# A problem with diode limitters

1. Nov 22, 2013

### bibo_dvd

hello guys

while studing the diode limitters

i found two examples i solved the first one but i couldn't with the second
1 -

2-

in the first i know that the diode while be forward biased in the negative half of the signal
so the negative half will be limitted to -0.7 v
and the V peak output will be = [(RL/(RL+R1)]*Vinput = 9.09v

but i don't know what will be the difference between the first and the second problem !!

2. Nov 22, 2013

### Staff: Mentor

In the second one, for + input signal, will the diode be forward or reverse biased? What does that mean about current through the resistor?

For - input signal, ask the same questions...

3. Nov 22, 2013

### bibo_dvd

okay in the + input the diode will be reverse biased so the current will not pass , but in the case of - input the diode will be forward biased and the current will pass

am i right ??

4. Nov 22, 2013

### Staff: Mentor

Correct. And V=IR, so you know when you will have an output voltage. Good job!

5. Nov 22, 2013

### bibo_dvd

okay i know that , but i still don't know how will i calculate Vout

i think Vout will be = -10+0.7= -9.3v .. is that right ??

6. Nov 22, 2013

### Staff: Mentor

That would be the negative peak voltage. The output waveform will follow the input waveform (minus the diode drop) when the diode is conducting.

7. Nov 22, 2013

### bibo_dvd

you mean that the output waveform will equal zero in the positive half and in the negative half it will equal -10v-0.7v ?? or what ??

8. Nov 22, 2013

### Staff: Mentor

In the negative half cycle, the output voltage equals the input voltage, minus the diode drop. Sketch that on top of the input waveform plot...