Post by thread:[EE:] Dropping the battery voltage

I am using a 3.6V lithium battery in one of me designs. In order to save battery life, I need to go 2.8V, at *most* 2.7V. How could I achieve this? I think I can just throw a diode, but this will give me .6V or .7V drop. I want to drop to between 2.7V and 2.8V. Is there a way to achieve this? Any ideas are welcome. -- http://www.piclist.com hint: To leave the PICList spam_OUTpiclist-unsubscribe-requestTakeThisOuTmitvma.mit.edu

At 08:04 PM 10/3/2003 +0300, =?iso-8859-9?B?1m1lciBZYWxo/Q==?= wrote: >I am using a 3.6V lithium battery in one of me designs. In order to save >battery life, I need to go 2.8V, at *most* 2.7V. How could I achieve this? >I think I can just throw a diode, but this will give me .6V or .7V drop. I >want to drop to between 2.7V and 2.8V. Is there a way to achieve this? Interesting one. Very low currents are the domain of specialty switchers. A burst mode switcher might do what you want. A linear reg, wether a series diode or whatever, is just going to spend the voltage differential as heat, like always. -- http://www.piclist.com hint: To leave the PICList .....piclist-unsubscribe-requestKILLspam@spam@mitvma.mit.edu

What is a switcher? Voltage regulator? -----Original Message----- From: pic microcontroller discussion list [PICLISTKILLspamMITVMA.MIT.EDU]On Behalf Of David VanHorn Sent: Friday, October 03, 2003 8:09 PM To: .....PICLISTKILLspam.....MITVMA.MIT.EDU Subject: Re: [EE:] Dropping the battery voltage At 08:04 PM 10/3/2003 +0300, =?iso-8859-9?B?1m1lciBZYWxo/Q==?= wrote: >I am using a 3.6V lithium battery in one of me designs. In order to save >battery life, I need to go 2.8V, at *most* 2.7V. How could I achieve this? >I think I can just throw a diode, but this will give me .6V or .7V drop. I >want to drop to between 2.7V and 2.8V. Is there a way to achieve this? Interesting one. Very low currents are the domain of specialty switchers. A burst mode switcher might do what you want. A linear reg, wether a series diode or whatever, is just going to spend the voltage differential as heat, like always. -- http://www.piclist.com hint: To leave the PICList EraseMEpiclist-unsubscribe-requestspam_OUTTakeThisOuTmitvma.mit.edu -- http://www.piclist.com hint: To leave the PICList piclist-unsubscribe-requestspam_OUTmitvma.mit.edu

At 12:18 AM 10/4/2003 +0300, Omer Yalhi wrote: >What is a switcher? Voltage regulator? Switchmode voltage regulator. There are many types, but for your particular problem, I think a burst mode switcher might be the only efficient answer. http://www.eetasia.com/ARTICLES/2002APR/2002APR08_ICD_PL_PD_MPR_TAC.PDF -- http://www.piclist.com hint: To leave the PICList @spam@piclist-unsubscribe-requestKILLspammitvma.mit.edu

> I am using a 3.6V lithium battery in one of me designs. In order to save > battery life, I need to go 2.8V, at *most* 2.7V. How could I achieve this? > I think I can just throw a diode, but this will give me .6V or .7V drop. I > want to drop to between 2.7V and 2.8V. Is there a way to achieve this? > > Any ideas are welcome. You won't save any power by doing this... Using a linear component to drop voltage will just waste the power in the linear component. If you use a switch mode DC-DC regulator chip from linear or maxim (for example), you can do this with a better efficiency. You could also build one yourself using a PIC. Its not too hard if you have an A/D and PWM output to spare. Regards, Bob Monsen -- http://www.piclist.com hint: To leave the PICList KILLspampiclist-unsubscribe-requestKILLspammitvma.mit.edu

