Can someone please tell me, whats the value of this capacitor?

Find value of capacitor, to have a DC volte acagross load equal to 9.5 volts.

 

You could make it nearly anything you want. There's no unique answer unless you state the problem more carefully.

 

Can someone please tell me, whats the value of this capacitor?

Are you given a ripple voltage by chance?

 

no ripple voltage

 

The capacitor value required will depend on how much ripple you can put up with, and how much current is drawn (in this case you would work that out from the voltage across the 330Ω resistor).

The capacitor value can be found from the formula C = I/(2*f*Vpp), where I is the load current, f is the mains frequency and Vpp is the peak to peak ripple voltage.

See this link: http://en.wikipedia.org/wiki/Ripple_%28electrical%29

Edit: To find I, you assume the voltage across the load resistor is the peak half-secondary voltage, less a diode drop. This simple formula assumes that the load current is constant, which is fairly accurate provided that the ripple is a small percentage of the total.

 

no ripple voltage

"Generally for DC power supply circuits the smoothing capacitor is an Aluminium Electrolytic type that has a capacitance value of 100uF or more with repeated DC voltage pulses from the rectifier charging up the capacitor to peak voltage. However, their are two important parameters to consider when choosing a suitable smoothing capacitor and these are its Working Voltage, which must be higher than the no-load output value of the rectifier and its Capacitance Value, which determines the amount of ripple that will appear superimposed on top of the DC voltage. Too low a value and the capacitor has little effect but if the smoothing capacitor is large enough (parallel capacitors can be used) and the load current is not too large, the output voltage will be almost as smooth as pure DC. As a general rule of thumb, we are looking to have a ripple voltage of less than 100mV peak to peak."

You can read more about the significance of the capictor value here: http://www.electronics-tutorials.ws/diode/diode_6.html

They demonstrate it's effects in a Bridge Rectifier, similiar to a Full Wave Rectifier.

Hopefully this helps.

 

I left out information. The question says....

Find the value of capacitor, to have a DC voltage across load equal to 9.5 volts.

 

"Generally for DC power supply circuits the smoothing capacitor is an Aluminium Electrolytic type that has a capacitance value of 100uF or more with repeated DC voltage pulses from the rectifier charging up the capacitor to peak voltage. However, their are two important parameters to consider when choosing a suitable smoothing capacitor and these are its Working Voltage, which must be higher than the no-load output value of the rectifier and its Capacitance Value, which determines the amount of ripple that will appear superimposed on top of the DC voltage. Too low a value and the capacitor has little effect but if the smoothing capacitor is large enough (parallel capacitors can be used) and the load current is not too large, the output voltage will be almost as smooth as pure DC. As a general rule of thumb, we are looking to have a ripple voltage of less than 100mV peak to peak."

You can read more about the significance of the capictor value here: http://www.electronics-tutorials.ws/diode/diode_6.html

They demonstrate it's effects in a Bridge Rectifier, similiar to a Full Wave Rectifier.

Hopefully this helps.

@ jegues: You really cannot say that you should generally have less than 100mV pk-pk ripple. In many cases this would result in an uneconomic capacitor value, but as importantly it would result in very high peak currents in the transformer, diodes and capacitor.

We can illustrate this by calculating the value of capacitor needed to achieve 100mV ripple in the OPs case. We now know that his output voltage is supposed to be 9.5V, so we can estimate the load current.

Load current I = 9.5V / 330Ω = 28.8mA. Ripple voltage Vpp = 100mV, so C = 0.0288A/(2*60Hz*0.1V) = 2.4E-3F or 2400μF.

That's quite a big capacitor for just 29mA of load current. At this rate, a 1A supply would need over 80,000μF, which is pretty enormous.

@ wil321: You may be able to find the capacitor value given a (mean?) output voltage requirement, but you would also need more information, such as the transformer secondary voltage and the diode voltage drop.

 

@ jegues: You really cannot say that you should generally have less than 100mV pk-pk ripple. In many cases this would result in an uneconomic capacitor value, but as importantly it would result in very high peak currents in the transformer, diodes and capacitor.

We can illustrate this by calculating the value of capacitor needed to achieve 100mV ripple in the OPs case. We now know that his output voltage is supposed to be 9.5V, so we can estimate the load current.

Load current I = 9.5V / 330Ω = 28.8mA. Ripple voltage Vpp = 100mV, so C = 0.0288A/(2*60Hz*0.1V) = 2.4E-3F or 2400μF.

That's quite a big capacitor for just 29mA of load current. At this rate, a 1A supply would need over 80,000μF, which is pretty enormous.

@ wil321: You may be able to find the capacitor value given a (mean?) output voltage requirement, but you would also need more information, such as the transformer secondary voltage and the diode voltage drop.

Thank you Adjuster, you make a very good point. I am still learning on such a topic and your comments are always provide new insight.

As for the questions you asked wil321, can't we simply compute the voltage across the secondary? (The OP does not indicate if the voltages given are RMS values but I assume so)

\(v_{s} = \frac{100\sqrt{2}}{5} = 20\sqrt{2} V\)

So the voltage across the secondary would be,

\(40\sqrt{2} V\).

As for the diode drop, is it unreasonable for us to assume a voltage drop of 0.7 when conducting?

I look forward to your input.

Thanks again!

 

Yes, I missed the 5:1 ratio. You would still want to be sure whether this was the whole or the half secondary voltage.

Depending which, if we assume 0.7V lost in the rectifier, and RMS sine-wave voltages, the peak output from the would be either

20V*√2 - 0.7V = 27.58V or 10V*√2 - 0.7V = 13.44V. (This circuit gives the half-secondary voltage, minus one diode drop.)

Whichever was the case, to get an average output of 9.5V would require a huge amount of ripple to be present. I'm not sure this is a reasonable question if these assumptions are correct. Certainly a solution could not use simple methods, as both exponential and sine wave shapes would need to be taken into account. Maybe a Spice transient simulation?

If on the other hand the voltages were peak values, with 10V peak the diode drop would take the voltage down to about 9.3V even before allowing for ripple. I wonder if the OP has stated the problem correctly - or maybe I'm missing something?