Calculating the DC value of the output voltage for a full wave rectifier

[JC2000]

Homework Statement

In a full wave bridge rectifier circuit having 230V, 50Hz at the primary input of a transformer, the peak value of the output voltage without filter is found to be 25 Volts. Calculate the (a) DC value of the output voltage with filter and (b) Value of capacitance needed for a ripple factor of 0.004 at IL= 0.5 mA

Relevant Equations

Without filter:
$V_{dc} = \frac{2*V_p}{ pi }$
$V_{rms} = \frac{V_p}{\sqrt{2}}$

With filter :

$V_{dc} = \frac{4fCR_L*V_p}{1 + 4fCR_L}$

I need help with part (a)... I know that the root-mean-square voltage is the dc-equivalent voltage for an AC waveform and what my book labels "$V_{dc}$ is actually the average voltage. Hence I am assuming the question is asking me to find $V_{rms}$ ...

Is my assumption that the root-mean-square voltage and average voltage of a full wave rectifier with filter are the same correct?! If so, solving the problem becomes easy since we can find $V_{rms}$ without filter using the peak voltage, which will remain the same when a filter is introduced to the circuit...

For part (b), since we now know $V_{rms}$ and the load current, we can apply Ohm's law and find out the load resistance. Then use the following equation to find the capacitance :

$Ripple Factor = \frac{1}{4\sqrt{3}fCR_L}$

 

[Tom.G]

(Hope I'm not giving away too much here.)
Hint: calculate the output characteristics based on the peak voltage.

 

[DaveE]

I know that the root-mean-square voltage is the dc-equivalent voltage for an AC waveform and what my book labels "Vdc is actually the average voltage.

No, RMS isn't "equivalent" to DC, I'm sorry someone led you to believe that. The RMS value of a waveform is a way of calculating the amount of heat generated in lossy components (resistors). It is commonly used to characterize AC waveforms.

RMS = "root mean square" $= \sqrt{ \frac{1}{T} \int_0^T{v(t)^2} dt} = \sqrt{ave(v(t)^2)}$

DC = average value $= \frac{1}{T} \int_0^T{v(t)} dt = ave(v(t))$

If v(t) is a constant "DC", then the two calculations have the same result. Purely AC waveforms have a DC value (average value) of 0. But, for example, a unipolar square wave (pulses between 0 and 1) has different values for DC and RMS.

 

[JC2000]

(Hope I'm not giving away too much here.)
Hint: calculate the output characteristics based on the peak voltage.

If I understand correctly, there are equations that relate the dc output voltage to the peak voltage, however I am not sure if the peak voltage value when the circuit does not have a filter can be used to calculate the dc output voltage when there is a filter...

I simulated two circuits with and without a filter (keeping everything else the same) and noticed that peak voltage seems to differ between the two hence I am not sure if it would be correct to use peak voltage (without filter) to calculate dc output voltage (with filter)...

 

[JC2000]

No, RMS isn't "equivalent" to DC, I'm sorry someone led you to believe that. The RMS value of a waveform is a way of calculating the amount of heat generated in lossy components (resistors). It is commonly used to characterize AC waveforms.

RMS = "root mean square" $= \sqrt{ \frac{1}{T} \int_0^T{v(t)^2} dt} = \sqrt{ave(v(t)^2)}$

DC = average value $= \frac{1}{T} \int_0^T{v(t)} dt = ave(v(t))$

If v(t) is a constant "DC", then the two calculations have the same result. Purely AC waveforms have a DC value (average value) of 0. But, for example, a unipolar square wave (pulses between 0 and 1) has different values for DC and RMS.

I see! Tbh, the distinction was fuzzy when I posted the question, thank you for clearing things up!
Just to be clear, this would mean that the question wants me to find out $V_{DC}$ for the ripple in the output?

 

[DaveE]

I see! Tbh, the distinction was fuzzy when I posted the question, thank you for clearing things up!
Just to be clear, this would mean that the question wants me to find out $V_{DC}$ for the ripple in the output?

a) Find the DC (average) output voltage of the filtered rectifier.
b) Find the capacitor value required to achieve the stated ripple factor. Ripple factor is the ratio of the AC only part of the waveform (the waveform with the average value subtracted), expressed as RMS, to the DC value of the waveform. Note that the RMS value of the waveform is different than the AC RMS value since the latter has the DC component removed before the RMS calculation.

 

[DaveE]

AC RMS = RMS after the DC component is subtracted $= \sqrt{ \frac{1}{T} \int_0^T{(v(t)-ave(v(t)))^2} dt} = \sqrt{ ave[(v(t)-ave[v(t)])^2]}$

 

[Tom.G]

I simulated two circuits with and without a filter (keeping everything else the same) and noticed that peak voltage seems to differ between the two hence I am not sure if it would be correct to use peak voltage (without filter) to calculate dc output voltage (with filter)...

That is because:
1) the filter capacitor charges only when the peak rectifier voltage exceeds the capacitor voltage
2) this results in a high peak current into the capacitor, instead of the current being spread out over the whole waveform
3) the high peak current thru the rectifier and transformer impedance causes a higher voltage drop in them

Therefore, the peak voltage into the filter is smaller.

Cheers,
Tom

 

[DaveE]

That is because:
1) the filter capacitor charges only when the peak rectifier voltage exceeds the capacitor voltage
2) this results in a high peak current into the capacitor, instead of the current being spread out over the whole waveform
3) the high peak current thru the rectifier and transformer impedance causes a higher voltage drop in them

Therefore, the peak voltage into the filter is smaller.

Cheers,
Tom

Yes, I agree. But I suspect that for students they will use some simplifying assumptions, like the capacitor charging without loading the transformer. Back in the day when Spice was a pain to use, I used the appendix in the old National semi audio/radio handbook. I still think it's one of the best references for this sort of design. In the real world, it's a difficult circuit, very non-linear.