From dual to single supply op amp circuit

[_Bd_]

Hi, I was just curious, I am no EE, I am an ME and I am trying to make something smaller if you have read my other posts you might know what's it all about, anyways I've been reading a lot about single supply op amps and that's probably the best choice to reduce the amount of power my circuit will take, I want to try to run it off batteries if possible.

Here are 2 circuits, The one on the top is the original op amp using an LM324 but using it with double supply +/- 5V, I am trying to achieve the same thing but using a single +5V supply and a VREF (VEE) that's half of that (roughly 2.5 V) or maybe less?

What I would like help with is finding out the values that will give me an equivalent circuit? How do I find the values of all C's and R's such that the signal is processed in the same fashion as the original (top) circuit?

Thanks for all your help so far
http://img41.imageshack.us/img41/7928/0fph.png
NOTE: for the top circuit, pins 4 and 11 are connected to +/- 5V respectively


as you can see I used the fourth op amp (of the LM324) to make a virtual ground, or reference voltage VEE which should be half of VCC (I don't know what resistor values I should use tho)

 

[sophiecentaur]

Afaik, the only reason for using +/- supplies for an OpAmp is so that you can have input signal values on either side of the system Ground. You can provide a reference which is around half way between the two terminals of a single supply and an AC coupled OpAmp circuit should work without a problem. If your input signal voltage is at or near the single ended power supply ground (DC coupled, as shown) then you could have a problem. Your buffered mid-rail reference may not be needed.
If you really need DC coupling then you have to provide some DC level shift of the input signal.

 

[meBigGuy]

VEE shorted to ground may be an issue.

Basically the circuit will behave the same around VEE as the original behaves around ground. If you are OK with that, and clipping at 0 and 5V, then you are good to go.

For a complete answer, you need to specify the input range and output characteristics you need.

 

[analogdesign]

One potential issue is the common-mode level of the input. It's not clear from your diagram but is it DC coupled? If so can you ensure that the input signal is within the common-mode range of your op amps?

 

[_Bd_]

Ok, let me re-state my question, I think I have a clearer understanding of what my problem is. (A problem well stated is a problem half solved right?)

INPUTS:
I am doing something similar to a pulse-oximeter (IR-LEDs + IR-Photodiodes), this is the circuit that will process the signal coming from the photodiode, from what I've read in multiple articles circuits all have a transimpedance amplifier for the diode, a low pass filter (in this case a butterworth), and an inverting amplifier, I am not sure about the order, as I've seen them mixed, (some switch the inverting and transimpedance first and last)

My attempt:
Anyways, I am using LTSpice, and AFAIK to model a photodiode you can do it with a regular diode and an AC current source, and that should model something like a photodiode right?

Ok, so I did the original circuit, as intended with a dual supply, +/- 3.3V and here is what I got when I modeled it:
http://imageshack.us/a/img29/1781/feua.png Now here's the thing, since that's the original circuit left untouched, I want to get a similar output with a single supply Op amp circuit, I understand that instead of negative voltages I should get zeros because the circuit itself is limited by my +Vcc and zero, but still I expect something similar to that without the valleys. . . right?

Anyways, here's what I get when I copy that exact circuit, without the dual supply, from what I've understood of switching from dual to single, you just need to reference the chip's -V to ground and instead change the other leg of the op amps to a Vref (or virutal ground) which is usually half your Vcc, anyways here's what I got which I believe is not what I should be getting, my question, what's wrong :(
http://img706.imageshack.us/img706/7433/5z82.png

Or is that a good graph?
the thing is that I don't really know what to expect, and what's "good" or "bad"?

 

[sophiecentaur]

Did you mean to post that? I can't spot any differences. Am I just being dumb?
[Edit - there have been some changes since my post - I couldn't be THAT dumb!]

 

[analogdesign]

I don't really know LTspice schematics, but it looks to me like the output of reference buffer U4 is grounded. This grounds all the non-inverting inputs of the op amps. Also, is the bottom of your reference divider floating (R2)? It doesn't seem to be connected to ground. I would probe both the inverting input of U4 and its output to see what's going on.

As an aside, you're burning 10 mA in your reference divider. Is that on purpose?

 

[_Bd_]

fixed,

BTW U4 on the single_supply is just a voltage splitter that gives me that virtual ground (vcc/2) or vref to be able to power the chip using single_supply isntead of double.

yes thank you for the reference to R2, I already fixed that but I still get pretty much the same graph, just that the first bump is a little thicker

 

[sophiecentaur]

Is that reference amp right? You have it connected as an inverter with positive feedback.

 

[analogdesign]

Yes that is the intention for U4. Its output should be Vcc/2 = 2.5V. But it looks like it is grounded (the output is connected to ground so it will be forced to 0V instead of 2.5V.). Can you probe the output terminal of U4 please?

 

[analogdesign]

Is that reference amp right? You have it connected as an inverter with positive feedback.

I didn't notice that but Sophiecentaur is right. The output is 1. Not connected as a unity gain buffer and 2. The output seems to be pegged to ground.

