Diode Clipping

[Legion]

I built these circuits:


The top circuit works as expected, dropping about 0.8V in each direction and clipping the rest of the top and bottom of the signal. This is as measured at the indicated test point and ground tied to terminal B of the FG.

I tried many variations of the second circuit. Resistors on the A side of the FG, resistors on the B side (as shown in the schematic), resistors on the anode side of each diode, resistors on the cathode side, etc.

My expectation was that the resistors would drop more of the voltage in addition to the 0.8V in each direction, in effect clipping less of the signal. Instead the entire signal comes through unaffected once the resistors are added. Why is that?

[Simon123]

Try adding resistor  after signal generator .

[ConKbot]

Add some series resistance to your source.  Current your voltage source is ideal, zero source impedance.  If there was a source impedance then it would behave as you expect. Once the diodes turn on, the source impedance and 100 ohm resistors make a divider, so it no longer clips, but instead reduces amplitude once it exceeds the threshold.  In audio equipment terms, its a signal compressor.

http://www.mediacollege.com/audio/processing/compression/

However your compression ratio will vary with the source impedance.

If you had tried this on a bread board, with a beefy AC source, then you would have let some magic smoke out of some diodes.  Just for giggles, in your simulation, see what the currents in the diodes are in the first case. 

[Legion]

Add some series resistance to your source.  Current your voltage source is ideal, zero source impedance.  If there was a source impedance then it would behave as you expect. Once the diodes turn on, the source impedance and 100 ohm resistors make a divider, so it no longer clips, but instead reduces amplitude once it exceeds the threshold.  In audio equipment terms, its a signal compressor.

http://www.mediacollege.com/audio/processing/compression/

However your compression ratio will vary with the source impedance.

If you had tried this on a bread board, with a beefy AC source, then you would have let some magic smoke out of some diodes.  Just for giggles, in your simulation, see what the currents in the diodes are in the first case.

There is source impedance. It's an FG with 50R output impedance.

[josip.budimir]

The diode always wants to keep the required voltage for itself
So if you have a 1V supply, the diode will measure 0,7V and the resistor 0,3
5V would make the resistor measure 4,3V and the diode 0,7
To measure current in a circuit with a diode in series, you have to take out the 0,7V drop voltage from the input voltage in the formula I=U/R
Lets say your supply is 0,8V the resistor is 10ohm's and a diode in series the current is:
I=(U-0,7)/R
I=0,1/10
I=0,01A
So if you have a supply under 0,7V, youre gonna get very low or no current