0
\$\begingroup\$

Charge Pump Schematic Signal Output

TL:DR
Can a short voltage spike / delay from a inverting charge pump cause any issues when it is used as the negative voltage for an op-amp?

Background
I have a project where I get an AC input of 24V from a transformer, which then goes through a full wave rectifier, voltage regulator and filters to the rest of my electronics. However, I need a -5V for an op-amp. I designed a Inverting Charge Pump that uses one of the AC signals before the full wave rectifier and then goes through a Zener diode to get -5V on the output. The schematic for my Inverting Charge Pump is attached.

Problem
The problem is, seeing as the capacitor takes a while to charge up and the diodes have a forward voltage that needs to be reached, it has a 9.5mS delay before the voltage drops down to -5V.

Question
Can this voltage spike cause any problems for me further down the line?

\$\endgroup\$
2
  • 2
    \$\begingroup\$ That's not a spike (a transient excursion beyond a baseline), that's just a startup ramp. We have no way to ascertain whether it's suitable for your circuit, without knowing that circuit and the components it uses. \$\endgroup\$ Commented Aug 28 at 22:15
  • 1
    \$\begingroup\$ Is your transformer voltage really 16.97V RMS? \$\endgroup\$ Commented Aug 29 at 2:57

1 Answer 1

0
\$\begingroup\$

There's no inherent danger to the op-amp, as long as its two inputs are never exposed to potentials that fall outside the range defined by its own its supply potentials at any given instant.

That's generally not a problem if the input signals come from sub-systems powered from the same supply rails, since they can't produce potentials outside of that range either.

Problems arise when there's a possibility to apply an input potential that lies outside the op-amp's supply range. For example, if at some given instant of time the supplies are +15V and -2V (the negative rail being slow to arrive at the full -5V), but somehow the source of op-amp input signal is trying to apply -4V to an op-amp input, then the input protection diodes of the op-amp will clamp the input to -2V, causing damaging levels of input current to flow. Here I've moved those protection diodes (boxed in blue) outside the op-amp to illustrate, with the current path shown in red:

schematic

simulate this circuit – Schematic created using CircuitLab

V3 here is a source of 4V negative with respect to ground, but if that source is powered from the same +15V and -2V supply, it will be unable to produce a voltage outside that range -2V to +15V, and the op-amp will be safe. However, if that source V3 has its own independent power supply (or comes from a completely separate system altogether) that settles faster than V1 and V2, then there's a possibility for this dangerous condition to exist.

For devices without input protection diodes, whether this condition is dangerous or not depends on the device itself, for which you must refer to its datasheet. In general, though, you should assume that devices won't survive.

So the question is, is it possible for any op-amp (or other device) powered by your \$V_{OUT}\$ to have an applied input potential lower (more negative) than \$V_{OUT}\$ during the time it takes \$V_{OUT}\$ to reach and settle at -5V? If not, then you're good.

On a side note, if you're concerned that the negative supply takes an age to reach its target potential, then you should be equally concerned that the positive supply suffers the same lethargy, but for some reason you aren't concerned about that. I very much doubt that the positive supply is any better in this respect, but typically it only becomes an issue if external modules with independent supplies are applying signals with potentials outside the instantaneous supply potentials of the recipient.

The positive supply, whether it's a DC-DC converter or a linear setup, will have a slow output rise too. Consider this linear design:

schematic

simulate this circuit

Here's the output \$V_P\$ rising slowly after AC power is applied:

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ Thank you for the explanation and the graphs. I did not think about the Positive side also taking it's time to reach it's max. This answers my question. Thank you \$\endgroup\$ Commented Aug 29 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.