > >Use a small resistor and a curent to voltage converter. >example: 0.1 ohm , 25mV to 250mV drop-out, >convert this potential with a good operational amplifier to 0-5V >Vasile > >On Thu, 5 Apr 2001, Thomas N wrote: > > > Hi everyone, > > > > I am building a robot and I want to monitor the current drawn by the >motor. > > The motor can draw from 250mA up to 2.5Amp! What do I have to do to >measure > > this current? > > > > I was thinking of puting a 1-ohm resistor probe in series with the >motor, > > and feed the voltage drop across the resistor into a ADC, but at 2.5A, >the > > voltage drops across the resistor is too high and it leaves nothing for >the > > motor! > > > > Is there any current sensor circuit out there that have very low loading > > resistance? Is there any IC doing this job? > > > > Thanks > > > > Thomas > > >_________________________________________________________________________ > > Get Your Private, Free E-mail from MSN Hotmail at >http://www.hotmail.com. > > > > -- > > http://www.piclist.com hint: The PICList is archived three different > > ways. See http://www.piclist.com/#archives for details. > > > > > >-- >http://www.piclist.com#nomail Going offline? Don't AutoReply us! >email KILLspamlistservKILLspammitvma.mit.edu with SET PICList DIGEST in the body > > _________________________________________________________________________ Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com. -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email RemoveMElistservTakeThisOuTmitvma.mit.edu with SET PICList DIGEST in the body

The resistor *is* your current to voltage converter. Not sure how you're controlling the motor (PWM? FET/BJT? simple on/off?), but if one side of the resistor is tied to ground, you can use an op-amp (non-inverting configuration) to scale the small voltage drop across the resistor into 0-5V for A/D conversion. Beware, though. The motor current will have a lot of noise. Some low-pass filtering before the A/D wouldn't hurt. I like the SenseFET for this application. International Rectifier IRCZ44 comes to mind, although you might not get much accuracy because that's a 50A device. It's a 5 pin MOSFET. The extra two pins are an integrated 'current tap' This current tap reflects a fraction of the drain-source current. You convert this scaled down current with a resistor. The advantage is that you don't have the current-sensing resistor wasting power. You only have the on-state resistance of the FET to worry about. For the IRZC44, that's 0.028 Ohm. Only drawback is that since the motor is 2.5A and the FET is 50A, you might not get much accuracy because the load is such a small fraction. You'd need a large gain on the op-amp, and possibly correlate some actual current readings with output to get good results. Or you could use a different version of this FET (one with less range). DigiKey now carries these types of FET's, so you might start there to see what you can easily get. {Original Message removed}

I will use PWM to control the speed of the Motor. If I put the sensing resistor in series with it and feed the resistor's voltage drop to the ADC, then the input to the ADC will be a square wave! (has the frequency and duty cycle of the PWM) which is bad. I think I will need some circuit to average it out... Thomas {Quote hidden}>From: "Eisermann, Phil [Ridg/CO]" <spamBeGonepeisermaspamBeGoneRIDGID.COM> >Reply-To: pic microcontroller discussion list <TakeThisOuTPICLISTEraseMEspam_OUTMITVMA.MIT.EDU> >To: RemoveMEPICLISTTakeThisOuTMITVMA.MIT.EDU >Subject: Re: [OT]:Current Sensor >Date: Thu, 5 Apr 2001 13:04:44 -0500 > > The resistor *is* your current to voltage converter. Not sure how >you're controlling the motor (PWM? FET/BJT? simple on/off?), but if one >side >of the resistor is tied to ground, you can use an op-amp (non-inverting >configuration) to scale the small voltage drop across the resistor into >0-5V >for A/D conversion. Beware, though. The motor current will have a lot of >noise. Some low-pass filtering before the A/D wouldn't hurt. > > I like the SenseFET for this application. International Rectifier >IRCZ44 comes to mind, although you might not get much accuracy because >that's a 50A device. It's a 5 pin MOSFET. The extra two pins are an >integrated 'current tap' This current tap reflects a fraction of the >drain-source current. You convert this scaled down current with a resistor. >The advantage is that you don't have the current-sensing resistor wasting >power. You only have the on-state resistance of the FET to worry about. For >the IRZC44, that's 0.028 Ohm. Only drawback is that since the motor is 2.5A >and the FET is 50A, you might not get much accuracy because the load is >such >a small fraction. You'd need a large gain on the op-amp, and possibly >correlate some actual current readings with output to get good results. Or >you could use a different version of this FET (one with less range). >DigiKey >now carries these types of FET's, so you might start there to see what you >can easily get. > > >{Original Message removed}

