Ideas for input/output overload protection

Discussion in 'General Electronics Chat' started by pvh1987, Jun 17, 2014.

  1. pvh1987

    Thread Starter New Member

    Nov 20, 2012
    7
    0
    I have been working on some circuits for overload protection of sensitive inputs and outputs. These circuits are to be used for example in measurement equipment such as function generators and voltmeters where the user of the equipment could place nasty voltages at external terminals either intended or by mistake. I am currently designing a digital curve tracer and I need to protect the inputs and outputs.

    I came up with a circuit for input protection that not only protects the input from overvoltage conditions but also provide indication of overload. The circuit consists of two parts: A clamp and a limiter. The limiter is based on a well known circuit: A diode bridge rectifies the voltage so only one zener diode is needed to limit the voltage for both polarities. This bridge includes a LED as well. This LED will light up when the zener is limiting the voltage. This LED could also be the input of an optocoupler. One can imagine the output of the optocoupler could make an interrupt in a micro controller that disconnects the input with a relay in the case of overvoltage.

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    Large version

    Since the whole thing is buffered, resistance in the path from the input to the operational amplifier should not be a problem since the input impedance of the op amp is very high. Of course low values of R1 and R2 should be selected since this makes the limiting more precise. Care should be taken not to overheat R1 or R2 or any of the diodes. The limit voltage of the zener bridge should of course be lower than the supply voltages.

    The clamp makes it much easier to dimension the bridge component. The voltage at the bridge will never be larger than V+ + 0.7V or V- - 0.7V.

    Using an optocoupler with a forward voltage of about 1.4V, 1N4148 diodes, a 10V zener, the voltage limit can easily be calculated: 1.4V + 2*0.7V + 10V = 12.8V. I then find that R2=200Ω is the lowest resistance that will not destroy the optocoupler I intend to use. I chose R1 to be 100Ω rated for 4W is enough to protect the input up to at least +/- 35V. The inputs of the op amp will never get above +/- 13V which is well below the supply.

    I tested this circuit today and it worked very well :)

    So, what about output protection? That is a totally different story. The output needs to be a low impedance path so this is much harder. Suppose the op amp is outputting close to -15V and the user connect the output (at the other side of the resistor R) to +35 V. Now the resistor R has 50 V over it and if R is, say 10 Ω to keep a low impedance, that will put a current of 5 A through the resistor and op amp. Even if the op amp will limit the current, it won't like the voltage. However, it might be possible to insert a clamp like this:

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    Large version

    It does not solve the problem. If V+ and V- are +15V and -15V respectively, the op amp output and inverting input could be at +/- 15.7 V - it will most likely not tolerate that. Furthermore, there is no LED or anything else to indicate an overload condition. Using LEDs instead of regular diodes in the clamp are not a good idea in my opinion (higher voltage drop).

    So, while it seems to me that I have found a solution for the input protection circuit, I would like your view on a suitable output protection circuit with overload indication. I do not care if it is simple or a complicated. In my application I will not put the output resistor R in the feedback loop or make it larger than 10 Ω, due to stability considerations and to keep a low impedance. The resistor is rated for 10 W.

    Thanks in advance for any ideas that could lead to suitable output protection circuit :)
     
  2. eetech00

    Active Member

    Jun 8, 2013
    648
    110
    Hi

    I think most modern opamps have built in protection against an output short circuit to ground or either supply, or
    choose one that does.

    What if you used "R" also as a current sense resistor for input to a current monitor? It could be used to drive an LED for overload indication..(?)


    eT
     
    Last edited: Jun 20, 2014
  3. pvh1987

    Thread Starter New Member

    Nov 20, 2012
    7
    0
    Thank you, eetech00. Yes, I actually use the resistor "R" as a shunt monitor to measure the current going through the load at the output terminal. This is done with an instrumentation amplifier. To make things more clear, I have made a new circuit drawing:

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    Large version

    The gain resistor for the instrumentation amplifier is not included in the drawing but present in the real circuit. I choose OPA1612 since it can supply 20-30 mA without problems and still have relatively low offset and stuff like that.

    In my opinion, there is two potentially dangers with this circuit.
    1. The inverting input of the instrumentation amplifier is directly connected to the output terminal where a high voltage could be present if the user misconnects the terminal.
    2. The op amp output and inverting input is not protected and could go well beyond the supply levels.

    While I could use the instrumentation amplifier to drive an optocoupler to indicate "too much current" it might indicate an overload even if it is not dangerous to the op amp. This could be solved by selecting a "trigger level" that is higher than the current the op amp would normally supply. That would be something like 50-60 mA in the case of OPA1612. Although this will indicate a problem, it will not prevent the op amp from destruction. Also the in amp has limited common-mode input range which might be a problem in this case.

    In normal use, the output terminal will be connected to a load and the op amp will not be destroyed even if the load is 0 Ohm. According to the data sheet of OPA1612, the op amp can tolerate being shorted to ground (and therefore also being shorted through the 10 Ohm resistor) with no time limit. I do not want the overload indication to indicate a problem in this case.

    The only situation that could destroy the op amp or the in amp would be an external voltage source or another kind of active load that might put a voltage back into the circuit through the output terminal. Ideally I want the indicator to activate only in this situation. Of course some protection circuit is needed to protect the circuit until the load is disconnected by a relay.

    I came up with a quite complicated circuit that I do not think is good enough:

    [​IMG]
    Large version

    Two Schottky diodes will limit the op amp output and inverting input from going over the supply by 0.3-0.4 V. OPA1612 is rated to 0.5 V beyond the supplies. The two diodes between the inputs are included inside the op amp and with the 10k resistor, the circuit driving this circuit (shown as voltage source "Vi") will not be destroyed while the non-inverting input follows the invert-input by 0.7 V in the case of an overload.

    The in amp inverting input is no longer in direct connection with the output terminal. A diode bridge (with zener and optocoupler) will limit the voltage going to the in amp and at the same time indicate if there is a problem.

    I feel this circuit is not good enough and still quite complicated. While it might protect the op amp and the in amp, the resistor "R" is now at even greater risk at being burned because the Schottky-diodes will provide a low-impedance path to the power supply and therefore the op amp will not limit the current in this situation.

    I will consider eetech00's great indication idea and try to work this into a circuit that at the same time offers protection for the shunt resistor "R" and the op amp and in amp. I might end up with two indicators, one for "too much current" and another one for "unexpected voltage at the output terminal". Two indicators might be better than one. Thanks so far.

    If you have an idea or solution to this problem, do not hesitate to post it. I am still open to any ideas. Thanks in advance :)
     
  4. dougc314

    Member

    Dec 20, 2013
    38
    11
    You could consider using a PTC thermistor for the series R (re-settable fuse) or for the utmost in protection put a quick blow real fuse in series with R.
     
  5. pvh1987

    Thread Starter New Member

    Nov 20, 2012
    7
    0
    Well, a PTC will not be a good idea. The impedance will change with temperature and at the same time make the circuit less linear.

    A resettable fuse might be a better idea, if it can react fast enough and if it is possible to reset electronically rather than pushing a button.

    Maybe I do not need to worry about the resistor "R". It is only 10 Ohm, but it is rated for 10 W and is large with a big heat sink. If I can make the relay at the output terminal disconnect within 5-10 milliseconds, the resistor will not get a chance to get hot and it will not be a problem at all. Right?

    I just need to make sure that the op amp and in amp is protected - they will burn up a lot faster than the resistor.
     
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