Engineering Full wave Bridge rectifier and smoothing circuit

[Jupiter_10]

L

Homework Statement

A Transformer with a stepdown ratio of 10:1 is supplied with a 230 RMS voltage at 50hz connected to a four diode full wave bridge rectifier and smoothing circuit. If the smoothing capacitor value is 1000uF and the load resistance R_L is 25 ohms, find:

the secondary winding rms and peak voltage,
the peak voltage across the capacitor
the peak to peak and rms ripple voltages
the load dc voltage and current

Homework Equations

The Attempt at a Solution

Ok the peak secondary winding voltage is 23V +1.4V(assuming 2 diodes at 0.7v each)=24.4V

The secondary rms is 24.4/sqrt2= 17.253V

Peak capacitor voltage V_CP is secondary rms - 1.4V = 17.253-1.4= 15.853V

Peak to Peak ripple voltage is V_RPP= (1/(2fCR_L))(V_CP-(V_RPP/2). Solving for V_RPP, V_RPP=0.4(15.853-V_RPP/2),
V_RPP=5.28V

RMS ripple voltage V_R(RMS) is 5.28/(2*3sqrt)= 1.52V

DC load voltage is V_CP - (V_RPP/2) = 15.853 - (5.28/2)= 13.213V

Load Current I_L is V_L/R_L= 13.213/25= 0.528A

Can someone verify the methods I have used.

 

[Merlin3189]

attempt at a solution
Ok the peak secondary winding voltage is 23V +1.4V(assuming 2 diodes at 0.7v each)=24.4V Can't see any reason for this.

The secondary rms is 24.4/sqrt2= 17.253V So how does this square with, Transformer is 10:1 and primary is 230 V rms

Peak capacitor voltage V_CP is secondary rms - 1.4V = 17.253-1.4= 15.853V Why would the peak V_C be diode drop from an rms value?
...

The following work looks ok, though I'm not familiar with these ripple approximations. (Perhaps you should quote the relevant equations to help people see what you are doing? I think you can find them in Hyperphys here.)

 

[Jupiter_10]

The peak secondary voltage is the desired output V_O + the diode voltage drops. i.e 23V + 1.4V. The question has not given me the values of the forward voltage drops across the diodes, I've assumed them. I know that when 2 are conducting, the other pair are reversed biased or "off."

The secondary RMS is the secondary peak voltage didvided by the square root of 2 I think.

The secondary voltage has to be higher than one diode drop (or in this case, 2). Unless its the peak VC is secondary peak voltage - 2 diode drops.

 

[collinsmark]

The peak secondary voltage is the desired output V_O + the diode voltage drops. i.e 23V + 1.4V. The question has not given me the values of the forward voltage drops across the diodes, I've assumed them. I know that when 2 are conducting, the other pair are reversed biased or "off."

The secondary RMS is the secondary peak voltage didvided by the square root of 2 I think.

The secondary voltage has to be higher than one diode drop (or in this case, 2). Unless its the peak VC is secondary peak voltage - 2 diode drops.

I have a couple of comments.

1) Shouldn't you be considering the voltage drop of the diodes when considering the peak voltage of the rectifier's output, rather than its input? I'm unclear as to your reasoning why the rectifier will affect the voltage on the transformer's windings directly (assuming an ideal transformer).

2) The diodes produce a voltage drop. "Drop" being the key word here. Shouldn't you be subtracting rather than adding?

[Edit: I'm assuming the circuit topology is 230 RMS voltage source --> transformer --> rectifier --> smoothing capacitor & load resistor (the load resistor and smoothing capacitor are connected in parallel). Correct me if I am mistaken on this assumption.]

 

[collinsmark]

Oh, and welcome to PF!

 

[Merlin3189]

The peak secondary voltage is the desired output V_O + the diode voltage drops. i.e 23V + 1.4V. The question has not given me the values of the forward voltage drops across the diodes, I've assumed them. I know that when 2 are conducting, the other pair are reversed biased or "off."

I see where you're coming from now. This would be true, if you were designing a power supply to provide a desired output. But

L

Homework Statement

A Transformer with a stepdown ratio of 10:1 is supplied with a 230 RMS voltage at 50hz connected to a four diode full wave bridge rectifier and smoothing circuit. If the smoothing capacitor value is 1000uF and the load resistance RL is 25 ohms, find:

here you are given a power supply and asked to work out the output.
You have to work forwards through the given components.
You can't say the transformer must provide 1.4V more than the peak capacitor voltage, because you don't yet know what that is.
You do know what the input of the transformer is, and you carry on from there.

Edit: changed output to input in last sentence.

 

[Jupiter_10]

Thanks for your responses guys, this is essentially the circuit if we assume the source input is actually a transformer with a 10:1 ratio at 230 rms at 50 Hz;

I can understand where your coming from about adding the diodes to the secondary rms voltage; the equation I used is a design equation indeed.

I did have original figures before I changed the equations as follows,

The secondary rms voltage is 23V as the primary is 230V rms at a ratio of 10:1. The peak secondary voltage is of course 23*√2= 32.53V


The peak capacitance voltage is always less than the applied peak voltage because of the voltage drops acrossed the diodes therefore Vc(peak) is peak secondary voltage - 1.4V(2V_D) (the equation from my textbook) which gives me 31.13V. I'm pretty confident about this seeing as though 0.7V drop per diode is the theoretical norm.

The peak to peak ripple voltage or V_PPR equation is 1/(2fCR_L)(Vc(peak)-(V_PPR/2))=V_PPR.
If I plug in the values and solve for V_PPR, I get 10.38V. And V_R(rms), because of a sawtooth wave form, is V_PPR/(2√3)= 2.99V

DC load voltage is Vc(peak) - half of the peak to peak ripple voltage or 31.13-(10.38/2)= 25.94V and so the load current is simply ohms law, 25.94/25=1.03A

 

[NascentOxygen]

Those values look good to me.

 

[Rafeng404]

Hi Guys,

I seem to be having trouble with the math here, I get the same values up until the 10.38V. Can you show me how you solved for Vppr please? The example in the book doesn't make sense either and I am pulling what little hair I have left out haha.

I feel like I have forgotten some basic maths principles

 

[collinsmark]

Hello @Rafeng404,

I recommend that you start a new thread.

Also, make sure you [follow the template and] show the work you have done so far. From there we should be able to point you in the right direction.

 

[Rafeng404]

Hello @Rafeng404,

I recommend that you start a new thread.

Also, make sure you [follow the template and] show the work you have done so far. From there we should be able to point you in the right direction.

Thanks mate, I have done as you advised :) hopefully someone will put me and a few others out of our misery soon haha.

 

[Whiley]

I have a couple of comments.

1) Shouldn't you be considering the voltage drop of the diodes when considering the peak voltage of the rectifier's output, rather than its input? I'm unclear as to your reasoning why the rectifier will affect the voltage on the transformer's windings directly (assuming an ideal transformer).

2) The diodes produce a voltage drop. "Drop" being the key word here. Shouldn't you be subtracting rather than adding?

[Edit: I'm assuming the circuit topology is 230 RMS voltage source --> transformer --> rectifier --> smoothing capacitor & load resistor (the load resistor and smoothing capacitor are connected in parallel). Correct me if I am mistaken on this assumption.]

Voltage in the load = Voltage in secondary winding - diode drops
Therefor:
Vs=VL+Vdrop

is this correct?