# How to protect OpAmp output from short circuit to +Vcc?

I am using an OpAmp to amplify an input signal that ranges from 0-3.3V to a 0-5V range. The output must only be greater than 0 if the input signal is as well.

If I understand it correctly, if +Vcc were shorted to the output, the output would be constantly at 5V. Is there a way to protect against this error case?

I have tried to come up with something or find an existing solution, but I'm rather new at electronics and maybe I'm missing the right terminology.

EDIT: I changed the schematic according to Huisman's comment.
The possible reason for a short between +Vcc and the output could be that something(a loose wire, metal shavings, etc.) physically connects the two OpAmp legs, that there is a break between the tracks on the pcb, or anything else, really. For safety considerations I'm supposed to assume a short occurs, no matter how it actually happens.

simulate this circuit – Schematic created using CircuitLab

• If V1 = Vcc, please change your schematic accordingly. (While naming it Vcc, do also rotate V1/Vcc 180 degrees.) If +V1/+Vcc and it were shorted to the output, there is a big chance the opamp wil get damaged. Before suggestions can be made for protection, you'd clarify first what the cause is +Vcc is shorted to the output. Commented May 10, 2019 at 10:18
• If the supply is shorted to the output then, the supply is shorted to the output and, apart from detecting this with some other circuit and flashing a beacon or sounding a buzzer, it will remain shorted until someone unshorts it. It's like asking if there were a way of preventing punctures on a bicycle or vehicle - no there isn't. Commented May 10, 2019 at 10:36
• Opamps aren't used to amplify signals in the order of a few V but in the order of a few μV. Commented Apr 12, 2021 at 10:12

This one is not internally protected against shorts, so you should put a resistor in series with the output. Tune the resistor value so the maximum current and dissipation in the opamp (and also the resistor) is not exceeded in case of short.

The possible reason for a short between +Vcc and the output could be that something(a loose wire, metal shavings, etc.) physically connects the two OpAmp legs, that there is a break between the tracks on the pcb, or anything else, really. For safety considerations I'm supposed to assume a short occurs, no matter how it actually happens.

If that happens, then it can happen anywhere on your PCB, for example on the component that controls the laser, turning it on permanently.

A circuit on a PCB cannot protect itself against random pins being shorted, unless it is dipped in tamper-proof resin. But then, it cannot protect itself against random component failures either. If someone drops the device on the floor and a heavy component falls off the board, or your capacitors crack, or the PCB cracks, it's probably not going to display a helpful error message, except maybe smoke signals.

So, unless the output of this opamp goes through a connector or something that is much more exposed to short circuits than the PCB itself, it is pointless to protect it.

However a circuit can protect itself against expected failures. For example, in an audio amplifier, if an output transistors blows and welds itself short, the output will be the power supply voltage, with enough current to melt your loudspeakers. So you can have a separate output DC detector circuit and a relay that disconnects the output. This is possible because the most likely (and expensive) source of failure and lawsuits is expected and known.

In your case, if you want proper safety, the safety mechanism should be something like a dead man's switch that would protect against many more failure mechanisms.

For example, put a simple resettable monostable on the laser driver, that the microcontroller has to reset every couple of milliseconds. That way, if the firmware has a bug, or gets stuck in an infinite loop, the laser will turn off. Note firmware bugs are much more likely than hardware bugs.

Then, in your firmware, validate your inputs. This can also be done in analog. For example, if you expect the normal output voltage of your opamp to be 0...4.5V, but you expect 5V in case something fails, like a sensor fails, or the laser driver transistor burning and welding itself short, then an analog comparator can turn off the power supply, for example. If a vital safety mechanism depends on a sensor, it should make sure the sensor is operating correctly, and at least do the right thing if the sensor fails open or short.

Is there a way to protect against this error case?

Protect against what?

I guess you want that nothing breaks when that happens. What needs to be done is that the current that can flow in that situation (when the output is shorted to 5 V) is limited. If the current is kept small enough then nothing can break.

Many opamps have an output that cannot deliver so much current, some even have current limiting. Besides that a resistor can be added between opamp output and the point where the short to 5V can be made.

For the AD8605 it is important to let the current not exceed 30 mA, as that is the maximum current for which the output is designed (see page 6 of the datasheet).

• Thank you for your answer. I want to protect against the output being at 5 V when it's not supposed to be. The output controls a laser, which would be permanently turned on in the case of a short. That should not happen and I'd rather have some parts break and have to replace them, as long as that would ensure the laser is not turned on. Commented May 16, 2019 at 9:00
• In that case, you don't want to protect the output from short circuit but rather from being stuck. That's a whole other jar of pickles you just opened. Commented May 16, 2019 at 9:04
• @Janka I was afraid of that. Are there any starting points you could recommend to find a solution to this? Commented May 17, 2019 at 7:22
• At not supposed to be. You have to measure the intensity of the laser and check whether it's supposed to be that luminous in that very moment or not. And if not, cut the power supply. And you better do that in a different, simple circuit which fails open. Commented May 17, 2019 at 8:21