 

[_Bd_]

so how should I connect that op amp? I wanted to be the reference amp as you suggested

I probed as you said, and I tweaked a little by adding some resistors, here's waht I got, i think it looks a lot better, at least closer to the sinewave input. . .but yeah as you said the reference stays at 5V instead of the Vcc/2 as I would like

http://img5.imageshack.us/img5/1758/d5ba.png

(Blue line is the probe right outside the reference op amp)

 

[analogdesign]

The reference is DC so C6 is blocking it. Get rid of it. Also I don't think you want the different op amps to have different bias points so get rid of R10 and R11 too.

 

[sophiecentaur]

If you really want a reference amp then connect the op amp as a non-inverting unity gain amp. Your half rail volts go to the positive and the output goes to the negative. Simples. Or chuck it and bias each OpAmp separately to half rail.

 

[_Bd_]

well I tried removing R10 and R11 and it gives me some weird stuff:

Red line = AC-signal from current supply
Blue Line = probe after the vreference op amp
Green Line = Signal out

http://imageshack.us/a/img826/7752/901q.png [/PLAIN]
thats with the capacitor removed and with the leads changed to make it vcc/2
thanks for thatand this is the weird one:
http://imageshack.us/a/img689/5107/ph6h.png
just removed those 2 resistors.

BTW I added those 2 resistors because of a paper i read from TI that says how to design single supply op amps and that's how they show an inverting op amp.

 

[sophiecentaur]

I would think it better not to drive the circuit with a rail - to - rail input signal swing. Did you consider the DC gain of the input stage?
Electronics is very unforgiving (like Computer Programming). A circuit only does what you tell it to do and not what you want it to do - just like a computer. Looking at your circuit, I am not at all surprised at what the simulation is giving you.
I can sympathise with your confusion, to some extent but I feel it may be better for you to launch out on a slightly more basic circuit than this, for what must be a new experience for you. Look at each stage individually, if you want more insight.

 

[_Bd_]

So, would you then say that that single supply circuit I designed (The first one on my previous post) will not work for my purposes?

I was honestly pleased with those results, it does look relatively similar to the original (the first picture on my second post),
a "squared" sine wave that oscillates between +3.3 and zero with a center somewhat in the 1.65V,

I mean I showed this to a Bioengineer (not an electrical), and he said he thinks that signal will suffice for my purposes, but Id trust an EE over a BE for this stuff anyday :P,

Again, Let me re-state this in the same post,
do you think this:

http://imageshack.us/a/img826/7752/901q.png is close enough to this:http://imageshack.us/a/img29/1781/feua.png

If your answer is "depends on the application"

The application as I said is something similar to a pulse oximeter, but not for medical uses (therefore accuracy is not such a high priority). I just want something that can sense when you move your finger (so this sensor will be on your knuckles) and when you move your finger your blood flow changes, that change will (hopefully) be detected by this circuit and then I can use it for whatever other purposes.

 

[sophiecentaur]

You have achieved the power supply change. That's good.
What I don't understand is why you have followed the first, squaring stage with the other stages which are filters. They are achieving nothing afaics as they are linear designs. What did you want those filters to do? Engineering is all about defining your requirement.

 

[analogdesign]

If the biomedical engineer thinks the circuit is good enough, it probably is. Remember fitness to purpose. Sophiecentaur makes a good point though. Obviously the output is saturating (you have a limiting amplifier) so the value of the filtering is not clear. If this is a done deal and it meets specs it doesn't matter though, but something to think about.

 

[_Bd_]

I would like it not to be squared, TBH I would love to get a perfect sine wave as my output but that's as good as I could get.

My original purpose was to get the same circuit but with single supply as opposed to dual, but after everything was done according to all those articles I've read about single supply that's the signal I got, which is still a million times better looking that the first attempts .The purpose of each stage:

First: (transimpedance amplifier)
This is just a current to voltage converter, AFAIK is to make it "easier" to read the outputs from the photodiode.

Second: 39 Hz Butterworth filter:

A low pass filter with a cutoff frequency of 39, TBH I don't know how to set the cutoff frequency but I assume is set by the values of the resistors/cap in that op amp, which is why I did not change the values on those resistors/caps.So first you amplifiy the signal, then you filter it, then you re-amplify it to make it easier to read? (that would be my assumption, again I am an ME not an EE so I am not too sure about this)

But thanks for all the help you guys have been extremely helpful!

 

[sophiecentaur]

Your first stage isn't doing what it should, clearly. The red line on your graphs shows a massive 500mA swing(?) (If the right hand scale is really the input current). I see the input source is written 5m, on the diagram. Is that 5mA? Have you actually set it up as a current source? I can't be sure what the legend on the graph means: I[I1] ? I am not familiar with that way of simulating a photodiode. A Google search throws up more complicated models- but you are only working at low frequencies so it may be OK. It will require some playing about with actual values in that part of the model, I think.
Reasons for my comments:
Vout of the first stage should be 4700*Iin so, for less than 5Vpk-pk out (to allow some headroom, you should be hitting it with much less than 1mA pk-pk (by my reckoning. If the source really is 5mA, then you are driving the circuit five times too hard, and that could explain the limiting behaviour. The DC Voltage gain of the filters is in the order of 3X, by my eyeballing so you want even less current input swing. That would be easy to try and it may give your filter a chance to do something useful (poor little thing, with all that signal belting through it!)