# How to plot output graphs of circuits involving diodes and capacitors?

• Engineering

## Homework Statement

Given that the diodes are ideal for the circuits a) and b), plot each of their output for the input shown. Label the most positive and most negative levels. Assume CR >> T

## The Attempt at a Solution

I am having a really hard time understanding the role of the capacitor since we are not given any values for both capacitors and resistors. I think it charges when Vin is positive and the resistor somehow discharges it when Vin is negative. Is there some equations I can use to know what Vout is?

For the circuit a) if Vin is negative, the diode is reverse bias thus should Vin=Vout?

Also does the graph change depending on where the Vin graph starts? Like if it starts off at +10 instead of -10 wouldn't the output graph be quite different?

Any help would be very much appreciated, I really just want to understand this underlying concept.

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CWatters
Homework Helper
Gold Member
I am having a really hard time understanding the role of the capacitor

What is the impedance of a capacitor (any capacitor) at very high frequencies?

Is a step voltage a low or high frequency event?

Impedance should be 0 at high frequency. I'm not sure if it's high or low frequency but I'm guessing step voltage is high frequency because of its sudden changes in voltage.

CWatters
Homework Helper
Gold Member
Correct. So during the fast rising and falling edges (the edges only) the capacitor behaves like a short circuit.

Try assuming that Vi = -10V for a long time before the first rising edge occurs. What will happen to Vo during that time?

Then what happens when the first rising edge occurs?

The voltage over the capacitor + the voltage at Vin will equal output voltage?

I understand the first one now. How do I go about doing the second one? When the resistor is in parallel I expected to see a the capacitor discharging/charging but my stimulation shows a rectangular waveform for the output

CWatters
Homework Helper
Gold Member
Have you set RC >> 1ms or RC << 1ms ?

If RC >> 1ms I think you will get a square wave. For example.. If both R are very large they might as well not be in the circuit so remove them. In that case Vo = Vi which is a square wave).

If RC << 1ms you will get alternate +ve and -ve steps/spikes each followed by an "RC curve" towards 0V. The curve will be faster in one direction than the other (eg either RC or 2RC).

I'm not quite understanding what it means when RC << T or RC >> T?

NascentOxygen
Staff Emeritus
I'm not quite understanding what it means when RC << T or RC >> T?

Those relations describe the two extremes of possibility. In the first, the capacitor is given plenty of time to discharge considerably during the half-period T, so expect to see exponential wavehapes around an RC circuit driven by square waves.

But if RC >> T then the capacitor has insufficient time to discharge during time T, meaning that the voltage across the capacitor plates barely changes during interval T. So, if there is a square-wave on one side of the capacitor, then where RC >> T you are going to see a nice square-wave on the other side of that capacitor.

The above holds true for RC circuits, but the presence of nonlinear elements (e.g., diodes) may change the effective R value at some voltage point.

CWatters