Damaging an unpowered IC by applying an input signal

Thread Starter

PeteHL

Joined Dec 17, 2014
473
One of my books on op amps advises to never connect an input signal to an op amp if the op amp lacks power supply voltages. However, that is what I would like to do, if it can be done without damaging the input buffering op amps of a signal processor, as this would eliminate the need for a switch to disconnect the input signals. In some situations, I would like to eliminate the output signal of the processor, and turning off the power of the processor would be the easiest way to do that. The schematic below shows the input stage of the processor.

Certainly under some circumstances I suppose damage could occur by applying an input signal to an IC and the IC is not powered. But I am unclear as to under what circumstances and how damage could occur.

-Pete

INPUT-&-NPWR.jpg
 

LowQCab

Joined Nov 6, 2012
4,023
Some Top-Quality Op-Amps have Protection-Diodes on the Inputs.
These Diodes work very well at protecting the Inputs,
( as long as you have the correctly sized Current-Limiting-Resistors on the Inputs ),
but, if the Power-Supply-Pins are floating, who knows what might happen.

It's a much better plan to Attenuate the Input to the Op-Amp-Circuit,
and it's extremely simple when compared to switching-Off the Power to the Op-Amp.

If You would supply a Schematic, you'll probably get 4 or 5 much better solutions.
.
.
.
 
Some Top-Quality Op-Amps have Protection-Diodes on the Inputs.
These Diodes work very well at protecting the Inputs,

Some devices have ESD protection diodes, BUT

they way an IC is made you get a shotkey diode by accident. If the inputs are restricted to negative supply -0.3V and the positive supply of +0.3V you have a max of +-.3V when unpowered UNLESS that input current is restricted. You can find that number in the datasheet. Drawback: The resistor adds noise.
 

Thread Starter

PeteHL

Joined Dec 17, 2014
473
It's a much better plan to Attenuate the Input to the Op-Amp-Circuit,
and it's extremely simple when compared to switching-Off the Power to the Op-Amp.

If You would supply a Schematic, you'll probably get 4 or 5 much better solutions..
The signal processor is a finished unit that when normally operated does not need adjustable attenuation of the input signal. So I would prefer not to add attenuation as it is a finished unit and I would like to keep the circuit design as simple as possible. The schematic in my post #1 shows the input stage of the processor. That is, the input to the processor is a buffering stage with the op amp configured as a voltage follower.

Thanks,
Pete
 

ericgibbs

Joined Jan 29, 2010
18,766
hi pete,
As that circuit is a high impedance Non inverting buffer, you could add say a 4k7 in series to pin #3 from the junction of the 33n and 47k
Many device d/s show the maximum permissible input current for the Input pins, check the d/s.
E
 

Irving

Joined Jan 30, 2016
3,843
What is the active line level signal (standard 1v rms)? What is source impedance? (standard 600ohm?)

Are power pins actually floating or are they connected to powered down PSU? What else is on those supply rails?
 

MisterBill2

Joined Jan 23, 2018
18,167
Feeding a signal to the input of an unpowered opamp, or other IC is usually a poor choice if the signal is great enough to bias any of the interior junctions into conduction. So at the very least use an external opamp so that if you damage it you will not need to replace your processor. And use a socket for the opamp to allow easy replacement.
 

Irving

Joined Jan 30, 2016
3,843
According to the data sheet:

"Input pins are diode-clamped to the power-supply rails. Input signals that may swing more than 0.5 V beyond the supply rails must be current limited to 10 mA or less "

So - impedance of 33nF @ 20kHz = 241ohms, so, assuming your signal source has zero output impedance then as long as your input <1.7v rms @ 20KHz it'll be ok. If your signal source has 600ohm output impedance then 6v rms @ 20kHz is OK.

Edit: all assuming power supply impedance to ground is 0ohm.
 
Last edited:

Thread Starter

PeteHL

Joined Dec 17, 2014
473
What is the active line level signal (standard 1v rms)? What is source impedance? (standard 600ohm?)

Are power pins actually floating or are they connected to powered down PSU? What else is on those supply rails?
The active line level input signal comes from an audio pre-amp, so yes the maximum level would be about 1Vrms. By "no voltage", I mean that there is no voltage at the output of the power supply of the processor wired to the op amps of the processor.

The processor has multiple op amps connected to the PSU of the processor.

Thank you,

Pete
 

Irving

Joined Jan 30, 2016
3,843
The active line level input signal comes from an audio pre-amp, so yes the maximum level would be about 1Vrms. By "no voltage", I mean that there is no voltage at the output of the power supply of the processor wired to the op amps of the processor.
1v rms should be ok. What do you know about the preamp output impedance? Do you have any info on it? eg chip etc.
 

Thread Starter

PeteHL

Joined Dec 17, 2014
473
According to the data sheet:

"Input pins are diode-clamped to the power-supply rails. Input signals that may swing more than 0.5 V beyond the supply rails must be current limited to 10 mA or less "

So - impedance of 33nF @ 20kHz = 241ohms, so, assuming your signal source has zero output impedance then as long as your input <1.7v rms @ 20KHz it'll be ok. If your signal source has 600ohm output impedance then 6v rms @ 20kHz is OK.

Edit: all assuming power supply impedance to ground is 0ohm.
Actually my pre-amp supplying the input signals (stereo pair) to the processor according to the specs has an output impedance of 47k Ohms (!). However I think that must mean that devices connected to the pre-amp should have input impedance of 47k Ohms.

The processor's PSU is a pair of linear regulators with 1N4001 protection diodes from the output terminals to ground.

The rule of not applying input signals to disabled active devices must be to cover all bases, as determining whether or not that might be unacceptable in a given case isn't easily determined (as your analysis shows).

-Pete
 
Top