Best Way for Precision +/- Rail - Page 1
Forum section: electronics / projects

author: aw_rw1
author_rank: Contributor
timestamp: August 01, 2022, 02:42:16 pm
content: I need a precision +/- rail. I tried using a non-inverting amp with a ref for the positive side, and I tried using an inverting amp with the same ref for the negative side. Is this the best way to go about a +/- rail?I saw in a text book somewhere that linear regulators were the way to go. Anyone know a good way to get a +/- rail with +/-10mV precision?

author: Zero999
author_rank: Super Contributor
timestamp: August 01, 2022, 04:00:04 pm
content: How much current at what voltage?If you want to use a plain linear regulator, then you need both a positive and a negative rail for the input.What's it for? Why do you require 10mV precision? That's very tight.

author: aw_rw1
author_rank: Contributor
timestamp: August 01, 2022, 04:56:15 pm
content: How much current at what voltage?Voltage: +/-18V and +/-15VCurrent: ~10mAIf you want to use a plain linear regulator, then you need both a positive and a negative rail for the input.I have a CUI PQEW-Q24-D24-T that would provide +/-24V.What's it for? Why do you require 10mV precision? That's very tight.It is to power a precision measurement instrument. The instrument is basically a voltage follow with a potentiometer wiper at the input.

author: KaneTW
author_rank: Frequent Contributor
timestamp: August 01, 2022, 05:04:59 pm
content: Quote from: Zero999 on August 01, 2022, 04:00:04 pmIs it? Like every modern chip I'm working with is "900mV +- 25mV from DC to 100 MHz" with a current draw of anywhere between 1-100A

author: aw_rw1
author_rank: Contributor
timestamp: August 01, 2022, 05:07:54 pm
content: QuoteI am not sure the exact opamp. But it is not modern.

author: KaneTW
author_rank: Frequent Contributor
timestamp: August 01, 2022, 05:19:35 pm
content: Quote from: aw_rw1 on August 01, 2022, 05:07:54 pmOh nah, just a general remark regarding "very tight"

author: Zero999
author_rank: Super Contributor
timestamp: August 01, 2022, 06:21:27 pm
content: Quote from: KaneTW on August 01, 2022, 05:04:59 pmWhat processors do you have which require a dual power supply? I think your post is a little off-topic.25mV of 900mV is a tolerance of 2.78%, which isn't tight at all. 10mV of 18V is 0.056%, which is very close tolerance.Quote from: aw_rw1 on August 01, 2022, 04:56:15 pmDoes the voltage of both rails have to be within 10mV of the specification, or do both rails have to be within 10mV of each other?Do you have a schematic?

author: aw_rw1
author_rank: Contributor
timestamp: August 01, 2022, 08:26:13 pm
content: QuoteEach top and bottom of each rail needs to be +/-0.01V. E.g. 18V needs to be within 18V +/- 0.01V.QuotePlease see attached. U1 is powered by +/-15V, and POT is powered by +/-18V.

author: TimFox
author_rank: Super Contributor
timestamp: August 01, 2022, 09:58:44 pm
content: Normally, when needing a precise positive/negative voltage out of a precision potentiometer, the precision voltage supply feeds only the potentiometer, which needs far less current than the op amp.The op amp itself will tend to reject any changes on its supply voltages (PSRR and CMRR from the data sheet).Also, if you put reasonable fixed resistors in series with each end of the pot, you can add a trimpot to one of the two ends in order to adjust the end-points to equal but opposite voltages to the extent required.Also, placing low-leakage capacitors at the junctions of the fixed resistors and the pot ends gives extra filtering of any noise present.If you really need +/- 18 V output (from a pot excited with +/- 18 V), the op amp won't work anyway if powered by +/- 15 V.It would be better to connect the op amp for a reasonable gain (1.5:1 or 2:1?) and excite the pot with a lower voltage, but you still need the power on the op amp to be higher voltage than the pot output range.For the low currents you need at the pot, you could use a pair of LM431 shunt regulators or better parts operating at only a few mA to stabilize the voltages to the pot, stepping down from the op amp supplies.

