Why does my diode limiter circuit output ±3.7V and ±4.7V?

  • Topic: Engineering 
  • Thread starter francisg3
  • Start date
  • Tags Tags
    Circuit Diode
1 reply · 5K views · last reply
francisg3
Messages
31
Reaction score
0
Hello,
I have recently had a lab on diode limiters and am a bit confused by the results. The circuit is seen in the picture below and I have obtained theoretical values for the Voltage output. The circuit is feed by a 16V sine wave with a frequency of 1kHz.

Now I know when the sine wave is positive, D1 conducts and D2 is cut-off. So the voltage output should be 3V + 0.7V = 3.7V right?

When the sine wave is negative I know that D1 is cut-off and D2 conducts. This means that the output voltage should be -4V + 0.7V = -3.3V correct?

I need to sketch the input and output so would the graph would be a sine wave clipped at +3.7V and -4.7V ?

Thanks for the help!
 
Attachments
  • Schematic of a diode limiter circuit: Vs1 sine source feeds R1 1K, then D1 with B1 = 3 V and D2 with B2 = 4 V, output labelled Vout.
    diode.jpg
    19.5 KB · Views: 470

Discussion

would the graph would be a sine wave clipped at +3.7V and -4.7V ?

Yes, it would be like this.