I want to simulate VDD rise. For example I want to raise 1.2V VDD to 4-5 Volts for tens of microseconds and lower back to 1.2V. Suppose I have a high rise and fall time pulse generator (200-500 picoseconds). My concern is that some digital ICs (like CPUs) have some capacitors on them to filter supply parasitics and serve as charge reservoir. Do these capacitors hinder the high rise and fall time of the signal generator because of their time constants? Can I still get a fast changing VDD in spite of the capacitors? Thank you.
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\$\begingroup\$ Why exactly do you want to do this? \$\endgroup\$– K HCommented Jan 28, 2021 at 9:34
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\$\begingroup\$ Know the value of those capacitors, and calculate the current you need to get your desired dV/dt. Then achieving that current is just a matter of engineering (possibly not an easy one!) Or, modify your expectations of dV/dt... \$\endgroup\$– user16324Commented Jan 28, 2021 at 13:46
1 Answer
Do these capacitors hinder the high rise and fall time of the signal generator because of their time constants?
Yes, the power supply decoupling capacitors will limit the rate in which voltage can rise unless you have a perfect pulse generator that has zero output impedance. It's a simple formula for most applications (no time constant): -
$$I = C\cdot\dfrac{dv}{dt}$$
So, if you have current drive of 1 amp into a 1 farad capacitor, you can expect the voltage to rise with a slope of 1 volt per second. If current is 1 amp and capacitance is 1 μF, then voltage will rise at 1 million volts per second or 1 volt per microsecond.
Can I still get a fast changing VDD in spite of the capacitors?
Not really because that is what those capacitors are there for. That is the purpose for which they are fitted.