author: jbb
author_rank: Super Contributor
timestamp: August 01, 2022, 11:45:36 pm
content: I understand that the above schematic may be an example rather than a detailed plan.I note that CR1 and CR2 look like protection diodes so the ‘ends’ of the pot can’t go (much) beyond the supply rails.I also note that there are fewer options for opamps above +-15V supply rails.How about:- normal regulated +- 15V rails (given worst case tolerances of 10% these could fall between 16.5 and 13.5V)- expect opamp output needs some headroom, so consider a max output of 12V (15V * 90% - 1.5V headroom)- voltage reference (2.5V, 5V, whatever available)- feed high side of pot by opamp doing k*Vref- feed low side of pot by opamp doing -1 * high side to get symmetrical output

author: Someone
author_rank: Super Contributor
timestamp: August 02, 2022, 02:06:54 am
content: Modern solution is to have a single ended reference voltage applied to the "mystery measurement sensor/pot" and (scaled) to the ADC reference so they track.... assuming this goes pretty much straight into an ADC.

author: David Hess
author_rank: Super Contributor
timestamp: August 02, 2022, 08:10:05 am
content: I have used the circuit shown below to do exactly what you are interested in. I used a 7805, PNP transistor instead of the JFET, a precision operational amplifier, and an LM329 reference for low noise and low drift. In your application a trimmer will be required to get exactly 18.000 volts. (1)A regulator is used instead of a pass transistor to add current limiting, safe operating area protection, and thermal protection. Since the regulator is inside the feedback loop, it does not contribute noise to the output. I used a 7805 to stay within the operational amplifier's output range, but a 317 could be used also if this is taken into account. In a low current application, I might use a 78L05/79L05, or 317LZ/337LZ. In a precision application the output of the precision operational amplifier should *not* be used to drive the load directly because load regulation will suffer from self heating of the operational amplifier.I used an LT1007 when I did it, but an OP-07, LT1001, or LT1012/LT1097 is a better choice because the low noise of an LT1007 class operational amplifier is overwhelmed by the reference noise anyway, even with the low noise of an LM329 zener reference.The same circuit could be duplicated and modified to deliver a negative output as well based on either the positive reference or the +18.000 volt output.(1) Do *not* use the operational amplifier's offset null adjustment to trim the output voltage! Doing so will ruin the operational amplifier's input offset voltage drift. In theory the offset null adjustment could be used to trim the temperature coefficient of the output voltage.

author: Zero999
author_rank: Super Contributor
timestamp: August 02, 2022, 10:59:24 am
content: Quote from: aw_rw1 on August 01, 2022, 08:26:13 pmThe power supply isn't shown in that schematic.Assuming +V goes to E7 and -V to E9, the op-amp is powered off a diode drop less than +V and -V.Use a precision reference such as the , REF70 with an non-inverting amplifier for the +18V rail and an inverting amplifier for the -18V rail. The resistors should be better than 0.05% tolerance.The op-amp needs to be low-offset, but nothing exotic. The jellybean OP07 will do. The non-inverting amplifier can be run off a single +24V supply and the inverting amplifier from +3V to +5V and -24V. The OP07 can output 10mA with no problem, so there is no need for pass transistors.

author: aw_rw1
author_rank: Contributor
timestamp: August 02, 2022, 02:48:11 pm
content: QuoteSorry, I mixed the +/-18V with +/-15V and vise versa.

author: aw_rw1
author_rank: Contributor
timestamp: August 02, 2022, 03:06:47 pm
content: QuoteYes, those are are protection diodes.QuoteI found that out right before ordering some opamps thinking they could handle +/-24V power supply. Instead of +/-24V, I am using +24/0V and 0/-24V to power opamps. Could you guess what happens when you use 0/-24V with an inverting amp?