> You won't save any power by doing this... Using a linear component to > drop voltage will just waste the power in the linear component. Actually you will save a little power because the target circuit will draw less current at the lower voltage. However, a reasonable switcher will still be more efficient. > You could also build one yourself using a PIC. Its not too hard if you > have an A/D and PWM output to spare. You don't need a A/D, just a comparator. If output voltage below threshold do pulses as you can, if above voltage don't do pulses. A 12F629 can with its built in comparator can do this job nicely. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: To leave the PICList RemoveMEpiclist-unsubscribe-requestTakeThisOuTmitvma.mit.edu

David thank you for the input. I looked over the app note. They are suggesting a design with the LTC3440. This might not be the best solution for my case. Currently when in sleep mode with pic, I am down to 24uA @3.6V and I need to be about 10uA (if at all possible). The quiescent current for this unit is between 25uA to 40uA which is *way* too high for my purposes. However, I got an idea and curious if it would work. The pic is in sleep mode for 1500ms, then wakes up for about 20ms (currently this is as low as I could go) and sleeps. There is the MCP2120 (min. 2.5V) and TFDU4100 (min 2.7V) on the board. I only need to power these devices during the 20ms interval. What I might be able to do is to power the pic from 3.6V battery with two diodes, voltage drops to about 2.2V which 16LF628 can handle. Connect the LTC3440 outputs (with 2.8V) to MCP2120 and the TFDU4100. I can then enable the LTC3440 to power the devices and cut the power when done. This raises two issues for me though: 1. I do not know what will happen between the 2.2V powered pic and 2.8V powered MCP2120. Could they still communicate? 2. MCP2120 needs 30ms (max) time when powered down (as opposed to only 1000 Tosc when enabled from a disabled state). Which is too much time anyways. So using a voltage regulator would not solve my problem it seems. My problems would be solved if I only could get the TFDU4100 and MCP2120 to sense the dummy wake up characters that I am sending, within a 1ms period (over 9600 baud). Then I would not have to worry about going as low as 2.8V! I appriciate if anyone is *even* reading this long message... -- http://www.piclist.com hint: To leave the PICList spamBeGonepiclist-unsubscribe-requestspamBeGonemitvma.mit.edu

Omer wrote: > Actually I am using 16LF628. I didn't want to use another chip as I am very > comfortable with the 628. However, It might make sense to look at those as > well. Hi Omer, You might want to check Microchip's 'webinar' on nanoWatt devices: NanoWattApr08.zip It's around 6 megabytes to download and extracts to a Windows Media Audio/Visual file -- it might just help you make your decision. As always, weigh in the errata notes on any processor you might switch to -- it's all about risk mitigation. Good luck. Best regards, Ken Pergola -- http://www.piclist.com hint: To leave the PICList piclist-unsubscribe-requestEraseME.....mitvma.mit.edu

From: "Vmer Yalh}" <EraseMEoyalhiTEKSAN.COM.TR> > Well I am thinking if the battery voltage was dropped the pic would be using > less current during sleep and DC-DC regulators use to much current (for this > purpose). I am at 24uA currently during sleep, and hope to go even lower. > This will make the difference between using a AA or C size battery. About > $5 difference for each unit that will be sold in qunatites of ten thousands. > I believe the PIC will use less power, but consider that if you use a linear regulator or diode to drop the voltage, you waste P = VI across the device. Thats how the device gets away with using less power at lower voltage, I'm guessing. Look over the switching power supply design given in AD216 from microchip, which uses a comparator and an output to drop voltage more. Regards -- http://www.piclist.com hint: To leave the PICList RemoveMEpiclist-unsubscribe-requestEraseMEEraseMEmitvma.mit.edu