No, it won't :) well, it depends on the off time, i suppose. The motor current can't decay instantaneously to zero because the motor is inductive. The current will decay somewhat during the off period, but it's not going to be zero unless you have a *very* slow PWM frequency. I don't suppose you have access to a scope, so you? If you do, look at the current waveform of the dropping resistor during PWM. But then, if you had a scope, you probably would have done that already. You should add a low pass filter anyway. Obviously, you want to set the break frequency lower than your PWM frequency. That will go a long way towards removing noise. Or you could add a peak-detect circuit if you want the peak current instead of the average. But I don't think that is necessary if your PWM frequency is reasonable (depends on the motor). How fast are you running (or planning on running) the PWM? {Original Message removed}

I am a bit confused. Let's say I have: +5V---0/ 0--------(0)-------/\/\/\------------| +5V switch motor Sense Resistor Ground Let's say the switch is closed and the motor is running, then suddenly I open the switch, should the voltage drop across the sense resitor be at 0 mA? If not, where is the current comming from? I am planing to use PWM (open and close the switch) at around 10Khz. How to I measure the average current flowing thru the motor while it is running accurately? Thomas {Quote hidden}>From: "Eisermann, Phil [Ridg/CO]" <peisermaEraseME.....RIDGID.COM> >Reply-To: pic microcontroller discussion list <EraseMEPICLISTMITVMA.MIT.EDU> >To: RemoveMEPICLISTEraseMEEraseMEMITVMA.MIT.EDU >Subject: Re: [OT]:Current Sensor >Date: Thu, 5 Apr 2001 14:05:44 -0500 > >No, it won't :) well, it depends on the off time, i suppose. The motor >current can't decay instantaneously to zero because the motor is inductive. >The current will decay somewhat during the off period, but it's not going >to >be zero unless you have a *very* slow PWM frequency. I don't suppose you >have access to a scope, so you? If you do, look at the current waveform of >the dropping resistor during PWM. But then, if you had a scope, you >probably >would have done that already. > >You should add a low pass filter anyway. Obviously, you want to set the >break frequency lower than your PWM frequency. That will go a long way >towards removing noise. Or you could add a peak-detect circuit if you want >the peak current instead of the average. But I don't think that is >necessary >if your PWM frequency is reasonable (depends on the motor). How fast are >you >running (or planning on running) the PWM? > >{Original Message removed}

The trick is that the motor is inductive. I don't know what your background is, mathematical or otherwise, so i'll try to offer a simple explanation without any math. I hope this makes sense. Its been a while :) If I omitted something, or goofed it up, I hope someone will point it out and correct me. An ideal inductor behaves according to v=L(di/dt). In non-math speak, that means the voltage (v) is proportional (via the inductance, L) to the time rate of change of the current through the inductor. In other words, the faster the current is changing, the higher the voltage across the inductor. Similarly, if the current isn't changing (e.g a constant current through the inductor), there is no voltage across the inductor. Of course, that's the ideal. A real inductor has to be made of wire, and wire has resistance. But you get the idea. Now, consider what would happen if the current were to somehow change from 2A to 0. Never mind why. What's the voltage across the inductor? It depends on how fast the current changes, as stated above. The faster it goes to zero, the higher the voltage that develops across the inductor has to be. So what would happen if it were to change instantaneously? The voltage across it would have to be infinite! Can this happen? Only in a textbook. (This is the principle behind spark coils, btw). In the motor cicuit, once you open the switch, the current has got to go somewhere. It cannot instantaneously dissappear (because that would require the voltage to be infinite). Where does it go? back to ground, through your resistor. It decays exponentially, not linearly. The faster your PWM frequency, the less the motor current decays in the off period. So, does that make any sense at all? Also, your 10kHz sounds like a good place to start. {Original Message removed}

