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# Single supply opamp cascade

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#### vaka85

##### Advanced Member level 4
Hi,

I was searching the net for this topic but it's hard to find...

I have a sine signal obtained by an active opamp filter from a square wave.
This opamp has single supply, and so the reference for input and output signals is Vcc/2.

But if I have to add a second amplifier stage (another opamp), how can I connect it to the output of the first stage?
In other words, how can I refer it to the virtual ground (Vcc/2)?
Because if i connect it directly to the input of the second stage it doesn't amplify anything...

Thanks

In many cases, you can simply "copy" the DC bias to the second stage. But without knowing your circuit, it's hard to say why it fails.

Wherever you are supposed to connect the second opamp to grond, connect it tothe same vcc/2, except its power pin. Remember to bypass the vcc/2 to ground unless you enjoy motorboating.

---------- Post added at 23:04 ---------- Previous post was at 22:58 ----------

Wherever you are supposed to connect the second opamp to grond, connect it tothe same vcc/2, except its power pin. Remember to bypass the vcc/2 to ground unless you enjoy motorboating.

I'll assume you're applying the sine wave to the non-inverting input of the second op amp.

For experimentation, connect the wiper contact of a potentiometer at the inverting input. Put a scope on the output. Use the pot to sweep between V+ and ground. Eventually this should bring the signal into view.

You've just found the average voltage of the waveform coming from the first amp.

There's a chance that vcc/2 does not match this. Offset errors and whatnot.

Your reference at the inverting input of the second amp has to match the average voltage of the waveform coming from the first amp.

Last edited:
vaka85

### vaka85

Points: 2
Your reference at the inverting input of the second amp has to match the average voltage of the waveform coming from the first amp.

wow! now I understand!
thank you all guys, tomorrow I try if it works and then I'll let you know...

it doesn't work...

here is the schematic...

the first opamp is an active filter, to make some filtering on the received signal from the antenna. This stage works fine.

The second one doesn't! I've tried a lot of resistor value, a lot of potentiometers, but without results...
I have a sine wave on the output, but it isn't amplified...

If I am right: if the first stage output has 0 as average value, the reference voltage for the second stage should be 0... Right?

I don't know how to proceed... how the first stage works and the second don't??

thanks

it doesn't work...

the first opamp is an active filter, to make some filtering on the received signal from the antenna. This stage works fine.
...............

I am a bit surprised - the 1st stage works fine? Are you sure? How do you know? To me, it seems to be impossible because of incorrect dc bias.

It works because I've tried it with a scope.
The input is a sine wave 500 mV peak to peak, and the output is a sinewave with an amplitude of about 1 V peak to peak...

Why are you saying that the dc bias is wrong?
How it should be?
with a single supply, the reference for all the signal should be Vcc/2? doesn't it?

thanks for your help...

The dc bias voltage at the non-inv. input is zero (ground), right?
The dc bias voltage at the inv. input is finite and unknown since the dc output is undefined.
Thus, the opamp does not work in its linear region.
Did you measure the dc output?
Perhaps the presented drawing does not reflect the real circuit?

---------- Post added at 14:36 ---------- Previous post was at 14:17 ----------

Most probably, the output is around zero - and the inv. input is biased with VDD/4.

vaka85

### vaka85

Points: 2
The dc bias voltage at the non-inv. input is zero (ground), right?
yes

The dc bias voltage at the inv. input is finite and unknown since the dc output is undefined.

yes. I think it's around Vcc/2 since the potentiometer is positioned at his half...

Thus, the opamp does not work in its linear region.
Did you measure the dc output?
Perhaps the presented drawing does not reflect the real circuit?

the dc output is zero.
The schematic is like the real circuit...

However, I've tried the circuit with dual supply and it works fine... I think I'll keep this version..

Anyway I'd like to learn how to identify the correct bias to apply to the inverting input...

thanks

yes
However, I've tried the circuit with dual supply and it works fine... I think I'll keep this version..
Anyway I'd like to learn how to identify the correct bias to apply to the inverting input...
thanks

If you can live with dual supply - fine.
Moreover, solutions for single supply operation have been discussed several times in this forum; you also can perform an internet search ("single supply") and you will get a lot of references.
But why do you prefer to bias the inverting input? Normally, this input is connected with the feedback path and the bias is provided at the non-inv. input.

I've found a very useful guide on single power supply from Texas Instrument, it says only to connect the reference to Vcc/2, that's why I didn't understood the dc bias issue...

But why do you prefer to bias the inverting input?
Isn't this the only way to do a non-inverting amplifier?

Isn't this the only way to do a non-inverting amplifier?

No. Look into your application notes and you will see that Vdd/2 is applied to the non-inv. input.
Then, of course, the signal has to be applied via a coupling capacitor.

Apart from biasing the 1st stage +ve input correctly, you can ease the bias setup by amplifying AC only. You won't need a potentiometer to adjust the bias.

It's very hard for me to understand this issue... I'll search in the forum for some discussions..

Thank you very much for your help

It would be easier with a basic understanding how an OP works. A visual explanation can be found here Lessons In Electric Circuits -- Volume III (Semiconductors) - Table of Contents Chapter 8 Operational Amplifiers

Of course, there are much more text books and tutorials.

To make an OP amplifier circuit work, the feedback network must be able to achieve zero input difference voltage. Analyzing your circuit, you'll realize, that this can't ever happen for the first stage (with single supply). The +ve input is tied to ground, but the -ve input can't go below Vdd/4, as LvW already mentioned.

vaka85

Points: 2