# Design of a rectifier

1. ### Elisabeth_91

8
1. The problem statement, all variables and given/known data

A classical rectifier is to be designed to provide 5V +/- 1% for a logic circuit, with output power of up to 12W. The input is conventional 220V, 50Hz source.
i) Choose a transformer ratio
ii) Choose a diode configuration
iii) Calculate the capacitor value to meet the requirements
iv) Draw the load voltage, the load current and the source current (secondary side) waveforms for maximum output power (12W).

2. Relevant equations

3. The attempt at a solution

I am a complete beginner in this. I have no clue how to solve. Can you please give me a hint?

### Staff: Mentor

It's a meaty topic. Start by a google search on: rectifier "filter capacitor"

3. ### Elisabeth_91

8
I need to use a full wave rectifier bridge with a Capacitor parallel to my load. Is this correct?

What I don't understand is, do I put a normal transformer before the bridge?

4. ### technician

You are correct. Use a full wave bridge arrangement and you need a transformer to step down 220V to 5V

5. ### Elisabeth_91

8
Would it also be possible to use just a Capacitor Filter, this seems to be the easiest way to solve the problem.
Like shown here: http://www.tpub.com/neets/book7/0249.GIF

What is the advantage of the bridge?

6. ### technician

A single diode allows 0ne half of the AC current to flow. Imagine an AC wave which goes + and -. A single diode will cut off the - sections (or the +sections!!). This gives the familiar 'half wave rectification waveform'... can you picture it?
The full wave combination of diodes needs 4 diodes connected in a 'bridge' and this allows all of the AC current to flow in the same direction....it is smoother.

7. ### Elisabeth_91

8
i) transformation ratio

In the end I want a DC Voltage of 5V. Using a full-wave bridge this means
$V_{dc}=\frac{2 V_{2,max}}{π} V_{2,max} = \frac{π V_{dc}}{2} = 7.85V$

Since I need a transformer that is giving this peak voltage
Transformer ratio =
$\frac{V_{1,max}}{V_{2,max}} = \frac{V_{1}*\sqrt{2}}{V_{2,max}} = \frac{220V*\sqrt{2}}{7.85V} = 39.6$

Can you tell me if this is correct?

8
9. ### technician

The output voltage is specified as 5V +/-1%
This means that the max change in voltage allowed is 0.05V
The capacitor stores electric charge so that current can be maintained between the peaks of the voltage, this is when the voltage is decreasing.
The power that needs to be supplied is 12W .
Can you calculate thew current which must be supplied?
The time between peaks is 0.01s (half a cycle)
Can you calculate the charge that flows in this 0.01s?
This charge comes from the capacitor. If you know the voltage change is 0.05V and also know the charge that must be provided, can you calculate the capacitor value.
Get as far as you can and if you get stuck come back!!!

10. ### technician

I put my numbers in my procedure and got C = 0.5F
That formula you have gave 0.48F
Looks like you we on the right track. The formula works and I hope that my steps show how the formula comes about !!!

11. ### Elisabeth_91

8

So you mean:
I,load = P / V,dc = 12W / 5V = 2.4 A
Q,c = I,load / (0.5*T) = 2.4 A * 0.01s
C = Q/U = 0.024C / 0.05V = 0.48F

Is this the correct way?
Is it coincidence that is the same result as with my first approach?

12. ### Elisabeth_91

8
Thank you very much!

13. ### technician

Yes, you have done more than I would have!
I would just say ratio = Vin/Vout = 220/5 = 44 to 1
One thing to realise that these calculations, in practice, only need to be 'near enough'.
To produce a supply of exactly 5V +/-1% would probably involve using a further electronic circuit.... but that is another matter.
Also the capacitor value we calculated is a minimum value anything bigger than 0.5F would do a better job.
Well done

14. ### technician

also...your post #11 is spot on. I think it is better to work through the physics to get the answer, once you can do this then it is OK to use the formula.
well done

### Staff: Mentor

Don't forget to account for the diode voltage drops! The 5V +/-1% specification puts a pretty tight constraint on the allowed peak voltage (Vp), which will follow the peak voltage supplied by the bridge rectifier.

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Last edited: Nov 19, 2011
16. ### technician

To calculate C I did this:
charge Q = I x t where t is the time for 1/2 cycle (full wave rectification) =1/2f
Q = I/2f
C = Q/V (V is the ripple voltage)
C = 1/2fV

You equation is V = 1/2fC

Both methods are exactly the same

### Staff: Mentor

I'm guessing that you averaged the full wave rectified waveform to determine this DC value? That analysis only applies where there is no filter capacitor, so is not applicable to your circuit. In your circuit, the big filter capacitor charges to the peak value of the AC and holds this voltage until the next peak. So the DC output will be roughly equal to the peak of the AC (less diode drops).

You calculated 0.5F. That's 500,000 microfarads so is a very large capacitance.

18. ### Elisabeth_91

8
I know 0.5F is huge.
But can you tell me how to calculate it correctly? I thought that in this case the transformer ratio question can be solved independent from the question of the capacitor.

19. ### cmb

628
As a 'mathematical' question, it's fair to say there is no need to examine the spec of the transformer in detail. However, transformers will output different voltages according to their state of load, so you need to make sure 12W would have no influence on the level of saturation in the transformer, else I expect it would run high if the 12W load was removed.

(Suffice to say, in the real world you'd pick a, say, 10V transformer, use a small cap that might allow, say, 3V of ripple, and stick a voltage regulator in. So your values for capacitance will look artificially high, due to the artificial nature of the question.)

Last edited: Nov 20, 2011

### Staff: Mentor

Yes, transformer and filter capacitor are determined separately.

The method you used looked right (though I didn't check the figures). Q=CV where Q is the charge drained by supplying the load current for half the AC period, and V is the allowable droop in voltage. It gives an adequate estimate.

In a practical sense, electrolytic caps are not precision elements. Typically, they may have tolerances of +100%, -40% so you don't have to worry about 3 sig figs.