0
\$\begingroup\$

I have a question about an idea of mine, concerning the split up of a 5 V supply voltage into a 3.3 V and a -1.7 V voltage. My circuit has to be as efficient as possible and therefore I don't want to use a combination of a simple 3.3 V voltage regulator and a voltage conversion IC (like the ICL7660). The negative voltage is only used as negative supply voltage for a precision operational amplifier whose output voltage will never be under -100 mV, but for sure will be negative too.

My idea was to use something like this, to stabilize the important 3.3 V voltage for the microcontroller on my board and leave whatever difference between the supply voltage (normally 5 V, but can vary because of battery usage) and the 3.3 V remains as negative reference voltage: Sorry, I don't have a better picture at the moment, but I hope you get the idea

(The used parts are only symbolic, I didn't take the time to find exact components for my circuit and needs, before I know the concept even works)

I don't need high currents, like I said it is a fairly low power circuit which this supply-circuit should power. Would this be a feasible option or did I miss/overengineer it too much? I tried simulating it and the results looked promising in the sense, that I could vary the supply voltage and load resistance (not in the picture) to a certain degree and everything looked stable.

So my questions now are:

  1. Can this concept work or what flaws can you think about?
  2. Will this be more efficient as the voltage regulator/voltage inversion-approach? Like I said I don't really care how big my negative voltage reference voltage is, because the op-amp will never go rail-to-rail on the output.
  3. Building on the last question, is it even a good idea to supply an op-amp with an unsymmetrical supply voltage, like 3.3 V and -1.7 V in my case?
  4. Any tips and tricks you could help me out with or do you have a better or easier approach in mind? Please let me know if that is the case.
\$\endgroup\$
5
  • \$\begingroup\$ since you did not draw a real circuit i did not bother to analyze the circuit :) but yes, the concept of splitting a power supply by creating a new rail in the middle is fine. And doing it with an op-amp buffer is fine. I think there are better ways to create the Vcc-3.3V reference voltage for the input to the buffer. \$\endgroup\$ Commented Nov 1, 2021 at 14:50
  • \$\begingroup\$ You can set the circuit ground point anywhere you like, but it will not be very efficient if you use linear regulation to adjust the voltage levels. \$\endgroup\$
    – tobalt
    Commented Nov 1, 2021 at 15:10
  • 2
    \$\begingroup\$ Building on the last question, is it even a good idea to supply an op-amp with an unsymmetrical supply voltage, like 3.3 V and -1.7 V in my case? To the opamp, it looks like you're powering it with 0 V and + 5 V. Which is not an issue for an opamp that can work with a 5 V single supply. My point: the opamp doesn't "know" you're making +3.3 V and -1.7 V and the opamp doesn't care either. So yes, this will work. \$\endgroup\$ Commented Nov 1, 2021 at 15:11
  • \$\begingroup\$ @user253751 thanks for the comment. may you elaborat about a better way to create the vcc-3.3v-reference? I know that it is for sure not perfect, because using all those different op-amps is for sure not the most efficient way. I couldn't think of a better way but would appreciate it, if you could provide an approach :) \$\endgroup\$
    – Rata
    Commented Nov 1, 2021 at 15:56
  • \$\begingroup\$ You can make it work; but it won't have any effect on energy efficiency. \$\endgroup\$
    – user16324
    Commented Nov 1, 2021 at 16:08

1 Answer 1

1
\$\begingroup\$

Can this concept work or what flaws can you think about?

Yes, it's fine.

Note that if you connect 3.3V stuff between the +5V and +1.7V rails, current goes into the 1.7V rail towards ground. You need to be aware of this because most voltage regulators are only designed for current to go out of them. Op-amps work both ways so you don't have this problem.

Note that all of that current goes through the op-amp. Make sure the op-amp is rated to handle all of the current that goes through your 3.3V stuff. This is a linear type of regulator - excess voltage is dissipated by creating heat. So make sure the power dissipation of the op-amp is okay too.

Will this be more efficient as the voltage regulator/voltage inversion-approach? Like I said I don't really care how big my negative voltage reference voltage is, because the op-amp will never go rail-to-rail on the output.

