# Reversed biased diode polarity in circuit

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

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:

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

• 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. – Andy aka May 30 '15 at 21:50
• @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? – Dimitri C May 30 '15 at 21:53
• It's only reverse (or forward) biased by 1.5 volts and the peak of the sinewave is 4 volts. – Andy aka May 30 '15 at 21:54
• + 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? – Dimitri C May 30 '15 at 21:58
• 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. – Andy aka May 30 '15 at 22:04