# What is the problem with connecting a negative voltage to an op amp's negative lead while the possitive one is grounded?

I'm reading a manual on op. amps that presents the following schematics:

And says:

[Context: Vin goes from -0.1V to -1V]

When Vcc is powered down while there is a negative voltage on the input circuit, most of the negative voltage appears on the inverting op amp input lead. The most prudent solution is to connect the diode, D1 , with its cathode on the inverting op amp input lead and its anode at ground. If a negative voltage gets on the inverting op amp input lead, it is clamped to ground by the diode.

So, while Vcc is connected, the positive terminal will be positive, therefore the negative one will necessarily be so as well. Then, there is no problem with having Vin (negative) connected to it.

What I don't understand is the situation that the paragraph above describes. First, it tells that Vcc is powered down, in other words, the positive lead would be grounded. Therefore, Vin, which is negative, would be connected to a virtual ground and, for some reason, this requires to protect the op. amps. with D1. What exactly would happen in the negative lead if Vcc is 'powered down'?

Therefore, Vin, which is negative, would be connected to a virtual ground and, for some reason, this requires to protect the op. amps. with D1.

First thing, if Vcc is powered down there cannot be a virtual ground so yes, there could be a situation where a negative voltage appears at the inverting input.

Secondly, for every op-amp I've ever used, if $$\V_{IN}\$$ is fed via a resistor (i.e. $$\R_G\$$ in the circuit) then it is highly current limited. So, providing that the negative voltage from $$\V_{IN}\$$ (via $$\R_G\$$) cannot push more current into the op-amp input pin than what the data sheet tells you is the absolute maximum, then it's a non-event and the diode is not needed.

Typical maximum currents are about 5 mA and, to get 5 mA flowing into the op-amp inverting input pin requires: -

• (a) For the pin to be about 0.3 volts below the negative power rail (i.e. starting to cause the chip parasitic diodes to conduct) and...
• (b) $$\V_{IN}\$$ to be -50.3 volts when feeding via the 10 kΩ resistor, $$\R_G\$$.

Basically, in all but a really small number of applications, you don't need the diode. Here's how much current you can push into an LM324 input pin: -

It's a whopping big 50 mA (an old device) but more modern devices are somewhat less like the AD8605: -

For the MCP6004 it's 2 mA (taken from data sheet): -

If relying on this current as a means of protection, always check the latest data sheet and don't push it to the limit.

Most parts (not just opamps) require that their inputs do not go beyond the power rails by a small amount (usually about 0.3V to 0.6V).

With the device powered down, the normal feedback mechanism will not operate because the internal circuits that would normally drive the output to keep the inverting input equal to the non-inverting input are themselves powered down.

Without the diode, the inverting input could exceed the rating (probably an absolute maximum rating). With the diode, the most negative the inverting input can go is one diode drop which will keep the input terminals away from the absolute maximum rating. The forward voltage of the diode will be quite low as the current is limited by Rg.

Although the current will be low, whether the diode is necessary or not depends on just how long the amplifier can 'tolerate' the inverting input being beyond the specification.

Update: If Vin were positive, the diode would need to be in the opposite direction. For a bipolar input, parallel diodes in opposing directions would provide protection.

• > Without the diode, the inverting input could exceed the rating [...] ---- that would happen as well if Vin were positive, right? In such case, D1 should be 'inverted'? Feb 4, 2021 at 16:45
• @Martel answer updated. Feb 4, 2021 at 16:51
• If that is the case, that diode should be necessary in almost all op. amp. designs, since Vcc could be always 'powered down'. Is this correct? Feb 4, 2021 at 16:54

Assumptions:

• power to Op Amp changed from Vcc=10V to 0V
• and Vee from 0V to -10V for input from -0.1 to -1V with a gain of -5.
• Vin+ reference was 300k/302k * 10V must also be shifted down 10V so all gnd symbols become Vee = -10V (EXCEPT INPUT GND REFERENCE)

There are a few ways to power down,

1. open circuit single supply ( in this case now it is -10V).
2. Vee decays up to 0V low impedance
3. Open Vcc =0, while -10V is still active ( not cool )

First, it tells that Vcc is powered down, in other words, the positive lead would be grounded. Therefore, Vin, which is negative, would be connected to a virtual ground and, for some reason, this requires to protect the op. amps. with D1. What exactly would happen in the negative lead if Vcc is 'powered down'?

First of all Gnd = 0 V by definition so the input does not change and it is still connected to your arbitrary 0V node. Everything will still work with the above changes, but your real question is what happens to a circuit if you power down the ground side of a single supply using method 1. Open circuit.

Essentially nothing works as expected, but there may be residual node voltages since power is still on. Inputs will start to bias internal ESD protection diodes to start biasing the open Ground connection. As long as Input currents do not exceed Absolute Maximum limits, no damage.

The virtual ground is created by Vin+ and when both the inputs are inside the Vcm range and output is not saturated the negative feedback drives Vin- to the same threshold as Vin+ output.

In order for your circuit to work the Op Amp must work with Vcm = Vcc which only occurs of specific types that work to one rail or both. BJT types often do not unless specified. So level shifting methods are sometimes done. As already stated the diode is only for emergency protection depending on OA specs and input transients.