
#1
Nov1711, 02:33 AM

P: 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? 



#2
Nov1711, 07:44 AM

HW Helper
P: 4,708

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




#3
Nov1911, 03:40 AM

P: 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
Nov1911, 04:02 AM

P: 1,506

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




#5
Nov1911, 04:05 AM

P: 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
Nov1911, 04:29 AM

P: 1,506

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
Nov1911, 05:15 AM

P: 8

i) transformation ratio
In the end I want a DC Voltage of 5V. Using a fullwave bridge this means [itex] V_{dc}=\frac{2 V_{2,max}}{π} V_{2,max} = \frac{π V_{dc}}{2} = 7.85V [/itex] Since I need a transformer that is giving this peak voltage Transformer ratio = [itex] \frac{V_{1,max}}{V_{2,max}} = \frac{V_{1}*\sqrt{2}}{V_{2,max}} = \frac{220V*\sqrt{2}}{7.85V} = 39.6 [/itex] Can you tell me if this is correct? 



#8
Nov1911, 05:21 AM

P: 8

ii) diode configuration
FullWaveBridge iii) capacitor value http://en.wikipedia.org/wiki/Ripple_...domain_ripple Can I use this simplification here? [itex] V_{ripple} = \frac{I_{load}}{2fC} [/itex] [itex] P = I_{load}*V_{dc} [/itex] 



#9
Nov1911, 05:37 AM

P: 1,506

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
Nov1911, 06:16 AM

P: 1,506

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
Nov1911, 06:48 AM

P: 8

Thank you very much for your quick reply.
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
Nov1911, 06:51 AM

P: 8

What about the transformer ratio, was this thought correct?
Thank you very much! 



#13
Nov1911, 06:58 AM

P: 1,506

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
Nov1911, 07:00 AM

P: 1,506

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 



#15
Nov1911, 08:45 AM

Mentor
P: 11,415

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.




#16
Nov1911, 09:06 AM

P: 1,506

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 



#17
Nov2011, 05:00 AM

HW Helper
P: 4,708

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



#18
Nov2011, 05:05 AM

P: 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. 


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