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# Diode ON/OFF dilema

Discussion in 'Homework Help' started by MartinV, Mar 2, 2012.

1. ### MartinV Thread Starter New Member

Jun 2, 2011
22
0
In this RC circuit at one moment I get to the polarization shown on the picture below. At that point the voltage of the diode is opposite of the condition (to work +anode - cathode), but yet the current flows first through the anode (which should work). This is because of the voltage inherited from the previous stage of the input voltage. I find this contradictory, but I may have my theory learned only the half way P). So, the diode should be OFF in this situation but I don't really get why.
The diode is ideal.
Here's a picture of the circuit:

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790

The presence of the ideal diode means that the capacitor / output voltage can never be negative.

To actually know what the output waveform looks like you would have to know the R & C values.

3. ### MartinV Thread Starter New Member

Jun 2, 2011
22
0
The output waveform is not a problem. And in moment of the picture the capacitor has +5 V, and the new -10V go to the resistor, that's understood. Maybe this picture will be more clear:
The question is: In this situation will the diode be ON or OFF? And why?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
The diode can't conduct whilst the capacitor voltage is greater than zero. With the instantaneous value @ +5V the capacitor will begin to discharge when the source switches to -5V. Once the voltage falls to zero volts the diode can begin to conduct.

5. ### MartinV Thread Starter New Member

Jun 2, 2011
22
0
So the answer is : diodes work only at positive charge (+ at anode) regardless of the current flow?

6. ### Vahe Member

Mar 3, 2011
75
9
I will start from t=0. As the input voltage goes to 10V, the output will increase exponentially towards 10V as well. Since the output voltage is positive in this case, the diode is reverse biased (OFF). When the input changes to 5V, the output voltage decreases exponentially toward 5V; however, since the output voltage is still positive, the diode will remain reverse-biased (OFF). When the input voltage changes to -5V, the output will start decreasing exponentially toward -5V. At the instant that vout crosses from positive to negative, the diode will become forward biased (ON) and the output will be clamped to 0V. The output will remain at 0V until the polarity of the input voltage is changed to a positive value in order to reverse-bias (turn OFF) the diode. The rate at which the output voltage exponentially increases and decreases depends on the time constant (the product of R and C). Since these were not given, the answer can only be given in qualitative terms as described above.

Best regards,
V