From: "Vmer Yalh}" <RemoveMEoyalhispam_OUTKILLspamTEKSAN.COM.TR> > Actually I am using 16LF628. I didn't want to use another chip as I am very > comfortable with the 628. However, It might make sense to look at those as > well. > If you can use the 675, you can decrease your current usage by quite a bit. I was seeing less than a uA through mine during sleep (at 2V). You need to turn off all the devices to make this happen, and depend on the watchdog to wake you up. I ran an LED flasher for almost a week on one 5V 1F capacitor. Regards. -- http://www.piclist.com hint: To leave the PICList RemoveMEpiclist-unsubscribe-requestTakeThisOuTspammitvma.mit.edu

> If you can use the 675, you can decrease your current usage by quite a > bit. I was seeing less than a uA through mine during sleep (at 2V). You > need to turn off all the devices to make this happen, and depend on the > watchdog to wake you up. Not if you want really low current consumption. The watchdog takes considerably more than a uA if I remember right. On a 16F630 design I did a few months ago, we created an external timer that woke up the processor every 50 to 100mS and only took about 250nA average current. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: To leave the PICList EraseMEpiclist-unsubscribe-requestspamspamBeGonemitvma.mit.edu

Ken, I can not find NanoWattApr08.zip file. If you have it, could you send it to me ?? I would apreciate it too much. Thanks, Gabriel Caffese.- -----Mensaje original----- De: pic microcontroller discussion list [RemoveMEPICLISTKILLspamMITVMA.MIT.EDU]En nombre de Ken Pergola Enviado el: Viernes, 03 de Octubre de 2003 20:04 Para: PICLISTSTOPspamspam_OUTMITVMA.MIT.EDU Asunto: Re: [EE:] Dropping the battery voltage Omer wrote: > Actually I am using 16LF628. I didn't want to use another chip as I am very > comfortable with the 628. However, It might make sense to look at those as > well. Hi Omer, You might want to check Microchip's 'webinar' on nanoWatt devices: NanoWattApr08.zip It's around 6 megabytes to download and extracts to a Windows Media Audio/Visual file -- it might just help you make your decision. As always, weigh in the errata notes on any processor you might switch to -- it's all about risk mitigation. Good luck. Best regards, Ken Pergola -- 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 Gabriel, Sorry I did not include the URL and I should have stated why I did not. After Microchip gave their web site a makeover, I checked the archived webinars and could not find this webinar anymore. I did download it onto my computer however. I use dial-up (I hear everyone laughing), so let me first go back to see if I can get you the URL and try to post it on this list for you and for the benefit of others who are interested as well. It's a 6.8 megabyte zipped file -- does your ISP impose any size limitations on your received e-mail file attachments? If you don't see another message within 10 minutes, I'm having trouble... Best regards, Ken Pergola -- http://www.piclist.com hint: To leave the PICList EraseMEpiclist-unsubscribe-requestEraseMEmitvma.mit.edu

From: "Olin Lathrop" <@spam@olin_piclist@spam@spam_OUTEMBEDINC.COM> > > If you can use the 675, you can decrease your current usage by quite a > > bit. I was seeing less than a uA through mine during sleep (at 2V). You > > need to turn off all the devices to make this happen, and depend on the > > watchdog to wake you up. > > Not if you want really low current consumption. The watchdog takes > considerably more than a uA if I remember right. On a 16F630 design I did a > few months ago, we created an external timer that woke up the processor > every 50 to 100mS and only took about 250nA average current. > Hmm... The PIC12F629/67 data sheet says typical current consumption of the WDT is about 300nA at 2V (See section 12.3, pg 88 of DS41190C). The total device current without anything enabled is typically 1nA. max 700nA, again at 2V. As I said before, I measured a current consumption of less than 1 uA for the sleeping PIC12F675 with WDT enabled, powered at 2V. This without ANY external components. Not bad. The only real problem is that the WDT period is dependent on temperature, so you can't get an accurate timeout. However, if you are just polling for an event, that isn't really a problem. You can get a 20mS poll out of the thing pretty easily. Regards, Bob Monsen -- http://www.piclist.com hint: To leave the PICList spamBeGonepiclist-unsubscribe-requestKILLspammitvma.mit.edu