A minimum of 1.7 5.0ths, which is 34%, of the power gets wasted as heat. Whether that's okay for you, I can't say. It's up to you. You might be able to do better with some type of switching regulator.

If the amount of power isn't very much, it might not be a problem.

Building on the last question, is it even a good idea to supply an op-amp with an unsymmetrical supply voltage, like 3.3 V and -1.7 V in my case?

Op-amps don't actually know where 0V is. No circuit knows where 0V is. Op-amps, though, don't even have a 0V pin. They need a voltage that's higher than all your inputs and outputs, and a voltage that's lower than all your inputs and outputs. As long as all your input and output voltages are between the + voltage and the - voltage, with some margin (check the datasheet), and the + voltage and - voltage aren't too far apart (check the datasheet) this isn't a problem.

Any tips and tricks you could help me out with or do you have a better or easier approach in mind? Please let me know if that is the case.

The op-amp to create the voltage is fine, however, the way that you get the 1.7V reference voltage for the op-amp input is unnecessarily complicated. You use a 3.3V regulator to get 3.3V above the ground rail, and then you use op-amp math to "flip it upside down" to get 3.3V below the +5V rail.

For one thing, these regulators are designed to handle reasonable amounts of current and you are only demanding a tiny trickle. Some regulators require a minimum amount of load or else they won't regulate properly.

For another thing, it's just overly complicated.

Make a voltage divider between +5V and ground. But instead of a resistor, in the top half put a 3.3V shunt voltage reference (like this one but not this exact one).

An alternative is a 3.3V Zener diode. However low-voltage Zener diodes actually aren't that accurate as the voltage changes with the current. If you use a Zener diode you can put a trimpot in the bottom half and adjust the current while checking the voltage with your multimeter.

Of course you can also just make a plain old resistor divider. 1.7V is 1/3 of 5.0V so the top resistor should be twice the bottom one. I just assumed you wanted to regulate the 3.3V in which case you can use one of the other two methods.

Anyway, now that you have a voltage divider with 3.3V in the middle, just connect the middle of the voltage divider to the input of your op-amp voltage follower.

\$\endgroup\$
8
  • \$\begingroup\$ Thanks for the extensive answer. About the 3.3V voltage regulator, I knew that it was overkill but it was the first thing that popped into my mind and I actually searched for easier possibilities to create the 3.3V reference. I just used it primarily because the model was preinstalled into LTspice and I wanted a proof-of-concept. But now that you gave me an idea for a better alternative and now knowing it can work, I will look into finding a better way. Just one last thing, what did you mean by "A minimum of 1.7 5.0ths", could you clarify? \$\endgroup\$
    – Rata
    Commented Nov 1, 2021 at 17:16
  • \$\begingroup\$ @Rata I mean 1.7 out of 5.0, 1.7 divided by 5.0, if you split up the thing into 5.0 parts then 1.7 of them. Of course I could write 1.7/5.0 but where's the fun in that? \$\endgroup\$ Commented Nov 1, 2021 at 17:20
  • \$\begingroup\$ I have to admit i can't follow you. I understand now what you meant, but why would 34% of the power go up in heat? I understand that a while using a voltage regulater 34% of the power go to waste and therefore I tried another approach. But should't in my case only minimal power go to waste, i still have the 1.7V voltage drop over the regulator but considering that the current should be absolutely low, there shouldn't be much power loss or what am I missing something? Same while using a shunt voltage reference/zener diode, do I have other power losses then the one in the diode/resistor? \$\endgroup\$
    – Rata
    Commented Nov 1, 2021 at 17:38
  • \$\begingroup\$ @Rata you have a 1.7V drop over an op-amp \$\endgroup\$ Commented Nov 1, 2021 at 17:39
  • \$\begingroup\$ So i won't have a better efficiency than going with a voltage regulator, but at least i don't have to introduce another component which draws power for generating a negative voltage reference, correct? \$\endgroup\$
    – Rata
    Commented Nov 1, 2021 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.