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How can an op amp be prevented from going into saturation if the feedback is intermittently disconnected?

For example, in this circuit (simplified case of a real-life problem), the op amp acts as a current source to a load, but the load may be disconnected sometimes.

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When the load is disconnected, the op amp output goes to the positive rail and the op amp goes into saturation. When the load is reconnected, the op amp takes extra time to start regulating current) and then slews to the expected current set point. Depending on the op amp, the time to recover from saturation may be very long. The current through the load is the maximum possible for that time (ouch).

How can saturation be avoided in this case? Are there some additional components to a feedback network that would do it? Perhaps some kind of input or output clipping circuit? Are there op-amps that would inherently limit their outputs (or inputs) some voltage away from the rails using built-in circuitry?

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  • \$\begingroup\$ Isn't it usually a bad idea to disconnect the load from a current source with an open circuit? Your circuit is trying to force a constant current through an infinite impedance, which requires infinite voltage, hence you get saturation. \$\endgroup\$ – DaveP Jan 28 '15 at 2:38
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    \$\begingroup\$ @DaveP: You sometimes have to. For example if it's a pluggable device,, or in my case if you need to switch the load between the current source and something else. \$\endgroup\$ – Alex I Jan 28 '15 at 10:07
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A zener diode connected from op-amp output to inverting input (possibly with a series std diode) and NOT switched by S1 plus a resistor from Vsense to inverting input will limit Vout+ excursion. If this is dual supply then back to back zeners will do the same thing symmetrically.

When Vout approaches Vzener negative feedback is provided. The resistor from OA- to Vsense needs to be large enough for the zener to dominate with minimal effect from Rsense.
A 1K should be fine but something like 100 x Rsense for low values of Rsense should be an OK compromise. Zener leakage at low output deviations should be "low". A more elegant solution implementing the same principle with more complex circuitry would yield truly minimal effect when the load is connected.

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Added:

The centre cannot hold!* I knew I should have added the extra :-). I thought about commenting about frequency response but didn't. As WhatRoughBeast has pointed out,the zeners have capacitance which may need to be accounted for, although in most cases the effect is probably minimal. eg with say Risol= 1k and if Czeners = 1 nF then the time constant is t=RC = 1000 x 10^-9 = 1 uS. With 100 R it's 0.1 uS. Whether this matters or matters much depends on the application.

Zener capacitance varies with (at least) model, applied voltage (forward or reverse) temperature, frequency. Actual values can vary widely but 1 nF seems a good rule of thumb to start with. Low capacitance versions are available.

The effect of the forward biased zener in series with the reverse biased zener at voltages << Vzener is left as an exercise for the student.

This 69 page RENESAS application note provides an excellent overview of zener diode charactyeristics. Pages 29-31 provide information on zener capacitance aspects - with numerous graphs showing examples of voltage versus capacitance.

Series:
.............. Capacitance at 0.1 V
HZS-LL ....1-10 pF
HZS-L .....10-40 pF
HZS ....... 30-200 pF HZ ......... 30-200 pF

BUT this older ONSEMI application note TVS/Zener Theory and Design Considerations indicates values in the 1 to 10 nF range in some cases. Capacitance starts on page 34.


These zeners are lower capacitance than many at 150 pF typical at 0V at 1 MHz. Capacitance falls with increasing reverse voltage.

Here are some ROHM zeners specifically designed to be low capacitance.

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    \$\begingroup\$ But also be aware that zeners have appreciable capacitance, and this may affect the frequency response of the amplifier. \$\endgroup\$ – WhatRoughBeast Jan 28 '15 at 1:16
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The only way to keep the OpAmp from saturating is to provide an inner feedback loop.

This could be done by changing the switch to from a SPST to a DPST type and adding a local feedback resistor Rfb.

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When SW1 and SW2 are open feedback would be provided by Rfb. Value of Rfb would be much larger than that of RLoad so that with the switches closed, RLoad and Rsense would dominate. For example if RLoad were 1k Ohm, Rfb could be 100k Ohm.

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You know when S1 is on, and when it is off (or open). Create a S1b (inverse of S1) signal, and using it in the following situation:

Your opamp is either differential (and you are providing a simple diagram), or differential-to-single-ended. In either case, you can short the

1 - differential outputs in a differential amplifier 2 - the single-ended output to the internal node in the differential branch.

Of course, this kills the gain, but everything is biased properly and the amplifier doesn't saturate.

We do this all the time in our circuits. It's simple, and it works.

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The easiest solution would be to parallel the load with some kind of nonlinear network such as two series Zener diodes, two back-to-back LEDs or two diodes. Of course the leakage would take current from the load, so that may or may not yield acceptable performance.

Op-amps that limit are available, but they're not all that common. You may also be able to find a conventional op-amp with a short recovery time.

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  • \$\begingroup\$ I see! In my case the load draws a significant amount of power, so paralleling it with anything (that would maintain meaningful feedback) is not practical. But paralleling Rsense with a diode looks possible :) \$\endgroup\$ – Alex I Jan 27 '15 at 23:21
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You should examine why you are disconnecting the opamp from its load in such a manner. If you want to shut down the current, driving the positive input to 0 would be better.

What are you trying to accomplish overall? Why do you think you have to break the connection between the load and the output of the opamp? Step back two levels and explain what's really going on.

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