Thread summary: goal is to reduce power consumption of PIC in sleep mode (one way is to switch to nanoWatt technology PICmicro) Hi Gabriel, I cannot locate the nanoWatt Technology presentation webinar in the archives on the Microchip web site -- perhaps they removed it from the archive temporarily. It is around 38 minutes audio/video presentation. Just e-mail me your e-mail address and I'll send the 6.8 MB file to you. Best regards, Ken Pergola -- http://www.piclist.com hint: To leave the PICList .....piclist-unsubscribe-requestspam_OUTmitvma.mit.edu

> Hmm... The PIC12F629/67 data sheet says typical current consumption of > the WDT is about 300nA at 2V (See section 12.3, pg 88 of DS41190C). The > total device current without anything enabled is typically 1nA. max > 700nA, again at 2V. This can vary between PICs, and the WDT current increases drastically with voltage. The incremental WDT current of a 16F630 at 3V is 2.7uA, and at 5V is 16uA. > As I said before, I measured a current consumption ... What you measured doesn't mean anything as long as it is at or below the maximum guaranteed by Microchip. Unless you're going to tell your customer "The battery should last a year, but might only last one week. If you happen to get some 1 week units, too bad", you need to stick to the worst case specs. > The only real > problem is that the WDT period is dependent on temperature, so you > can't get an accurate timeout. Our micropower external wakeup oscillator period also varied greatly with temperature, voltage, phase of the moon, proximity to a dead fish, etc. We got around this by occasionally leaving the internal oscillator on for a whole wakeup period to measure it. The external oscillator could then be used to track time to the accuracy of the internal R/C oscillator, which is a couple percent if I remember right. We did this measurement every 1000 external cycles, so it only added a small amount to the average power draw. You don't need to do it very often because the supply voltage, temperature, the moon, and dead fish just don't move that fast. > However, if you are just polling for an > event, that isn't really a problem. You can get a 20mS poll out of the > thing pretty easily. If you're polling for an event, the wakeup on change pins or the interrupt pin can be very useful. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

On Friday 03 Oct 2003 11:16 pm, you wrote: > Well I am thinking if the battery voltage was dropped the pic would be > using less current during sleep and DC-DC regulators use to much current > (for this purpose). I am at 24uA currently during sleep, and hope to go > even lower. This will make the difference between using a AA or C size > battery. About $5 difference for each unit that will be sold in qunatites > of ten thousands. > Not sure about the PIC but with most micros, sleep mode current is heavily dependent on the way the inputs and outputs are terminated. I once was involved in the development of a CO2 detector wich had a sleep mode design target of 15uA which should have been achieveable. When we got first masked parts it was 150uA!!. Turned out to be due to duff info about the best way to terminate ins and outs to minimise sleep current. Ian -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

so based on your experience, what's the best way to terminate ins and outs to minimize sleep current ? dld On Sat, 4 Oct 2003 09:23:50 +0100, Ian Bell wrote: {Quote hidden}>On Friday 03 Oct 2003 11:16 pm, you wrote: >> Well I am thinking if the battery voltage was dropped the pic would be >> using less current during sleep and DC-DC regulators use to much current >> (for this purpose). I am at 24uA currently during sleep, and hope to go >> even lower. This will make the difference between using a AA or C size >> battery. About $5 difference for each unit that will be sold in qunatites >> of ten thousands. >> > >Not sure about the PIC but with most micros, sleep mode current is heavily >dependent on the way the inputs and outputs are terminated. I once was >involved in the development of a CO2 detector wich had a sleep mode design >target of 15uA which should have been achieveable. When we got first masked >parts it was 150uA!!. Turned out to be due to duff info about the best way >to terminate ins and outs to minimise sleep current. > >Ian > >-- >http://www.piclist.com hint: PICList Posts must start with ONE topic: >[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads David Dunn Digital Racing Products TakeThisOuTdave.....TakeThisOuTdigitalracingproducts.com http://www.digitalracingproducts.com -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