author: David Hess
author_rank: Super Contributor
timestamp: August 02, 2022, 03:08:48 pm
content: Quote from: Zero999 on August 02, 2022, 10:59:24 amBe really careful if you try that in a precision circuit. The OP-07's open loop gain is marginal for this level of accuracy, and such a heavy load will make it worse. In any precision application, the load on the operational amplifier's output should be minimized.

author: TimFox
author_rank: Super Contributor
timestamp: August 02, 2022, 03:32:12 pm
content: Quote from: David Hess on August 02, 2022, 03:08:48 pmAlso, the high output current will heat up the OP-07 and increase its DC errors. An external buffer (inside the feedback loop) will keep the heat away from the precision amplifier.

author: aw_rw1
author_rank: Contributor
timestamp: August 02, 2022, 03:51:29 pm
content: David Hess,Thank you for pointing out the LM109 regulator configuration to get precision +18V. I have to think about this to implement it. How did you get +18V at the output? The schematic has +10V at the output. Did you change the input or the resistor values?- RW

author: David Hess
author_rank: Super Contributor
timestamp: August 02, 2022, 04:53:13 pm
content: Quote from: aw_rw1 on August 02, 2022, 03:51:29 pmAny positive regulator can be used in place of the LM109.The feedback resistors, R2 and R3, to the inverting input set the output voltage based on the reference voltage, so these need to be precision resistors and one needs to be trimmed to adjust the output voltage if you want high accuracy. The Kelvin connections shown implement remote sensing, removing errors from the wiring resistance.When I did this circuit as I described it, the load regulation from 0 to 1 amps was about a microvolt. Of course your requirement is for much lower current, so a lower current regulator like a 78L05 or 317LZ should be used. A transistor can be used in place of the regulator to save cost but some changes have to be made to keep the voltage at the output of the operational amplifier within its output range.

author: Zero999
author_rank: Super Contributor
timestamp: August 02, 2022, 09:04:39 pm
content: Quote from: TimFox on August 02, 2022, 03:32:12 pmGood points. If the load is constant, a resistor can be connected to the output, to keep the current low.Another option is to use another op-amp to take most of the current. This has the advantage of providing over-current protection, which you don't get with pass transistors. It can be any crappy op-amp such as the old 741, or use the LM358 with both sections in parallel, via resistors.

author: David Hess
author_rank: Super Contributor
timestamp: August 02, 2022, 09:40:33 pm
content: Linear Technology designed their LT1010 buffer to be used inside the feedback loop of an operational amplifier to boost its current capability and preserve precision, but they cost more than an integrated regulator. A diamond buffer using 4 transistors can be used, or a single transistor, but all of these solutions require a level shift if the operational amplifier is powered by the precision output voltage, which an integrated regulator provides for free.Another option is to use a fast high current operational amplifier as a buffer, but again, this is more expensive than a 3 terminal integrated regulator and still needs a level shift.The level shift problem when a single transistor is used can be made somewhat simpler by using a precision operational amplifier which has a rail-to-rail output, but such parts cost significantly more than an OP-07. I do not think it is worth it because the level shift can be done with a low voltage zener diode. Tektronix used this method when they had a 30 volt operational amplifier controlling the output of a 50 volt regulator.Why not use a common emitter output transistor, like a low dropout regulator? This can be done, but it complicates frequency compensation and is likely to respond poorly with some loads.To sum up my point, using a regulator as the pass element not only simplifies the circuit and provides better protection, but it is also a less expensive option. 78L05s and 79L05s cost less than 50 cents in single units these days.

author: Zero999
author_rank: Super Contributor
timestamp: August 03, 2022, 07:46:27 am
content: I don't understand the requirement for level shifting, fast high current operational amplifiers and linear regulators. The output current is only 10mA and it's powering a resistive load. Fair point, this may be too much to draw from the OP-07, whilst preserving the precision, but nothing a cheap LM358 can't handle.Here's how the negative side would look. I've shown the dual LM358 connected in parallel. There would need to be two LM358 chips, one for +18V and another for -18V, as the total power supply voltage would be too high to use one channel for the + and the other for the -, but they're cheap.

