how to limit voltage output of op amp to symetric range

[danjel]

I have an inverting op amp stage that is summing various control voltage sources. I would like to limit the resulting sum to a symmetric range (e.g. +/- 5v).

The op amp is powered off +/-12V.

One idea was to find a true rail-to-rail, dual supply opamp for which the output symmetrically swings withing mV of the supply. So far I have not been able to find something suitable.

Another option is to use schottky clamping diodes on the outputs and reference them to a + and - voltage (in which case I will need to add regulators to supply the reference voltages).

I am trying to solve this with the least amount of parts possible.

This seems like it would be a common problem. What are the most conventional solutions to this?

 

[ericgibbs]

hi,
Have you considered using say, two 4.7V zener diodes in opposing series across the OPA's feedback resistor.??

EDIT:
Something like this, add the 1N914 diodes.

 

[danjel]

Since the Zeners have small forward voltage drops you could just use Two Zeners (no need for the 1N914) in series with each other and in opposite polarity.

The Postive limit would be: Vz1+ Vd2
Negative limit: Vd1+Vz2

If I used 4.7V zeners and they had Vd of about 1V then I would get a limit of 4.7+1 = +/-5.7V

 

[ericgibbs]

Since the Zeners have small forward voltage drops you could just use Two Zeners (no need for the 1N914) in series with each other and in opposite polarity.

The Postive limit would be: Vz1+ Vd2
Negative limit: Vd1+Vz2

If I used 4.7V zeners and they had Vd of about 1V then I would get a limit of 4.7+1 = +/-5.7V

If you recheck the circuit I posted you will see that I used 4.3V zeners and 1N914 have a forward voltage drop about 0.7v

 

[MikeMl]

Here is a similar circuit which produces a "squared" output. Note what happens at node X.

 

[danjel]

If you recheck the circuit I posted you will see that I used 4.3V zeners and 1N914 have a forward voltage drop about 0.7v

Yup I saw it and it works too. All I am saying is that you can get the same result with only two diodes (just two zeners total) not four diodes.

These are two solutions I found. The second one reduces the negative effect of diode leakage current.

**broken link removed**

**broken link removed**

 

[crutschow]

If you want a near symmetric plus and minus clip, the use one zener diode in the center of a diode bridge circuit. That avoids the tolerance voltage difference between two zeners of the same nominal voltage. But with that you now have two forward diode drops added to the zener voltage.

The bridge circuit can be use in the feedback circuit as danjel showed, or clamping the output as Mike ML showed.

 

[danjel]

Do you have a link to a schematic of this solution? I can't quite work this out....

 

[danjel]

like this? and so the clamp voltage is Vz + 2* Vd ?

 

[ericgibbs]

Do you have a link to a schematic of this solution? I can't quite work this out....

hi,
This is that version.

 

[crutschow]

hi,
This is that version.

Eric, your version is non-inverting, which means that the gain is still 1 when the clamp is operating. That gives the rounded waveform clamp levels shown in the simulation.

An inverting version should give sharper clamp levels.

 

[danjel]

You mean inverting op amp configuration (like my diagram?)

This also explains why the clamp voltage seemed to be higher than what I calculated since the combine forward voltage drop should be 0.6+0.6+3.3 = 4.5V but his simulation is showing a clamping above 5v

 

[MikeMl]

...
This also explains why the clamp voltage seemed to be higher than what I calculated since the combine forward voltage drop should be 0.6+0.6+3.3 = 4.5V but his simulation is showing a clamping above 5v

Your assumption that the forward drop of a Si diode is 0.6V is far too simplistic.

 

[danjel]

Your assumption that the forward drop of a Si diode is 0.6V is far too simplistic.

care to elaborate? Obviously it is an approximation and I guess in reality it is closer to 0.7V. What calculations should I be doing then?

 

[MikeMl]

care to elaborate? Obviously it is an approximation and I guess in reality it is closer to 0.7V. What calculations should I be doing then?

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

 

[danjel]

Thanks that is very illustrative!!

btw what simulator program are you using? I need to start using this....

 

[crutschow]

Here's my simulation with a 3.9V zener and a 10kΩ feedback resistor for an un-clamped gain of 10, which gives a clamp voltage near 5V. The actual clamp voltage will vary slightly with the input and feedback resistor, since that varies the clamp current and thus the voltage drop across the diodes.

 

[MikeMl]

Thanks that is very illustrative!!

btw what simulator program are you using? I need to start using this....

LTSpice. What Spice should have been from day 1....

 

[ericgibbs]

Eric, your version is non-inverting, which means that the gain is still 1 when the clamp is operating. That gives the rounded waveform clamp levels shown in the simulation.

An inverting version should give sharper clamp levels.

hi,
As suggested, inverting OPA Gain *10.

 

[crutschow]

LTSpice. What Spice should have been from day 1....

Berkeley Spice day 1 was done with IBM punch cards and batch processed on a main-frame computer in the computer room with not a computer alpha-numeric terminal (much less a graphic terminal) in sight. It was great fun waiting for the half-day turn-around of your run to see if the simulation worked or not (printed on 15" wide continuous-fold paper with sprocket holes by a noisy impact line printer running at 10-20 pg/min).

But I imagine most of you are way too young to remember any of that.