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.
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:
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
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.
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:
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