author: aw_rw1
author_rank: Contributor
timestamp: August 03, 2022, 03:42:54 pm
content: Quote from: Zero999 on August 02, 2022, 09:04:39 pmYes, a buffer is needed. When I use a non-inverting amp for 15V and 18V with a ref, the voltages are fine. When I add an inverting amp for -10V with the ref, the voltages drift by 0.1V.Quote from: Zero999 on August 03, 2022, 07:46:27 amThanks for inverted version. For both versions, I don't get why the voltage followers are in parallel. Should they be in series?

author: Zero999
author_rank: Super Contributor
timestamp: August 03, 2022, 05:02:34 pm
content: The advantage with having the buffer op-amp(s) in parallel is it doesn't add an extra delay in the feedback loop, which would increase the likelihood of oscillation.Look up current divider to work out the output current through each op-amp's output.

author: aw_rw1
author_rank: Contributor
timestamp: August 03, 2022, 09:30:27 pm
content: Quote from: Zero999 on August 03, 2022, 05:02:34 pmThat makes sense. I'll try the buffers in parallel.

author: David Hess
author_rank: Super Contributor
timestamp: August 04, 2022, 12:39:06 am
content: Quote from: Zero999 on August 03, 2022, 07:46:27 amCheck the difference between the schematic I posted and yours. In mine the precision output is also the supply for the reference and operational amplifier, removing power supply rejection and line regulation errors. But this also means that the precision operational amplifier's output cannot rise above the precision output to drive an emitter follower or even a buffer.With an operational amplifier configured as a follower to buffer the output, a low voltage zener diode, a couple of normal diodes, or even a base-emitter junction in breakdown could replace the short from output to inverting into to make the follower have a positive voltage offset.In a less demanding application, power the OP-07 from the input supply and add a transistor emitter follower to buffer its output.

author: nctnico
author_rank: Super Contributor
timestamp: August 04, 2022, 12:57:43 am
content: I'd be wary of using a regular opamp to supply currents in the 10's of milli-amp range. Most aren't designed to do this. An OPA189 might be a better choice to create precission supply rails to deliver several 10's of mA as a buffer to a precission voltage reference.

author: Neomys Sapiens
author_rank: Super Contributor
timestamp: August 04, 2022, 02:39:49 am
content: Another possibility is to use a dedicated symmetric tracking regulator.Some ecamples: RC4194 (also XR..), SG1501/2501/3501,

author: Zero999
author_rank: Super Contributor
timestamp: August 05, 2022, 09:38:00 am
content: Quote from: David Hess on August 04, 2022, 12:39:06 amIt's clever but I thought it was overkill, given the original poster is already powering the op-amps from a +/-24V regulated power supply.https://www.cui.com/product/resource/pqde6w-t.pdf

author: David Hess
author_rank: Super Contributor
timestamp: August 05, 2022, 03:03:03 pm
content: Quote from: Zero999 on August 05, 2022, 09:38:00 amBut the original poster wants 10 millivolt precision at 18 volts.

author: Zero999
author_rank: Super Contributor
timestamp: August 05, 2022, 05:02:59 pm
content: Quote from: David Hess on August 05, 2022, 03:03:03 pmI don't see why the circuit I posted won't do that, when fed from a regulated +/-24V rail.

author: David Hess
author_rank: Super Contributor
timestamp: August 05, 2022, 07:54:13 pm
content: Quote from: Zero999 on August 05, 2022, 05:02:59 pmOh, it would meet the requirements. I just originally posted how I have handled this problem in the past, but with higher current and precision levels.

author: Terry Bites
author_rank: Super Contributor
timestamp: September 03, 2022, 09:07:52 am
content: You might want to look at a diff driver like a ADA4922-1