Thank you for your reply. It clears up the cloud a little bit for me. thanks! Thomas {Quote hidden}>From: "Eisermann, Phil [Ridg/CO]" <RemoveMEpeisermaspam_OUTKILLspamRIDGID.COM> >Reply-To: pic microcontroller discussion list <RemoveMEPICLISTTakeThisOuTspamMITVMA.MIT.EDU> >To: EraseMEPICLISTspamspamBeGoneMITVMA.MIT.EDU >Subject: Re: [OT]:Current Sensor >Date: Thu, 5 Apr 2001 16:10:10 -0500 > > The trick is that the motor is inductive. I don't know what your >background is, mathematical or otherwise, so i'll try to offer a simple >explanation without any math. I hope this makes sense. Its been a while :) >If I omitted something, or goofed it up, I hope someone will point it out >and correct me. > > An ideal inductor behaves according to v=L(di/dt). In non-math >speak, that means the voltage (v) is proportional (via the inductance, L) >to >the time rate of change of the current through the inductor. In other >words, >the faster the current is changing, the higher the voltage across the >inductor. Similarly, if the current isn't changing (e.g a constant current >through the inductor), there is no voltage across the inductor. Of course, >that's the ideal. A real inductor has to be made of wire, and wire has >resistance. But you get the idea. > > Now, consider what would happen if the current were to somehow >change from 2A to 0. Never mind why. What's the voltage across the >inductor? >It depends on how fast the current changes, as stated above. The faster it >goes to zero, the higher the voltage that develops across the inductor has >to be. So what would happen if it were to change instantaneously? The >voltage across it would have to be infinite! Can this happen? Only in a >textbook. (This is the principle behind spark coils, btw). > > In the motor cicuit, once you open the switch, the current has got >to go somewhere. It cannot instantaneously dissappear (because that would >require the voltage to be infinite). Where does it go? back to ground, >through your resistor. It decays exponentially, not linearly. The faster >your PWM frequency, the less the motor current decays in the off period. > > So, does that make any sense at all? Also, your 10kHz sounds like >a >good place to start. > >{Original Message removed}

> In the motor cicuit, once you open the switch, the current has got > to go somewhere. It cannot instantaneously dissappear (because that would > require the voltage to be infinite). Where does it go? back to ground, > through your resistor. It decays exponentially, not linearly. The faster > your PWM frequency, the less the motor current decays in the off period. And the circuit is probably completed by the nice bright arc on your switch contacts. Bob Ammerman RAm Systems (contract development of high performance, high function, low-level software) -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email RemoveMElistservKILLspammitvma.mit.edu with SET PICList DIGEST in the body

At 01:04 PM 4/5/01 -0500, Eisermann, Phil [Ridg/CO] wrote: {Quote hidden}> The resistor *is* your current to voltage converter. Not sure how >you're controlling the motor (PWM? FET/BJT? simple on/off?), but if one side >of the resistor is tied to ground, you can use an op-amp (non-inverting >configuration) to scale the small voltage drop across the resistor into 0-5V >for A/D conversion. Beware, though. The motor current will have a lot of >noise. Some low-pass filtering before the A/D wouldn't hurt. > > I like the SenseFET for this application. International Rectifier >IRCZ44 comes to mind, although you might not get much accuracy because >that's a 50A device. It's a 5 pin MOSFET. The extra two pins are an >integrated 'current tap' This current tap reflects a fraction of the >drain-source current. You convert this scaled down current with a resistor. >The advantage is that you don't have the current-sensing resistor wasting >power. You only have the on-state resistance of the FET to worry about. For >the IRZC44, that's 0.028 Ohm. Only drawback is that since the motor is 2.5A >and the FET is 50A, you might not get much accuracy because the load is such >a small fraction. You'd need a large gain on the op-amp, and possibly >correlate some actual current readings with output to get good results. Or >you could use a different version of this FET (one with less range). DigiKey >now carries these types of FET's, so you might start there to see what you >can easily get. Exactly right! So use the IRCZ24 (much lower current rating). We use a lot of them . . . The exact on resistance is not critical since it is a current ratio arrangement (assuming you use IR's recommended amplifier circuit which, unfortunately, requires a negative supply). dwayne Dwayne Reid <dwaynerSTOPspamspam_OUTplanet.eon.net> Trinity Electronics Systems Ltd Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax Celebrating 17 years of Engineering Innovation (1984 - 2001) * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Do NOT send unsolicited commercial email to this email address. This message neither grants consent to receive unsolicited commercial email nor is intended to solicit commercial email. -- http://www.piclist.com hint: To leave the PICList spamBeGonepiclist-unsubscribe-requestSTOPspamEraseMEmitvma.mit.edu

Go to Maxim's web site and check out the MAX471 and MAX472. http://dbserv.maxim-ic.com/quick_view2.cfm?pdf_num=1108 It's a single supply "high side current sense" IC that outputs a voltage proportional to the current through a very low value (0.035 Ohm) sense resistor. The sense resistor itself is built into the IC and can handle up to 3 Amps I believe. It comes in 8 pin DIP and SO packages, so it doesn't take up much space and it runs on microamps of supply current. Maxim also provides some good applications notes with the datasheet. Follow them and you should have very few problems. Matt Pobursky Maximum Performance Systems On Thu, 5 Apr 2001 00:37:33 -0000, Thomas N wrote: {Quote hidden}>Hi everyone, > >I am building a robot and I want to monitor the current drawn by the motor. >The motor can draw from 250mA up to 2.5Amp! What do I have to do to measure >this current? > >I was thinking of puting a 1-ohm resistor probe in series with the motor, >and feed the voltage drop across the resistor into a ADC, but at 2.5A, the >voltage drops across the resistor is too high and it leaves nothing for the >motor! > >Is there any current sensor circuit out there that have very low loading >resistance? Is there any IC doing this job? > >Thanks > -- http://www.piclist.com hint: To leave the PICList KILLspampiclist-unsubscribe-requestspamBeGonemitvma.mit.edu

