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I'm designing a circuit with a simple op-amp summer as part of it.

I've noticed a weird behavior of the output voltage, if I increase the resistor value at the output, the output voltage drops. This is undesirable behavior to me. I'm curious why it happens, and what design choices I can make to remedy the situation. The resistor change is something like changing R19 to 1M instead of 1k. The output is on the circuit's output, so I'd like to protect against a user plugging into a large load or something.

Circuit

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    \$\begingroup\$ Why do you want to increase the value of R19? What are you really trying to accomplish here? \$\endgroup\$ Commented Oct 16, 2020 at 22:10

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First, this is probably vast overkill for your stated purpose -- most op-amps will inherently current limit before they damage themselves, so your resistor scheme is unnecessary. If you are using a power op-amp that can destroy itself, then any resistor scheme that effectively protects it will ruin it's power output capabilities.

(It could, however, make sense if you were trying to protect whatever was being driven by the op-amp, be it some sort of actuator or some sensor that needed bias, but not too much).

Second, if you just have to do it, here's how. Bring R5 (your R19) into the loop. The op-amp will then swing as much as it needs to in order to force Vout to be the correct value.

Third, depending on the load, you may or may not need a bit of compensation to prevent oscillation. If the load is capacitive, you'll need to find a value of C1 that'll stabilize the amplifier. I'm not going to go into depth on how to do that, because, first, you probably don't need R19 at all.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Protecting against short-circuit is often not necessary, but a series resistor can also accomplish protecting against being connected to a large external voltage. Most of the voltage gets dropped by the resistor instead of forcing the opamp output to rise above supply voltage or below GND. \$\endgroup\$
    – jpa
    Commented Oct 17, 2020 at 20:09
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It's because you're connecting it to a load on the output which draws non-zero current. The bigger the current coming out of your op amp, the bigger the voltage drop on that resistor. The larger the resistance, the bigger the voltage drop. Without seeing the rest of your circuit I can't say for sure, but I'd guess that you can do without R19 altogether.

At the risk of sounding condescending... V=IR. Do a basic nodal analysis of that circuit with a resistor load, and you should understand what's going on.

If you absolutely must have overcurrent protection, a better design choice might be a resettable PTC fuse. However as stated in TimWescott's answer, it's likely overkill since most op amps can handle being shorted to ground without frying.

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  • \$\begingroup\$ The output is facing out of the circuit. So a user could connect a wire or whatever with a large resistance. Furthermore, wouldn't a large resistance lower the current and therefore the load the amp has to drive? \$\endgroup\$ Commented Oct 16, 2020 at 22:11
  • \$\begingroup\$ A large resistance will lower the current by increasing the voltage drop. Again, I can't really advise you on whether it should be there without context for what you're actually trying to do with this circuit. \$\endgroup\$
    – Ocanath
    Commented Oct 16, 2020 at 22:16
  • \$\begingroup\$ V does = IR (usually) \$\endgroup\$
    – Ocanath
    Commented Oct 16, 2020 at 22:16
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    \$\begingroup\$ If you're trying to protect the op amp from a user accidentally shorting it to ground, you could (probably) use a resettable, low resistance PTC fuse \$\endgroup\$
    – Ocanath
    Commented Oct 16, 2020 at 22:18
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If the output of the op-amp is going to a user, it's best to set R19 to be a high enough resistance to prevent an overcurrent from the op-amp.

If the max current of the op-amp is 5mA (have to check datasheet), and the max voltage was 5V, I would choose R19 to be 1k. That way if the user attached R19 to ground it would draw 5mA and not damage the amp.

This is an oversimplification, and more information is necessary to give you a complete answer.

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    \$\begingroup\$ That would protect the op-amp but the voltage supplied to the load would be lower than the op-amp output because you have created a potential divider. For example, if feeding a 1k load then it only gets half the expected voltage and current. Ocanath explains why moving the 1k resistor inside the feedback loop solves both the original and the problem you have introduced. \$\endgroup\$
    – Transistor
    Commented Oct 17, 2020 at 16:47

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