Clipping Circuit

[JasonMcG]

If you have the following clipping circuit, how do you analyse it?
How do you determine whether the diode is on or off?

Regards
Jason

 

[MrAl]

Hi,

If the diode is forward biased then it is considered to be 'on'.

There are variations on what is considered forward biased. Some texts make it simple and state that whenever the anode is more positive than the cathode the diode conducts, and other texts will impose a voltage such as 0.5 volts where the anode must be 0.5v more positive than the cathode to be considered 'on', and this voltage can vary from 0.5 to 0.7 so some texts might use 0.7 instead.

To find out what voltage to use (if any) you have to look at other problems from the same book or source where the circuit came from.

 

[JasonMcG]

Diode is on at 0.7 volts.
What does the source before the resistor do?

 

[KeepItSimpleStupid]

Your circuit almost effectively describes a Zener diode shunt regulator. I really would not call it clipping circuit because the source is DC.
I would need to put in the polarities of the batteries for sure. I never get the big and little line right.

For a Zener shunt regulator which this thing MIGHT be, the first resistor limits the current. The diode (usually a revered biased zener diode) has to be able to dissipate the required power.

 

[MrAl]

Diode is on at 0.7 volts.
What does the source before the resistor do?

Hello again,

Assuming Va and Vb are drawn correctly, Va and Vb both act to forward bias the diode, or at least help to forward bias it (turn it on sooner with a given Vin). Va makes it forward bias with a lower positive Vin, and Vb does the same plus also makes the clipping action voltage less. So with Vb the output will appear to be lower than without it when the diode clips.

For example, with Va=0.2 and Vb=0.4 and Vin=0.0 the total diode voltage is 0.6 but the diode 'needs' 0.7 to turn on so it is off when Vin=0.0 volts, so the output is 0.0+0.2=0.2v. When Vin goes up to 0.1 volts however the total diode voltage now is 0.1+0.2+0.4=0.7 so the diode starts to turn on and the output is 0.1+0.2=0.3v, and when Vin goes up to 0.2 volts the total is 0.8v so the diode clips and the output is equal to 0.7-0.4=0.3 volts. Anything over 0.2v for Vin and the output stays at 0.3v so it clips at 0.3v.

Note that if we did not use Vb and only used Va we would not clip until 0.7v, so Vb helps to lower the clip voltage point. This example was with Va=0.2 and Vb=0.4 but of course they can be other voltages, so you could try a few different values and see what you get.

The output is Vin+Va until the clipping action starts, and then it is Vd-Vb where Vd is the diode turn on voltage (we used 0.7v above for the diode turn on voltage).

In the attachment, the input is a slow triangle wave that runs from 0 to 0.5 volts shown in blue and the output voltage is shown in red. Note that the output follows the input but is 0.2v higher until it clips, and then it is clamped to 0.3 volts.