# Voltage across an resistor

In the figure shown, Vi is a 100 hz triangular wave having a peak to peak amplitude of 2 volts and an average value of 0 volts. Given that the diode is ideal, the average value of the output voltage Vo is ________.

My attempt.

I took an assertion that at the moment when the input voltage is 1v(it's maximum), the diode is reverse biased. Then using voltage divider, the voltage across the resistor turned out to be Vi/2 i.e 0.5 volts. Voila ! The assertion is indeed correct.

That means the output voltage will just be a copy of the input voltage dimnished by a factor of 1/2.

I.e a 100 hz triangular wave having a peak to peak amplitude of 1 volts.

Therefore, the average value of the output voltage is zero.

Is this answer correct ? And if yes, can we also take the opposite assertion assuming the diode to be forward biased and then prove it wrong. If yes, how ?

• It depends on how you define an ideal diode - without specifying the material (e.g. Si or Ge) you would take the forward drop as 0V rather than 0.6V giving a shift of the average. If the 0.6V cell shown in the circuit is actual (rather than simulating the forward drop) then you are correct, the diode will not conduct. – JIm Dearden Aug 24 '17 at 9:32
• Yes the cell is real. Can you please give me a feel for as to why the diode won't conduct. I am unable to get it. – user158942 Aug 24 '17 at 9:34
• The maximum voltage the diode 'sees' in the forward direction (anode to cathode) is 0.5V (as you have determined). The voltage across the diode is 0.1V but the anode voltage is lower (more negative) than the cathode by this voltage making the diode reverse biased, the diode cannot conduct and appears as an open circuit. On the negative half cycle the diode is always reverse biased. – JIm Dearden Aug 24 '17 at 9:41
• But the doubt remains, being that I determined the maximum forward voltage considering the diode to be reverse biased- hence open circuited. That is why I was able to apply voltage divider and determine the voltage across the resistor that turned out to be 0.5 v. What if I had not considered the diode open circuited. – user158942 Aug 24 '17 at 9:44
• I mean I could have taken an assertion the other way round too and in that case the diode would have been short circuited and the voltage across the resistor would have been 0.6V. – user158942 Aug 24 '17 at 9:46

It's funny how diode questions are ironically hard, they are often under-looked as simple devices, however they are too much non-linear and impose many paradoxal circuit problems.

This is rather a very pragmatic way of solving it. When solving diode circuits you use a un-analytical solving method. Assume each combination of conduction and not conduction.

1) Assume it is conducting.

simulate this circuit – Schematic created using CircuitLab

Its pretty straightforward that $$\v_o = v_d\$$.

However you will find a hidden inequation!

$$\I_3 = I_1-I_2>0\$$ this will imply that:

$$\\frac{v_i-v_d}{R_1} - \frac{v_d}{R_2} >0\$$ or $$\ v_i> v_d (1+\frac{R_1}{R_2})\$$

So keep it from now, $$\v_0 = v_d\$$ if $$\ v_i> v_d (1+\frac{R_1}{R_2})\$$.

2) Assume its not conducting, a simple voltage divider and:

$$\ v_o = \frac{R_2}{R_1+R_2} v_i\$$, however there is another inequation hidden and not $$\I_3 < 0\$$ this wouldn't help much...

It's $$\v_o -v_d < 0\$$ or $$\v_i <(1+\frac{R_1}{R_2})v_d\$$. YEAH, CONSISTENCY!

We have: $$\ v_0 = \frac{R_2}{R_1+R_2} v_i\$$ if $$\v_i <(1+\frac{R_1}{R_2})v_d\$$.

3) Conclusion:

$$\ v_o = \begin{cases} v_d,&\quad v_i > (1+\frac{R_1}{R_2})v_d\\ \frac{R_2}{R_1+R_2} v_i,&\quad v_i < (1+\frac{R_1}{R_2})v_d \end{cases}\$$

Numerically:

$$\ v_o = \begin{cases} 0,6V,& \quad v_i > 1,2V \\ \frac{1}{2} v_i,& \quad v_i < 1,2V \end{cases}\$$

a) The output is zero! Yes, because $$\v_i < 1 < 1,2\$$ always, therefore Vo will be half of Vi always having also 0V as mean.

b) Check previous inequations.

Edit: After @Granger Obliviate pointed out I did correct the last statement. Thanks!

• Yes this is the best way to solve it. I only think that you made a small mistake in the system, the diode is OFF for vi<1.2V, not vi<0.6V. Another tip I give for the OP and well, for you too, is to always check in the ending if we have defined vO for all possible values of vI, to make sure we didn't forget anything :) – Granger Obliviate Apr 3 '20 at 18:32
• Thanks for the observation! – Trevor Apr 3 '20 at 19:25

okay firstly, you said 'ideal' diode, which gets forward biased immediately after 0V rathher than 0.6V or 0.7V, now considering what you have said that the diode was reverse biased even at 1V it means that you are talking about the practical diode which has a constant voltage drop of about 0.6-0.7V (considering it as Si, otherwise germanium is at 0.2 but that would forward bias it).

now since the diode remains reverse biased in both the cycles we can see that the circuit is linear and yes we can apply simple VDR (as a triangle wave is a sum of sinusoids according to Fourier series so apply VDR to each sine wave) indeed what we will get is that the voltage amplitude across output resistor will be half the input and 'YES' the average will be 0V

Coming to the second part of your question that considering the diode to become forward biased (that is the diode not having that previous 0.6-0.7V drop) will it have an average voltage of 0V ? The answer is 'NO', for the wave form that you get, if you find its Fourier series, you will see that its DC component will not have a value of 0V

• How would you get to know seeing the circuit that the diode would be forward biased in both the cycles ? Applying voltage divider ? – user158942 Aug 24 '17 at 12:02
• I said that the diode will be reverse biased in both the cycles, if it becomes forward biased you cannot apply VDR – Syed Mohammad Asjad Aug 28 '17 at 9:23

Assuming an ideal diode, it is forward biased when Vin > 0.6V and reverse biased when Vin < 0.6V.

When Vin > 0.6V you then have Vout = 0.6V. When Vin < 0.6V, you then have a voltage divider: Vout = Vin/2.

Consider a period T = 1/100Hz = 10ms. Suppose Vin = -2V at t = 0. The equation of the first half of the triangle is Vin = -2V + t*(2V - (-2V)) / (5ms) so Vin = -2V + t * 4/5 (t in ms) Solve Vin = 0.6V you get t = 3.25ms. Hence for t = 0.. 3.25ms your diode is reverse biased, and for t = 3.25ms..5ms your diode is forward biased.

The equation of the second half of the triangle is Vin = Vin = 2V - (t - 5ms)*(4/5) with t in ms. Solve Vin = 0.6V you get t = 6.75ms. Hence for t = 5ms.. 6.75ms your diode is forward biased, and for t = 6.75ms..10ms your diode is reverse biased.

Hence for one period, your output is the following:

Vout = -1V + t * 4/10, t = 0 .. 3.25ms

Vout = 0.6V, t = 3.25ms..6.75ms

Vout = 1V - (t - 5) * 4/10, t = 6.75ms..10ms

Now you can compute the average value:

Average = 1/T * integral from 0 to T of Vout(t).dt

= (1/T) * (I1 + I2 + I3) where: I1 = integral from 0 to 3.25ms = -1.1375

I2 = integral from 3.25ms to 6.75ms = 2.1

I3 = integral from 6.75 to 10ms = -1.1375 (we could have immediatly find that I1 = I3 because of the symetry).

So you have:

Average = 1/10 * (-1.1375 + 2.1 + -1.1375) = -17,5mV