Opamp circuit for ADC input

[erikl]

Im looking to minizime the board, so im wondering if you could use the same "bias circuit" for all 4 Opamps, without damaging the effect.

Sure you can!

Same question for the protection voltage aswell, is it enough with 1 cap and 8 diodes, or should i use 1 cap for every oppamp aswell?

1 cap and 8 diodes is fine. Or use that more accurate clipping circuit below. Replace the upper branch by your 4 TLV2262 circuits, and use the lower branch (the actual clipping circuit) only once. Replace the 3.3V voltage source by a 2.8V one (resulting from the voltage divider + cap described in a former posting). Then you need only 4 diodes (D1) after the output series resistor (1 .. 10kΩ) - i.e. from the ADC inputs - to the output of the lower branch buffer.
Have fun!

 

[bt36]

Hi erikl, Tom-nor,

I enjoyed reading this post and Erik's great answers ... what a pleasure

Erik, I was wondering with what software are you producing those nice
schematics with ?

Thanks in advance,

Gal.

 

[erikl]

Erik, I was wondering with what software are you producing those nice
schematics with ?

Hi Gal,
I used Qucs on a Mac. See here or there.
Erik

 

[Tom-nor]

thnx for great help Erik!

I think i will stick to the 8 diodes for now, dident understand your circuit so gna look closer at it later on.

I attached a simple circuit, with two signal inputs and 2 opamps.
I skipped the "safety" circuitry at the "end", just to keep it simple.

In this case the bias is about 0.6V at the opamp output, but my question is how this type of "sharing bias" will affect the input impedance?

Can i use 4x Opamps with feedback circuits and 1x bias circuit. Like in the circuit attached, 2x Oamps and 1x bias circuit.

I think i made my self unclear in the question earlier, by asking if i could use the "same" circuit, i ment if they could "share" one bias circuit.

will it still be 10 M || Oppamp input resistance ?



Regards: Tom

 

[erikl]

I think i will stick to the 8 diodes for now,

Ok. Now I think you can even save the 4 diodes to GND, because the opAmps cannot output a negative voltage - as long as you use a common GND.

I attached a simple circuit, with two signal inputs and 2 opamps.

In this case the bias is about 0.6V at the opamp output, but my question is how this type of "sharing bias" will affect the input impedance?

will it still be 10 M || Opamp input resistance ?

In this circuit, you have connected (short-circuited) the 2 channels after the separation caps (C2 & C3). I'm sure that's not what you'd want! You must use an extra R7 for each channel - both (i.e. all 4) connected to the common node R5-R6-C1. In this case, all channels will see the 10 M || opAmp input resistance. For R6 better use a 5kΩ potentiometer, so you can adjust the opAmps' quiescent output voltage(s).

 

[Tom-nor]

yeah i figured that was wrong.

Awesome stuff bro!

 

[djsfantasi]

I have been working on a what I hope to be a similar problem, so I apologize if I am unintentionally hijacking this thread.

In my application, I want the input to come from a laptop audio/headphone jack. I am only interested in the envelope of the positive side of the signal. I also need to scale the signal from 0-1.55V to 0-5V ...or thereabouts; I have some 'slop' in my application. What I want to do is detect the level of the signal with a 0-5V ADC (converts 0-5VDC to 0-255 binary), in order to programatically determine how far to drive a servo. Would a Schottky diode be a better choice than a silicon diode here? I don't think so because I am not interested in precision scaling, but rather a coarse scaling.

I modified the circuit to run on 5VDC, supplied by the ADC power supply (it actually is part of LynxMotion's SSC-32 servo controller) to include a diode in the input line, to only process the postive side. Additionally, I have modified the values of R1 and R2 for a gain of 4.9. This way, the 1.55V peak signal is dropped to .95V by the diode, but with a gain of 4.9, I expect a peak output voltage of 4.7V


I am NOT trained in electronics, but can read a circuit, so I am sure I am missing something here. One issue is that I am not clear how the original requirement of an input offset voltage should be handled in my application; I do not think that requirement applies and do not know where to modify the circuit if that is the case.

Can someone assist me in developing a circuit for my requirements?
Thanks,
dj

 

[erikl]

... I am only interested in the envelope of the positive side of the signal. I also need to scale the signal from 0-1.55V to 0-5V ...or thereabouts; ... Would a Schottky diode be a better choice than a silicon diode here? I don't think so because I am not interested in precision scaling, but rather a coarse scaling.

Can someone assist me in developing a circuit for my requirements? dj

Hi dj,
you don't even need a diode, moreover you can save the bias devices R3, R4 & C3, see the schematic below. If you use an LM324 instead of the TLV2262, you'd have to scale down R5 about one unit of magnitude (i.e to 1MΩ) and increase C1 to 1µF.

Have fun! erikl

 

[djsfantasi]

Thanks a lot... This looks good.

What about the capacitor on the output side (C4 in my original diagram), to smooth out the signal spikes during the missing negative going pulses? What would be an appropriate value? The max frequency I would expect to process is 20kHz.

