# help understanding clipper circuits

#### justtrying

Joined Mar 9, 2011
439
While I understand the circuit in general, I have a problem figuring out the exact voltage at which the diode will turn on:

circuit 1: the diode will conduct during negative cycle and pea output will be Vin+3V. As far as turning the diode on, I look at the voltage difference between the DC and since voltage drop across diode is 0.7V, to me it makes sense that diode will turn on once -2.3V (i.e. a more positive voltage appears on the anode)

circuit 2: basically same logic and I get 3.7V to turn the diode on.

problem... this seems to be wrong as per multisim, with the answers reversed, I cannot make any sense of it.

Any explanation would help. Thanks in advance.

p.s. Vout is taken across 1k resistor...

#### Attachments

• 12.9 KB Views: 43
Last edited:

#### praondevou

Joined Jul 9, 2011
2,942
In the first circuit the diode will start to conduct when the AC voltage is still in it's positive cycle. 2.3V if you consider 0.7V forward voltage drop for the 4005. That's 3V minus the forward drop of the diode.

However, have a look at the voltage forward drop parameter in the 4005 database, it may not be 0.7V but less.

In the second circuit you will need -3.7V on the AC source to get the diode to conduct, 3V to overcome the DC source, 0.7V for the voltage drop.