1
\$\begingroup\$

Look at the following circuit: Circuit

Here the \$VPLUSE\$ has \$50\%\$ duty cycle with \$1ms\$ time period. The upper voltage level is \$10V\$ and the lower one is \$-5V\$.

My question is about the plot I found: Plot Why the capacitor (voltage of capacitor is plot with red line) is charging slowly (which is expected), but discharging so fast? Should the time constant not affect the charging and discharging equally?

\$\endgroup\$
0

2 Answers 2

Reset to default
4
\$\begingroup\$

When the capacitor is charging the diode is reverse-biased so the time constant is RC.

When the capacitor is discharging the diode is forward-biased so the time is much less since the diode forward biased junction is effectively in parallel with the 1kΩ resistor.

\$\endgroup\$
2
  • \$\begingroup\$ Perhaps I got your point. "A Capacitor doesn't want to change voltage instantaneously" was ringing in my head before I read your answer. As the capacitor voltage is changing very fast, sparking will occur, right? \$\endgroup\$ Commented Nov 24 at 9:50
  • 2
    \$\begingroup\$ Not necessarily sparking, but there will be substantial current through the capacitor and diode since the voltage is changing fast. \$\endgroup\$ Commented Nov 24 at 11:27
1
\$\begingroup\$

It is actually the other way around.

The capacitor is instantly charged through the diode, when the pulse is -5V. The capacitor voltage is: 15-0.6-(-5V) = 19.4V.

When the pulse is going up to 10V, the capacitor is discharged. The voltage over the capcitor cannot change instantly, so when the voltage on one plate is forced to 10V the other have to follow up to maintain the same voltage difference across the capacitor. The voltage at the diode cathode is going to be 19.4V + 10V = 29.4V.

Then the diode is reversed biased and will discharge through R4.

\$\endgroup\$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.