 

[erikl]

What about the capacitor on the output side (C4 in my original diagram), to smooth out the signal spikes during the missing negative going pulses? What would be an appropriate value?

An opAmp with such a high capacitive load directly at its output is prone to oscillation. Moreover, as the opAmp's output is very low-resistive, 47nF won't help a lot. Separate C4 with a ≈1kΩ resistor from the opAmp's output, then tinker with its size to get the right smoothing!

 

[djsfantasi]

I did some modelling using LTSpice. I found the following circuit seems to do what I want, but don't know why. Perhaps you could help explain it for me.

The circuit shown below is what I came up with.


Basically, it is your circuit (Thank You Very Much) with a simple envelope follower circuit at its output and before the input to my microprocessors built-in ADC.

I tried adding a capacitor after a 1K resistor, but the voltage drop in the resistor negated the gain in the op amp circuit and I couldn't get the signal to smooth without the full RC network of C3 and R3.

I also tried a resistor in place of the Schottky diode, but got no "smoothing" result. A standard silicon diode also caused too much voltage drop in the output stage.
So I think this circuit will do what I desire. I couldnt get LTSpice to simulate with varying amplitude levels on the AC signal, so will have to breadboard it to see how that works (wish I had an oscilliscope).

Do you see anything herein that I missed?

Thanks again, erik,
dj

 

[erikl]

... I found the following circuit seems to do what I want, but don't know why. Perhaps you could help explain it for me.

In my 1st answer to you, it was a simple (one-way) rectification circuit with an opAmp amplification circuit. The rectification occurs because the opAmp's input is related to GND (via R5), hence the negative part of the wave simply isn't processed. The opAmp's gain is set to (1+R2/R1)=3 , which should achieve a max. output voltage of 3*1.55V = 4.65V, close to the opAmp's maximum.

... with a simple envelope follower circuit at its output and before the input to my microprocessors built-in ADC.

By this you built a peak detector with a decay time constant R3*C3=0.66s. That means, without more or higher input signal peaks, the level decays by 63% after 1 time constant. This is the traditional automatic recording level control method (the time constant isn't standardized, it should depend on the type of audio signal). See the PDF below.

I tried adding a capacitor after a 1K resistor, but the voltage drop in the resistor negated the gain in the op amp circuit and I couldn't get the signal to smooth without the full RC network of C3 and R3. I also tried a resistor in place of the Schottky diode, but got no "smoothing" result.

You can get a true smoothing with a 1k serial resistor and C3, but you'd have to re-amplify this signal again by a factor 10 .. 100 . Smoothing, however, is quite a different application from peak detection.

A standard silicon diode also caused too much voltage drop in the output stage.

Sure, a Schottky diode definitely is better here.

So I think this circuit will do what I desire. I couldn't get LTSpice to simulate with varying amplitude levels on the AC signal, so will have to breadboard it to see how that works (wish I had an oscilloscope).

Do you see anything herein that I missed?

No, it's fine if you want automatic recording level control - and not smoothing.

 

[djsfantasi]

What I want is to be able to:

1) Amplify the amplitude from 0-1.5 V to 0-5 V to use the entire scale of the uP built-in ADC (the rectifier and amplifier portions of the circuit) Approximately. I want to use a greater portion of the input scale...

2) Be assured that the uP accurately gets the value of the envelope and doesn't miss it during the negative going pulse or close thereof. (the "smoothing" or "envelope follower" portion of the circuit)

I am using it to drive a servo motor for an animatronic mouth. When the volume/level of the sound is higher the mouth should open wider. When it is quiet, the mouth should close.

I did find when plotting the output against varying amplitude levels, that it was decidely non-linear with everything above ~.8V in, the output was flat (See attached graphs). When I reduce R@ to 50K ohms, the behavior is what I expect; What is the function of R1 and R2 in the circuit?

Thanks for all your help - I am learning much from you!

 

[erikl]

... I did find when plotting the output against varying amplitude levels, that it was decidely non-linear with everything above ~.8V in, the output was flat (See attached graphs). When I reduce R@ to 50K ohms, the behavior is what I expect; What is the function of R1 and R2 in the circuit?

I already explained this in my answer from Sun, 28 Mar 2010 18:41 (1st paragraph).

 

[djsfantasi]

Yes, you did. Sorry... But I do not understand why the results of my simulation do not agree with the results expected from calculations based on your explanation.

With a 50K resistor for R2, I would expect the gain to be (1+50/100)=1.5, which should achieve a max. output voltage of 2.33V (1.5*1.55 = 2.33V); however the simulation shows that I receive a max. output of 3.83V and the results are fairly linear from amplitudes of .1V to 1.5V (at audio signals of 85Hz, 110Hz and 155Hz). There is a slight bend near 1.3V, but if my experiments are correct, I might expect that reducing R2 to a slightly lower value will also straighten out this part of the graph as well.

Level....Va
0.1......0.136
0.3......0.593
0.5......1.133
0.7......1.693
0.9......2.263
1.1......2.838
1.3......3.416
1.5......3.821

So... the part that I do not understand is why the gain I see in the circuit simulation is different than what I should expect from your formula?