Thomas N wrote: > > I am a bit confused. Let's say I have: > > +5V---0/ 0--------(0)-------/\/\/\------------| | | ---------|<-------------- > +5V switch motor Sense Resistor Ground diode > Let's say the switch is closed and the motor is running, then suddenly I > open the switch, should the voltage drop across the sense resitor be at 0 > mA? If not, where is the current comming from? > > I am planing to use PWM (open and close the switch) at around 10Khz. How to > I measure the average current flowing thru the motor while it is running > accurately? First add the diode if you are using PWM on a motor that has inductance (that's all of them!). Thomas, search the net for "Jones on steppers" he has a very nice tutorial that covers motor controlling, and the "constant current" chapter applies to all motors, not just stepper motors. Well worth reading if you plan of getting good performance from any motor. -Roman PS. Start your PWM experiment at a SLOW speed first. This will be easier to control and measure. Once you speed it up to 10kHz you start to get quirky things like effects from switching speed limits of the transistor and diode, and switching surge losses get much higher. It might work great at 500Hz but cook parts at 10kHz. :o) -- http://www.piclist.com hint: To leave the PICList EraseMEpiclist-unsubscribe-requestEraseMEmitvma.mit.edu

On Thu, 5 Apr 2001 16:10:10 -0500 "Eisermann, Phil [Ridg/CO]" <@spam@peiserma@spam@spam_OUTRIDGID.COM> writes: > > In the motor cicuit, once you open the switch, the current > has got > to go somewhere. It cannot instantaneously dissappear (because that > would > require the voltage to be infinite). Where does it go? back to > ground, > through your resistor. It decays exponentially, not linearly. The > faster > your PWM frequency, the less the motor current decays in the off > period. It would with a catch diode! Without a catch diode, the energy in the inductor (LI^2/2) is dissipated in the switch as an arc. The voltage will increase until it is high enough to maintain the current at the instant the switch opened. It will then ramp down as that energy is dissipated. So... is a catch diode typically included in motor pwm circuits? Harold ________________________________________________________________ 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/tagj. -- http://www.piclist.com hint: To leave the PICList spamBeGonepiclist-unsubscribe-requestKILLspammitvma.mit.edu

Is it common to put a "catch diode" to ground after the switch in these applications? If the motor appears as an inductive load, the current will ramp up when the switch is closed and ramp down (through the diode) when the switch is open. The voltages present, the inductance, and the PWM frequency would determine the ripple in the current. Harold On Thu, 5 Apr 2001 19:24:30 -0000 Thomas N <.....thomasn101spam_OUTHOTMAIL.COM> writes: {Quote hidden}> I am a bit confused. Let's say I have: > > +5V---0/ 0--------(0)-------/\/\/\------------| > > +5V switch motor Sense Resistor Ground > > > Let's say the switch is closed and the motor is running, then > suddenly I > open the switch, should the voltage drop across the sense resitor be > at 0 > mA? If not, where is the current comming from? > > I am planing to use PWM (open and close the switch) at around 10Khz. > How to > I measure the average current flowing thru the motor while it is > running > accurately? > > Thomas > > > > >From: "Eisermann, Phil [Ridg/CO]" <TakeThisOuTpeiserma.....TakeThisOuTRIDGID.COM> > >Reply-To: pic microcontroller discussion list > <TakeThisOuTPICLISTKILLspamspamMITVMA.MIT.EDU> > >To: .....PICLISTRemoveMEMITVMA.MIT.EDU > >Subject: Re: [OT]:Current Sensor > >Date: Thu, 5 Apr 2001 14:05:44 -0500 > > > >No, it won't :) well, it depends on the off time, i suppose. The > motor > >current can't decay instantaneously to zero because the motor is > inductive. > >The current will decay somewhat during the off period, but it's not > going > >to > >be zero unless you have a *very* slow PWM frequency. I don't > suppose you > >have access to a scope, so you? If you do, look at the current > waveform of > >the dropping resistor during PWM. But then, if you had a scope, you > >probably > >would have done that already. > > > >You should add a low pass filter anyway. Obviously, you want to set > the > >break frequency lower than your PWM frequency. That will go a long > way > >towards removing noise. Or you could add a peak-detect circuit if > you want > >the peak current instead of the average. But I don't think that is > >necessary > >if your PWM frequency is reasonable (depends on the motor). How > fast are > >you > >running (or planning on running) the PWM? > > > >{Original Message removed}