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I have to find the operating interval of the diode in the following circuit..

enter image description here

NOTE: The DC source has its positive pole grounded.

My work so far is this..

\$V_i>-1.5 V\$: the diode does not conduct and we have open circuit which results in $$V_o=0$$.

\$V_i\le-1.5 V\$ the diode does conduct an it becomes short and we have voltage divider which results in $$V_{r2} = (V_i+E)*{R_2\over R_1+R_2}$$ and $$V_{o} = V_{r2}-E =>V_o={3\over4}V_i-{3\over8}$$

So the sinewaves of input an output are the following: enter image description here

Can somebody please tell me if I am right or if I am making some mistake I am missing??

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  • \$\begingroup\$ Hint - a perfect diode would conduct half the waveform of the input sinewave if the DC voltage source were 0V. So, subtract 1.5 volts from the voltage source and imagine E was replaced by a short.... The problem you face is understanding the polarity of the DC voltage source - as a symbol, the positive end is grounded but the + and - signs indicate it is the wrong way round. This is a problem in the question. \$\endgroup\$ – Andy aka May 30 '15 at 21:50
  • \$\begingroup\$ @Andyaka My confusion is a result from the diode being reverse biased. If it is reversed biased how can it conduct half the waveform of the input sinewave? \$\endgroup\$ – Dimitri C May 30 '15 at 21:53
  • \$\begingroup\$ It's only reverse (or forward) biased by 1.5 volts and the peak of the sinewave is 4 volts. \$\endgroup\$ – Andy aka May 30 '15 at 21:54
  • \$\begingroup\$ + sing is mapped to the cathode of diode and - sign is mapped to anode of diode. Same thing happens to the DC source as you said, and I still don't get when the diode conducts or not!! Can you please help a little more? \$\endgroup\$ – Dimitri C May 30 '15 at 21:58
  • \$\begingroup\$ Forget about the resistors, at what point in the sinewave does it go negative (forcing current thru the diode) when there is a 1.5 volt offset. It has become a math problem really. \$\endgroup\$ – Andy aka May 30 '15 at 22:04
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When the input voltage is greater than +1.5V the diode does not conduct so the output voltage is fixed.

When the input voltage is less than or equal to +1.5V the resistors act as a voltage divider. Consider the diode a short and should be able to calculate the output voltage for any input voltage less than 1.5V.

Hint: The change in output voltage will be 1/4 of the change in input voltage, but they are going to be equal at 1.5V in = 1.5V out.

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  • \$\begingroup\$ Thanks a million, quick question: the voltage applied on the anode of the dioded is fixed because of the battery an if yes what is it? If not, the voltage of the anode changes how? \$\endgroup\$ – Dimitri C May 31 '15 at 1:36
  • \$\begingroup\$ The voltage source supplies the 1.5V to the output through the 1K resistor. If it is not fixed, then replace 1.5V with V or v(t) in the above statements. \$\endgroup\$ – Spehro Pefhany May 31 '15 at 1:43

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