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capacitor value

In the above image a small part of a large circuit I'm trying to understand.

  • V1 : 220v @50
  • D1 : Any diode like 4007
  • R1 : 270k
  • R2 : 16k
  • C1 : 10u to 100u
  • M1 : Voltmeter

Vout = (Vs*R2) / (R1+R2)

According to the above formula, the voltage out from the resistors (voltage divider) should be 12.3v.

But the voltage after adding the capacitor is 4v

Where is the 4v out of the capacitor is coming from?

What's the formula to calculate it?

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  • \$\begingroup\$ Since you have a low pass filter with pulsating DC, you need to calculate the average voltage for your output which requires integration. \$\endgroup\$
    – qrk
    Commented Mar 4, 2022 at 22:00

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In this hypothetical question, the peak voltage is 220V , 50Hz which makes the RMS voltage 154.77 Vrms (non-standard)

The Thevenin equiv's are:

  • Rth = 15.1k = 16k//270k with an attenuation ratio \$\alpha\$= 5.6%
  • Vth = 12.75 Vp before diode which will have slightly less than 1mA peak current so assume Vf=0.6V thus Vth' after diode is Vp= 12.15V.

The average DC voltage without the capacitor is the best estimate.

\$V_{dc} =\frac{V_p}{\pi} = \dfrac{12.15}{\pi}=3.87~Vavg\$

Then adding the Capacitor 100 uF x 15.1k =C*Rth =T = 1510 ms time constant to 63% and will require at least 5T or 7.5 seconds to settle < 1%.

With 4V across the 16k load it is drawing a steady V^/R= 1mW during the discharge cycle but the charge cycle it is half sinusoidal so the power for charging must be greater for a shorter time. enter image description here

\$dV= \dfrac{4.0V*(50/3)ms}{15.1k*100uF}= 44 mVpp\$ That is derived from Ic=CdV/dt.

This is a sharp sawtooth estimate. But since it is sine driven the peaks are reduced by about 2/π or 64% of 44mV so the ripple is ~ 28 mVpp and the average is 2/3 roughly of the Vpp ripple so the Vdc comes out to about 4.02Vdc +/-14mV with about 1% margin of error..

How does this compare to your test result?

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  • \$\begingroup\$ I've simulated this design (tinyurl.com/yal74q3f) and your result seems to be correct. The main issue I have is that I'm not able to follow your explanation. What is Vth, and how did you get to 12.75V? And in the formula for dV, how did you determine dt = (50/3)ms? \$\endgroup\$
    – Velvet
    Commented Mar 5, 2022 at 1:20
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    \$\begingroup\$ Do you know how to do Thevenin transformation for a voltage divider? Rth, Vth for triangle what is the average ? 2/3 of peak. I estimated the charge time is 1/3 and discharge is 2/3 approx then rounded this sawtooth estimate by reducing the peaks using 2/pi to get sin shaped triangular ripple Vpp as an est. without messy integration \$\endgroup\$
    – Tony Stewart EE75
    Commented Mar 5, 2022 at 1:23
  • \$\begingroup\$ Thanks for the hint, I'll check it out. \$\endgroup\$
    – Velvet
    Commented Mar 5, 2022 at 1:24
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    \$\begingroup\$ You may recall integration of a triangle , or the average = 2/3 of the peak above a base. without proof... but the ripple is not sharp triangle so I reduced by averaging the triangle peaks \$\endgroup\$
    – Tony Stewart EE75
    Commented Mar 5, 2022 at 1:34
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    \$\begingroup\$ I see. Maths combined with the experience/intuition to know where to simplify. \$\endgroup\$
    – Velvet
    Commented Mar 5, 2022 at 1:46

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