# Full wave Bridge rectifier and smoothing circuit

• Engineering
• Jupiter_10
In summary: So if you have a transformer with a 10:1 ratio at 230 rms at 50 Hz, then the secondary rms will be 23.0*(1.4-0.7)*(1.4)=24.4.
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 RL 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

## 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 VCP is secondary rms - 1.4V = 17.253-1.4= 15.853V

Peak to Peak ripple voltage is VRPP= (1/(2fCRL))(VCP-(VRPP/2). Solving for VRPP, VRPP=0.4(15.853-VRPP/2),
VRPP=5.28V

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

DC load voltage is VCP - (VRPP/2) = 15.853 - (5.28/2)= 13.213V

Load Current IL is VL/RL= 13.213/25= 0.528A

Can someone verify the methods I have used.

Jupiter_10 said:
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 VCP is secondary rms - 1.4V = 17.253-1.4= 15.853V Why would the peak VC 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.)

The peak secondary voltage is the desired output VO + 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.

Jupiter_10 said:
The peak secondary voltage is the desired output VO + 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.]

Last edited:
Oh, and welcome to PF!

Jupiter_10 said:
The peak secondary voltage is the desired output VO + 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.

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(2VD) (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 VPPR equation is 1/(2fCRL)(Vc(peak)-(VPPR/2))=VPPR.
If I plug in the values and solve for VPPR, I get 10.38V. And VR(rms), because of a sawtooth wave form, is VPPR/(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

Those values look good to me.

Jupiter_10 and collinsmark
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

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.

collinsmark said:
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.

collinsmark said:
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?

## 1. What is a full wave bridge rectifier?

A full wave bridge rectifier is an electronic circuit that converts alternating current (AC) to direct current (DC) by using a combination of diodes in a bridge configuration. This allows for both positive and negative portions of the AC input to be converted into positive DC output.

## 2. How does a full wave bridge rectifier work?

A full wave bridge rectifier works by using four diodes arranged in a bridge configuration. During the positive half cycle of the AC input, two diodes are forward biased and conduct current while the other two are reverse biased and do not conduct. During the negative half cycle, the roles of the diodes are reversed, resulting in a continuous flow of current in the same direction.

## 3. What is the purpose of a smoothing circuit in a full wave bridge rectifier?

A smoothing circuit, also known as a filter circuit, is used in conjunction with a full wave bridge rectifier to smooth out the pulsating DC output. This is achieved by using a capacitor to store and discharge electrical energy, resulting in a more constant and stable DC output.

## 4. What are the advantages of using a full wave bridge rectifier and smoothing circuit?

A full wave bridge rectifier and smoothing circuit have several advantages, including a higher output voltage and lower ripple voltage compared to a half wave rectifier. The full wave rectifier also allows for a more efficient use of the AC input, resulting in less power wastage.

## 5. What are some common applications of a full wave bridge rectifier and smoothing circuit?

Full wave bridge rectifiers and smoothing circuits are commonly used in electronic devices that require a steady and stable DC power supply, such as computers, televisions, and power supplies for electronic circuits. They are also commonly used in charging systems for batteries and in renewable energy systems to convert AC power from solar panels or wind turbines into usable DC power.

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