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I've designed this circuit using Multism 10.1 :

This circuit is part of clock generator for 8088 uP:

Why does the wave of ch1 of Oscilloscope which represent the voltage across the capacitor take this form?

What is the relationship between that and the delay that is caused by the capacitor? Click here

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    \$\begingroup\$ Try to see what happens when you take that diode out. \$\endgroup\$
    – scld
    Dec 10, 2013 at 14:57
  • \$\begingroup\$ The capacitor doesn't discharge quickly? \$\endgroup\$
    – ammar
    Dec 10, 2013 at 15:04
  • \$\begingroup\$ Is that what happened? Now, try to flip the diode around, the other direction. \$\endgroup\$
    – scld
    Dec 10, 2013 at 15:14
  • \$\begingroup\$ It does the same in the other direction. \$\endgroup\$
    – ammar
    Dec 10, 2013 at 15:55

1 Answer 1

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When you charge the capacitor the 100k resistor limits the current so the voltage on the capacitor is:

$$v = V \left( 1 - \text{exp} \left( - \frac{t}{C \cdot R}\right) \right)$$

Where V is size of the input square wave and R is 100k.

The discharge current goes mainly through D1 and not through the 100k resistor. So the current is not limited by the resistor value and the discharge is much faster.

If you turn the diode around you should see a fast charge and slow discharge.

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    \$\begingroup\$ This is very true but if you tried it by just switching off the power supply it might make you think it isn't working because the discharge path relies on a low impedance to ground when the power supply is disconnected. \$\endgroup\$
    – Andy aka
    Dec 10, 2013 at 15:23
  • \$\begingroup\$ Thanks. But what is the relationship between that and the delay that is caused by the capacitor? electronics.stackexchange.com/questions/92686/… \$\endgroup\$
    – ammar
    Dec 10, 2013 at 15:56
  • \$\begingroup\$ And what do you mean by saying "turn the diode around you"? \$\endgroup\$
    – ammar
    Dec 10, 2013 at 15:57
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    \$\begingroup\$ So, the portion of the delay caused by the capacitor does not change. It is the same in both directions. The portion of delay caused by the resistor, however, does. When the current goes "against" the diode (when the cathode voltage is higher), the diode acts like an open circuit. So the RC constant uses the resistor value. When the current is going "in the direction" of the diode (that is, when the anode voltage is higher than the cathode), the diode acts like a small resistance (close to a short circuit for your purposes). So, you use the R value of the diode for the RC constant. \$\endgroup\$
    – scld
    Dec 10, 2013 at 16:07
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    \$\begingroup\$ @ammarx The rate at which the capacitor charges or discharges is the amount of current flowing through it. When the input voltage is switched high this current is limited by the 100k resistor and as it charges the voltage across the resistor falls so less current flows and the capacitor charges more slowly. When the input voltage switches low initially the current is only limited by the diode and so the cap discharges rapidly. Once the voltage falls to about 0.6 volts however the diode stops conducting and all the current has to go through the resistor hence the slow discharge tail. \$\endgroup\$
    – Warren Hill
    Dec 11, 2013 at 13:15

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