do both circuits have the same output waveform?

[PainterGuy]

Hi, :wink:

In the link given below circuit #1 and #2 are same except that they have opposite polarities for the battery V. On the left bottom there is output waveform after the clipping. In the beginning I was thinking that output for both circuits, #1 and #2, will be same. But now I am doubtful!:confused: The diode D1 starts conducting when its anode is more positive than cathode and there is potential difference across them of at least 0.7V. Let's assume the battery V is of 6 volts.

Link:--
http://img687.imageshack.us/img687/716/changingpolarityofbatte.jpg

Do both circuits have the same output which is given on left bottom? Tell me please. Much grateful for any help you can give me.

Cheers

[Antiphon]

No. Not the same unless battery is 0 volts.

[skeptic2]

As Antiphon says at a battery voltage of zero both circuits will behave the same. If however you apply a battery voltage to the cathode of the diode, the clipping voltage will change from the +0.7 V by the same amount as the battery voltage whether positive or negative

[PainterGuy]

As Antiphon says at a battery voltage of zero both circuits will behave the same. If however you apply a battery voltage to the cathode of the diode, the clipping voltage will change from the +0.7 V by the same amount as the battery voltage whether positive or negative

Many thanks Antiphon and skeptic2. :approve:

Okay. I think the output waveform I have drawn in the diagram on the below link is correct for only circuit #2. Approve it if you also find it not incorrect please.

Link: http://img687.imageshack.us/img687/716/changingpolarityofbatte.jpg

So I will focus on circuit #1. As I said in first post that let take the voltage of battery V 6 volts. Also take +Vp and -Vp equal to +10V and -10V respectively.

The cathode of the diode is at -6 volts (because -ve terminal of the battery V is next to it). For the diode to conduct there should be potential difference of at least 0.7V across the diode and anode should be more positive than cathode. This can be written as a equation: Anode Voltage minus Cathode Voltage = +0.7V.

Av - (-6) = 0.7
Av = -5.3V

So it means the diode will start conducting at -5.3 volts. It will not conduct between -5.3V > Vrb >= -10V. Vrb stands for the voltage when the diode would be reverse biased.

Here is diagram: http://img38.imageshack.us/img38/5811/voutforcircuit1.jpg

Is this all correct to this far?

If it is, I still don't understand how to draw a output waveform for this. Can you help me with output graph for circuit #1 please?

Cheers

[skeptic2]

Your output waveform will be the same as the input waveform except that there should be a straight horizontal line at -5.3 V between the two points on the sine wave.