 ### Network # Simple op-amp summing junction question

K

#### Kingcosmos

Jan 1, 1970
0
This question arose from the typical "what do I do to protect an op-
amp when the input voltage is higher than supply?"

In a typical non-inverting configuration the answer seems straight
forward since the input will be at the non-inverting pin of the IC.
If the input voltage is higher than supply a simple series resistor to
current limit can be suitable in most cases.

With a typical inverting configuration things seem to be a bit more
nebulous. The voltage at the inverting pin of the IC will be held at
virtual ground (ideally) so a higher than supply input voltage coming
in on the input resistor does not immediately seem to be an issue.

I am trying to think of a way to violate this; i.e., make the voltage
at the summing junction be a diode drop higher/lower than either of
the supply voltages, and simulate in Spice. My configuration is as
follows:
Vcc = +5V
Vee = -5V
Rin = 10k
Rf = 3k
Vin = +15V
Vo = -4.5V

If I assume that the amplifier is in this state for a while the
summing junction would be ideally held at virtual ground. Now, if I
have a step input down to -15V, I am imagining that the output would
STILL be -4.5V for some period of time (ns, us). It will take a
period of time for the output of the amplifier to slew to the proper
output thus forcing the summing junction to virtual ground to make all
is well in the universe.

During this transient, I am calulating that the summing junction will
be ~-6.9V. This is more than a diode drop "lower" than the negative
supply rail thus turning on the internal ESD protection diodes so
input current limitation techniques must be used.

I am having trouble catching this in simulation. Perhaps I do not
have my sim set up correctly? I only see a transient of ~-1.3V which
is well within the supply rails.

Am I correct or am I all wet?

http://www.geocities.com/knightofsolamnus/SummingJunction.pdf

A link to show the math I calculated. Hope I didn't make a mistake!

J

#### John Popelish

Jan 1, 1970
0
Kingcosmos said:
This question arose from the typical "what do I do to protect an op-
amp when the input voltage is higher than supply?"

In a typical non-inverting configuration the answer seems straight
forward since the input will be at the non-inverting pin of the IC.
If the input voltage is higher than supply a simple series resistor to
current limit can be suitable in most cases.

You may have to make sure that current goes somewhere that
does not cause trouble. Some opamps (like the LM324) do not
clamp positive voltages with any internal clamp diodes, so
the voltage might rise to where damage is done. Some opamps
clamp the voltage, but current through these clamps cause
the amplifier to draw excessive supply current. So you
might need to add an external clamp diode to the positive
rail, or a zener diode to ground.
With a typical inverting configuration things seem to be a bit more
nebulous. The voltage at the inverting pin of the IC will be held at
virtual ground (ideally) so a higher than supply input voltage coming
in on the input resistor does not immediately seem to be an issue.

The input voltage is held to a fixed voltage that matches
what is applied to the non inverting input, only as long as
the output voltage is not saturated or slew rate limited.
I am trying to think of a way to violate this; i.e., make the voltage
at the summing junction be a diode drop higher/lower than either of
the supply voltages, and simulate in Spice. My configuration is as
follows:
Vcc = +5V
Vee = -5V
Rin = 10k
Rf = 3k
Vin = +15V
Vo = -4.5V

If I assume that the amplifier is in this state for a while the
summing junction would be ideally held at virtual ground.

As long as the output is not saturates, there is not a thing
wrong with having the voltage applied to the input resistor
being outside the supply rail voltages. But if that input
voltage went to +40 volts, the inverting input would be
driven above the +5 rail.
Now, if I
have a step input down to -15V, I am imagining that the output would
STILL be -4.5V for some period of time (ns, us). It will take a
period of time for the output of the amplifier to slew to the proper
output thus forcing the summing junction to virtual ground to make all
is well in the universe.

Right. For a brief period, the virtual ground will not be
enforced because the output is in slew rate limit or waiting
for the propagation delay of the amplifier.
During this transient, I am calulating that the summing junction will
be ~-6.9V. This is more than a diode drop "lower" than the negative
supply rail thus turning on the internal ESD protection diodes so
input current limitation techniques must be used.

Yes. I would split the input resistor into two equal halves
and add a pair of clamp diodes to the rails from that node.
This makes sure that, no matter what, the input will never
see more than a diode drop outside the rails through 5k of
resistance.
I am having trouble catching this in simulation. Perhaps I do not
have my sim set up correctly? I only see a transient of ~-1.3V which
is well within the supply rails.

Am I correct or am I all wet?

I think you have a valid concern. This is usually not a
problem for signals created on the board from the same
rails, but for signals and static discharges from off the
board, such details can make the difference between a robust
and a fragile system.

K

#### Kingcosmos

Jan 1, 1970
0
I am glad that it was not a stupid question, coming from a junior
engineer. Excellent response John; I appreciate your insight.

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