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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.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ 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. \$\endgroup\$ – Huisman May 10 at 10:18
  • \$\begingroup\$ 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. \$\endgroup\$ – Andy aka May 10 at 10:36
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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).

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  • \$\begingroup\$ 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. \$\endgroup\$ – Eye of Horus May 16 at 9:00
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    \$\begingroup\$ 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. \$\endgroup\$ – Janka May 16 at 9:04
  • \$\begingroup\$ @Janka I was afraid of that. Are there any starting points you could recommend to find a solution to this? \$\endgroup\$ – Eye of Horus May 17 at 7:22
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    \$\begingroup\$ 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. \$\endgroup\$ – Janka May 17 at 8:21

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