On Saturday 04 Oct 2003 6:49 pm, you wrote: > so based on your experience, what's the best way to terminate ins and outs > to minimize sleep current ? > That's the problem - it is processor specific. We were using a Zilog part IIRC and the manufacturuer gave us specific advice on each pin. With unused inputs you can only pull them up or down. I would suggest you try each one each way in turn and measure the resultant current. You could post the results here as I am sure there are others who would be interested. One other tip, to reduce current on our project we took advantage of the fact that current draw is dependent on oscillator frequency. We ended up running at about 10KHz and achieved a 5 year life from an alkaline PP3. Ian -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> I would like to thank everyone who answered my question. Currently, I am > thinking of placing a zener diode Vz=2.7. I think this would consume the > least amount of current. Would I be terribly wrong with this assumption? That's not a good assumption.. YOUR ORIGINAL SPEC WAS: I am using a 3.6V lithium battery in one of me designs. In order to save battery life, I need to go 2.8V, at *most* 2.7V. How could I achieve this? I think I can just throw a diode, but this will give me .6V or .7V drop. I want to drop to between 2.7V and 2.8V. Is there a way to achieve this? I think a detailed reading of the replies would have given better ideas than that. SO A zener that is "just" not taking any current when the battery is at its lowest level will be wasting power for the entire rest of the time. If you want minimum energy loss overall then - Best Switching regulator Next Linear series regulator with ultra low dropout voltage Worst Shunt regulator (zener etc) If energy efficiency is not an issue and the zener's "soft" regulation curve is OK then it may do the job.. Russell McMahon. -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> >If energy efficiency is not an issue and the zener's "soft" regulation curve is OK then it may do the job.. > > > Russell McMahon. Often overlooked point: Zeners have weak knees. They shift their breakover point according to current and temperature, and aren't generally too accurate to start with. I've seen a number of designs where a high value pullup was used with a zener, expecting to get the voltage across the zener at the nameplate rating. It was significantly higher, of course. -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> If you want minimum energy loss overall then - > > Best Switching regulator > Next Linear series regulator with ultra low dropout voltage > Worst Shunt regulator (zener etc) I could be missing something here but the quiescent current for voltage regulators are too high for my purposes (200-300uA). I could find one low dropout regulator from Linear technology LT1512 with 12uA. Although this will be a perfect solution for another application that I have in mind, not for this one. The best thing would be for me to test what happens with a diode and another test with zener diode. Well, that's for tomorrow (9:32pm here). :) regards, Omer -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

Hi Omer, Omer, unless I'm missing the big picture, I was wondering: Is the PIC current consumption that much higher running directly from the 3.6V lithium (as opposed to 2.7V or 2.8V) that warrants going through all of this trouble? I'm not trying to mean or anything -- just trying to be a helpful devil's advocate. What is the PIC current consumption at Vdd = 3.6 volts? What is the PIC current consumption at Vdd = 2.7 volts? Have you determined the delta current consumption between these two scenarios? It would be interesting to see what it is in your case. Best regards, Ken Pergola -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

That is a very good question actually. I am about 24uA at 3.6V, and tested the system 3.0V Lithium battery, and got 19uA. Now, I have not tested, but assuming (hoping) to get 15uA at 2.7V. But how I get 2.7V is the big question for me. 15uA in this case will be all I need. regards, Omer -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> I've seen a number of designs where a high value pullup was used with a > zener, expecting to get the voltage across the zener at the nameplate > rating. It was significantly higher, of course. You meant to way "lower", I presume. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> Connect two lithiums cells in parallel and you'll double > the lifetime. It should more than double. Halving the current drain at that end of the scale will give a better Ah rating per cell than halving it at higher drains. That's been my experience with other batteries (and as told to me by manufacturers beforehand, although they don't seem to have charts to prove it. What they do have is mostly for the middle range) -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads