How can I supply −5 V to an op-amp?

Question

This is the circuit that I intend to implement on a breadboard.

I only have an ESP32 microcontroller, so I can supply the +5 V, but I have no idea how to supply the −5 V.

If I change Vee to 0 V (GND on the ESP32), the circuit will not work as intended.

Frame challenge: Do you NEED a negative rail? Or is it okay to just offset the inputs and outputs? If you can get away with the latter, then you can avoid the complexity of creating an opposite-polarity power rail that you don't already have. This is done surprisingly often with consumer audio: The analog I/O is centered on GND, which is the negative-most rail, while all of the processing is centered on half of a single supply. The translation is usually done with an R-C highpass at each end, but you can also use an active DC offset if you really need a DC response. — AaronD, Oct 09 '22 at 20:24

Andy aka · Accepted Answer · 2022-10-08T15:18:24.990

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This is the circuit that i intend to implement on a breadboard.

I wouldn't if I were you; The op-amp used has a maximum supply voltage of 7 volts so, if your new combined rail becomes + and - 5 volts, you'll exceed the op-amp maximum ratings by 3 volts.

So, this is the first thing you need to put right.

If you want a quick method of making a temporary source of - 5 volts use a 9 volt battery and a 7905 regulator. If you want a more permanent solution the ICL7660 looks a strong candidate: -

Image from ICL7660 CIRCUITS.

Or, you can use the pin-for-pin equivalent Linear Technology part: -

Image from Inductorless 5V to -5V Converter.

edited Oct 08 '22 at 15:18

answered Oct 08 '22 at 14:56

Andy aka

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Perhaps i can just change the supply to +-3V, it will still work. Thanks for warning me about that – Scipio Oct 08 '22 at 15:07
Why have you used two identical parallel circuit for processing V2? – Andy aka Oct 08 '22 at 15:15
I took this circuit from here: https://electronics.stackexchange.com/questions/373492/analog-analog-multiplication-part-of-a-hybrid-cpu-for-fun/373608#373608 , It's in the answers. Still need to figure out the transfer function but it probably will have logs because it has diodes as feedback in the op amps – Scipio Oct 08 '22 at 15:25
Isnt there a way to have negative voltages without buying more equipment? – Scipio Oct 08 '22 at 15:29
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He hasn't paralleled two circuits; one is fed from V1 and the other is fed from V2. You have circuit redundancy (good news for a change). – Andy aka Oct 08 '22 at 15:29
It's the way it is because i want the square of the input, so in reality there's only one signal not two. – Scipio Oct 08 '22 at 15:31
@G0tBlackOps isn't that the same as asking if it were possible to have more stuff without spending more money. I think you probably know the answers to that; spend or steal. – Andy aka Oct 08 '22 at 15:31
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You don't need both parallel circuits. – Andy aka Oct 08 '22 at 15:32
Are you saying i can just take one out? Also, do you know any circuit configuration that can work as an analog multiplier as this one but without have to supply negative voltages ? – Scipio Oct 08 '22 at 16:52
@G0tBlackOps You don't need both circuits. Run a simulation and see. The link you gave to where you took the original circuit has an analogue multiplier idea by me. – Andy aka Oct 08 '22 at 17:00
You're right, it works. Why did OP there added two equal circuits? – Scipio Oct 08 '22 at 17:03
He was driving one with V1 and the other with V2. They were separate inputs. – Andy aka Oct 08 '22 at 17:09
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So, I've saved you some money that you can invest in an ICL7660 or a LTC1046 but, post your circuit first (if you are not sure) and post it in a new question. This question is flogged to death now. – Andy aka Oct 08 '22 at 17:11

score 6 · Answer 2 · answered Oct 08 '22 at 21:05

You can make a simple negative voltage charge pump using a square wave output from your microcontroller. The following circuit assumes a 3.3 volt power supply for the microcontroller, and produces -2.7 VDC into a 10k load. You can reduce this by using silicon diodes (1N4148), and also use one to drop the 3.3 volts to about 2.6 VDC.

Higher current can be obtained by using an NPN/PNP buffer.

score 3 · Answer 3 · edited Oct 08 '22 at 16:03

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A very elegant approach to making symmetric voltages from a single 9 V battery is the following:

From Eliott Sounds on HowTo make a Soundcard DAQ

The purpose of the op-amp is to create a clean ground.

Update: @Andy Aka brought up a very important point. You cannot supply more than +/- 2.5 V to the power rails from this op-amp. Look for an alternative one with at least 15 V of supply voltage swing.

edited Oct 08 '22 at 16:03

ocrdu

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answered Oct 08 '22 at 14:56

Daniel Melendrez

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The op says this: *i can supply the +5V*. Not sure if he can provide a floating 9 volt supply for anything other than a temporary test. – Andy aka Oct 08 '22 at 15:54
@Andy aka I have used this circuit in multiple portable instrumentation circuits and it works like a charm. He wouldn't be supplying 9 V rater +/- 4.5 V. If he uses this supply's ground as the common ground, then what is your reasoning regarding the _temporary test_? – Daniel Melendrez Oct 08 '22 at 19:10
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I'm not saying the circuit is bad or doesn't work. I'm saying that the op has +5 volts and 0 volts; as far as I can tell, he doesn't have another supply voltage. If he wished to use a floating supply like a 9 volt battery then, the duration is limited in time (battery depleted) as I pointed out in my answer. – Andy aka Oct 08 '22 at 19:44
@Andyaka Is there some way i can just use this circuit while avoiding having to supply negative voltages? – Scipio Oct 10 '22 at 06:51
@G0tBlackOps needing a negative supply will depend on the nature of your input signal and the type of opamp that you use. If you need full voltage swing (between +/- 4.5 V) and if your opamp can work without the negative rail. Some amplifiers are not guaranteed to operate under optimal conditions without this negative rail supply – Daniel Melendrez Oct 10 '22 at 12:01

score 1 · Answer 4 · answered Oct 11 '22 at 02:40

How can I supply −5 V to an op-amp?

You don't - you don't have to do that.

Instead, run the amplifiers referenced to a small positive voltage - about 0.8V is plenty enough, since that's just the maximum forward voltage expected across the diode at the maximum currents expected in this circuit.

Also, U6 in the circuit you show is redundant. It doesn't change the voltage seen at the input of U2.

A fixed-up version of your circuit would look as below. The op-amps are supplied from 0V and 5V rails. They need to be rail-to-rail input/output types, such as MCP6021, or TLV6001, etc.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Incidentally, even though the small signal 1N4148 diodes are usually better in small signal circuits than the general purpose 1N4001, for a squaring circuit it's better to use the latter. That's because the 1N4001's non-ideality factor is closer to 1 than the 1N4148's.

The response of the circuit, that is V(Z)-VGND vs V(x) is plotted below.

How can I supply −5 V to an op-amp?

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