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I have the following circuit:

enter image description here

where Vin is waveform with the frequency of 1kHz and amplitude 10V. The diodes are ideal. I need to find the average value of Vout.

I found that when Vin > 0 then D1 is off, and D2 is on, so Vout = I1R1 = 1V. And when Vin < 0 then D1 is on and D2 is off, so Vout = 1 + Vin. Is this correct?

Now I think I need to use

enter image description here

I tried taking b = 1/f and a = 0 and writing the integral in two parts, one going from 0 to 1/(2f) and second from 1/(2f) to 1/f with. I used the Vout functions I got above in them, respectively, but I get a bad result. How do I do this correctly?

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  • \$\begingroup\$ What result did you get? (Yes, your setup seems okay but you don't show your exact process.) \$\endgroup\$ – jonk Nov 6 '17 at 22:27
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I found that when Vin > 0 then D1 is off, and D2 is on, so Vout = I1R1 = 1V. And when Vin < 0 then D1 is on and D2 is off, so Vout = 1 + Vin. Is this correct?

Yes, that's correct.

Now I think I need to use \$f_{avg}=\frac{1}{b-a}\int_a^bf(x)dx\$

\$f\$ in this sense means function, not frequency. Because this question will give the same answer if the frequency is 1 kHz or 10 kHz, the average will be the same since everything is ideal.

So in order to calculate \$f_{avg}\$ correctly we need to know what \$a\$ and \$b\$ are. I will write it in your way, and in my way.


Your way

Find out when the \$V_{in}\$ is less than 0, because that's the only time we have to care about it as you stated above in the first yellow box.

If we know that the frequency is 1 kHz and a sine wave, then the period time is 1 ms, from 0.5 ms to 1 ms it will be negative. This is \$a\$ and \$b\$.

\$f_{avg}=\frac{1}{(1-0.5)×10^{-3}}\int_{0.5×10^{-3}}^{1×10^{-3}} 10×\sin(2×\pi×1000×x)dx\$

If we continue we get that \$f_{avg}=-6.3662\$


My way

We want to get a half period of a sine wave, right? Then let's just integrate a half sine and change the sign.

\$f_{avg}=-\frac{1}{\pi}\int_{0}^{\pi}10×\sin(x)dx\$

In my opinion this one is much easier to calculate, I can even do it in my head.

\$f_{avg}=-\frac{1}{\pi}×10[-\cos(x)]_{0}^\pi = -\frac{1}{\pi}×10(1+1) = -10×\frac{2}{\pi} = -6.3662\$

Ahh! Now I see where those numbers came from. No wolframalpha here.


We know that half of the time \$V_{out}\$ will be \$1\$ V, other half of the time it will be \$1-6.3662\$.

So the total average for a whole period will be \$\frac{1+1-6.3662}{2}= -2.1830\$ V.

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  • \$\begingroup\$ Just curious, but did you note the OP writing "amplitude 10V?" \$\endgroup\$ – jonk Nov 6 '17 at 22:31
  • \$\begingroup\$ @jonk Gaaaaaaah! \$\endgroup\$ – Harry Svensson Nov 6 '17 at 22:35
  • \$\begingroup\$ Oh. And I may as well mention something else that's missing. But perhaps I'll wait until you fix things up and if I still see it missing, I'll mention it. \$\endgroup\$ – jonk Nov 6 '17 at 22:35
  • \$\begingroup\$ @jonk Hit me up Spock. \$\endgroup\$ – Harry Svensson Nov 6 '17 at 22:39
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    \$\begingroup\$ Nah. Looking so much better now. You have 3 areas, two of them \$1\times\pi\$ in area and the third one computed good now as \$-20\$. So that's \$\frac{1\cdot\pi+1\cdot\pi-20}{2\pi}\$. And I'm in the same place now. Great! \$\endgroup\$ – jonk Nov 6 '17 at 22:46

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