RC Network Spike Generator Confusion

[Confusedcat]

Hey all, bit confused here. I'll be referring to the "high pass" image, with the left circle as input and right as output.

Say you have an RC network with a 100nF capacitor and a 100k resistor set up as a spike generator. I understand that if 5v were to be given to the left hand plate of the capacitor, this causes a spike. Then, as this has charged, 5v then passes through the resistor and the right hand plate becomes 0v as well as at the output. However, when the left hand plate then receives 0v, -5v appears at the output. why is this? Sorry if this is too vague.

 

[(*steve*)]

The output of the high pass filter is controlled by the change of voltage that appears at the input.

The values of the components determine the time constant of the effect, but in essence the output voltage is proportional to the change in input voltage over some period of time -- similar to the difference between the current (instantaneous) voltage compared to some average voltage over some recent time period. (it's more complex than that, but stick with the simple explanation).

Imagine that the input has remained at 0V for a long time. The input voltage has not changed (the average = the instantaneous voltage), so the output voltage is zero.

Now let's assume the input voltage suddenly jumps to 5V. The instantaneous voltage is 5V, but the average over some period is still 0V. The output is the difference (5V).

As time goes on, with the input voltage no longer rising, the average voltage over time becomes closer and closer to the instantaneous voltage so the output falls (and the component values determine how fast this occurs).

Imagine that the input has now remained at 5V for a long time. The input voltage has not changed (the average = the instantaneous voltage), so the output voltage is zero.

Now the input voltage suddenly jumps to back to 0V. The instantaneous voltage is 0V, but the average over some period is still 5V. The output is the difference (-5V).

As time goes on, with the input voltage no longer falling, the average voltage over time becomes closer and closer to the instantaneous voltage so the output rises (the component values determine how fast this occurs).

Increasing the capacitance or the resistance in this circuit will make the circuit look at what in this simple analogy is a longer period of time.

The nature of the "average", and how the components affect this require some real maths, but in essence the average is an exponential function that places more weight on recent values, but which theoretically is affected by all voltages back to the beginning of the universe!

So in summary, the output goes negative because the output is related to the change of the input voltage, and when that change is from a more positive voltage to a less positive voltage (i.e. in the negative direction) the output goes negative.

 

[KrisBlueNZ]

The simple answer is that when the input has been at +5V and the output has stabilised at 0V (due to the action of the resistor), the capacitor is charged to 5V. The left hand plate is 5V more positive than the right hand plate. So if you suddenly bring the left hand plate down to 0V, the capacitor keeps the same voltage across it, and the right hand plate, and therefore the output, jumps down to -5V. The capacitor then discharges through the resistor, giving a negative pulse at the output that's a mirror image (reflected in the 0V line) of the pulse at the output when the input went from 0V to +5V.

Remember that the voltage across a capacitor cannot change instantaneously; it can only change at all if a current is present. The direction of the change depends on the direction of the current, and the rate at which it changes is proportional to the current, in conjunction with the capacitance.

 

[Confusedcat]

Thanks guys, this has cleared a lot. So would I be right in thinking that, as the capacitor receives 0V (after 5V), the right plate goes to -5V temporarily until it's discharged through the resistor as the capacitor want to keep the same voltage across it? thanks again

 

[KrisBlueNZ]

That's right.

 

[Confusedcat]

Thank you!