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To supply op-amp 1, I need a power supply voltage of +2.5V and -2.5V. To supply op-amp 2, I need a power supply voltage of +4V and -1V. The outgoing signal from the op-amp stage will be fed into an ADC, so the rails have to be stable. I thought about using a DC to DC converter, but I could not find one with a positive and negative output voltage.

  • How can I convert a 5V source to a bipolar voltage source?

I have this recommended circuit from Texas Instruments, where THS4509 is op-amp 2. They recommend a supply voltage of +4V and -1V to set Vcm as 1.5V.

schematic from TI datasheet

Image source: Texas Instruments ADS612x 12-Bit ADC family datasheet

My signal source (Vs) will be op-amp 1, the output signal from op-amp 1 can be between +1V and -1V.

  • By the way, what do I need for the resistors Rs and Rt?
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    \$\begingroup\$ This is a strange question to ask, for a person with "cuk" in their user name. \$\endgroup\$
    – The Photon
    Oct 15, 2020 at 16:12
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    \$\begingroup\$ More to the point, can you share your circuit design so far? How much current is needed by each op-amp? Could you just run both op-amps on +/- 5 V? \$\endgroup\$
    – The Photon
    Oct 15, 2020 at 16:13
  • \$\begingroup\$ No I cant run both on +-5V, because I need a Vcm of 1.5V for opamp2. I could probably run opamp1 from +-5V. So cuk converter is what I will look for \$\endgroup\$
    – electrococuk
    Oct 15, 2020 at 16:19
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    \$\begingroup\$ Just define "0V" for the first stage as 2.5V, and for the second stage as 1V. Level shift by 1.5V between stages. Then you can run them both off the same supply. \$\endgroup\$
    – user16324
    Oct 15, 2020 at 16:27
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    \$\begingroup\$ Having a common mode input or output voltage of +1.5 V doesn't mean you can't use +/- 5 V supplies. You'll need to share your circuit if you want us to be able to make sense of your requirements. \$\endgroup\$
    – The Photon
    Oct 15, 2020 at 16:41

3 Answers 3

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So you have +5V available. You can use a linear regulator to drop that to +4V and a linear regulator to drop +5 to 2.5V.

A DC-DC with isolated output will give you -5V or -3.3V, which you can then similarly regulate to -2.5V and -1V.

That's one possible overall topology. However the devil is in the details- the DC-DC will introduce noise into the +5V supply (on top of whatever other digital noise may be there) so you will probably want to add various filters to the input and output of the DC and to the inputs and outputs of the linear regulators, and note carefully the PSRR at various frequencies of the linear regulators. Usually they do better when they're not micropower regulators and when they have a bit of headroom to work with.

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  • \$\begingroup\$ Thanks! This question is off topic, but do you know for what Rs and Rt are needed, if needed at all? \$\endgroup\$
    – electrococuk
    Oct 15, 2020 at 17:07
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    \$\begingroup\$ @electrococuk It's explained in the datasheet: i.imgur.com/ebzA6hJ.png \$\endgroup\$
    – Spehro Pefhany
    Oct 15, 2020 at 17:09
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    \$\begingroup\$ @electrococuk They might be. Depends on your requirements. If your converter is actually only 12 bits it's easier than if you're shooting for 19 or 20 useful bits. \$\endgroup\$
    – Spehro Pefhany
    Oct 15, 2020 at 17:13
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    \$\begingroup\$ Thanks again, you are the best! \$\endgroup\$
    – electrococuk
    Oct 15, 2020 at 17:14
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    \$\begingroup\$ @electrococuk For driving low power opamps, I've had good luck with switched capacitor inverters like the LM2776 followed by a high bandwidth linear regulator. You have to remove the switching frequency, but if you pick the linear regulator carefully you can get it down extremely low. \$\endgroup\$
    – user1850479
    Oct 15, 2020 at 20:34
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I don't know how the real EEs feel about it (I really am just a software geek), but I've done this on small, low-frequency, breadboard projects in the past...

schematic

simulate this circuit – Schematic created using CircuitLab

The "Ground" for the rest of the circuit is the output of OA1, which it will hold at half way between the two power supply rails. I don't know what kind of improvements would be needed to turn this into a real circuit.

Probably works best with an isolated power supply (V1).

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  • \$\begingroup\$ Bypassing on the supplies for OA1 will likely make OA1 'sing'. \$\endgroup\$
    – Spehro Pefhany
    Oct 15, 2020 at 20:37
  • \$\begingroup\$ This is called a virtual ground, and is widely used. Thing is, the more your signals are differential, the more you want a true +/- source, for example in Audio applications. The biggest concern here is that the LM324 has typical sink ratings of 20-50uA. I don't know how this would function in real life with a LM324. \$\endgroup\$
    – Overrice
    Oct 16, 2020 at 8:44
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    \$\begingroup\$ If the tool allowed me to put down an op-amp symbol with no part number, then I would have done. I picked "LM324" because I dimly remembered that there are some of those in a box in my basement, which probably means they were inexpensive. I never knew that they could not sink significant current. (How can you even call that an "op-amp?") Oh well. Obviously, I am no EE. \$\endgroup\$
    – Solomon Slow
    Oct 16, 2020 at 13:19
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Again, as discussed in the comments and in your previous question: you don't need a bipolar supply at all. Set -V_s to 0 V, and +V_s to 5 V! There's nothing in your whole circuit that relates the absolute voltage between in- and output:

All signals going in are AC-coupled, all signals going out are AC-coupled, and all signal connections to ground are also AC-coupled through capacitors.

The capacitor doesn't care whether there's a static potential difference between its electrodes - it only conducts AC!

Therefore, there's nothing according to which your circuit would work differently with -+2.5 V and 5V. Notice how the specific circuit puts CM simply in the middle between -V_s and +V_s, and the difference to that voltage is all that matters. and it's the same with a bipolar or unipolar supply.

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  • \$\begingroup\$ You might be right for opamp1 where I could set +Vs 5V and -Vs 0V. But for opamp2 I need a Vcm of 1.5V. (4V-1V)/2 = 1.5V = Vcm. The ingoing signal to opamp2 will swing around positive and negative voltage. \$\endgroup\$
    – electrococuk
    Oct 16, 2020 at 10:10
  • \$\begingroup\$ That is not true. There's no DC path between the ADC's common voltage and Vcm of opamp 2, at all, nor should there be. Take out a red pen and mark all traces connected to you Vcm through components that aren't capacitors. Do you reach Opamp 2? \$\endgroup\$
    – Marcus Müller
    Oct 16, 2020 at 10:53
  • \$\begingroup\$ You are right. There is no DC path between ADC Vcm and opamp2 Vcm. Thanks for your foolproof explanation. \$\endgroup\$
    – electrococuk
    Oct 19, 2020 at 12:53

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