# Rectifier Circuit - Guidance requested

1. Apr 5, 2017

### sr0019

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

I have an unregulated power supply circuit consisting of the following:

Transformer, 10:1 ratio, 230VAC rms 50Hz incoming voltage
Full Wave Bridge Rectifier
Filter Capacitor, 1000μF

Find the following:

1) RMS & peak voltage across secondary winding of the transformer
2) Peak voltage across the capacitor
3) Peak to peak and rms ripple voltages
4) Load d.c. voltage & current

I am not after answers - just guidance & a push in the right direction. I want to learn, rather than just be told. The real problem I tend to have is confidence that I am doing the right thing. This applies to my answers for 2,3 & 4. Apologies if I have missed any information/not set out the query correctly!

2. Relevant equations
For 1) V2/V1 = N2/N1
For 2) VL pk = Transformer secondary V (V2) - 1.4V (V.D. due to rectifier)
For 3) Vr = IL/fC (remembering that the frequency will be double that of the incomer, as it is bridge rectifier)
For 4) To be determined

3. The attempt at a solution

1) V2/V1 = N2/N1. V2/230 = 1/10 . Therefore, V2 = 23V rms or 32.5V pk

2) VL pk = V2 - 1.4V. 32.5 - 1.4 . Therefore, VL pk = 31.1V pk

3) Assuming Vdc = VL pk , approx value of IL = VL/RL. Therefore, IL = 1.24A
Vr = IL/fC . Vr = 1.24/(100 x (1000 x 10^-6)) . Therefore Vr = 12.4V pk

4) I am yet to attempt number 4 or convert 12.4V to RMS, as I feel my answer for 3 is completely out - the ripple voltage seems too big.

2. Apr 5, 2017

### tech99

It looks as if 1000uF is nowhere near enough, at it gives a time constant with the load resistor of about 2.5 cycles. It would be better to be ten times that value.

3. Apr 5, 2017

### BvU

Hello sr,
You've done quite well in my view, so no need to apologize. Possible improvements:
mention the 0.7 V per diode when conducting.
It's a good habit to briefly explain the meaning of ALL symbols used (unless 'everybody' knows that ...). Especially Vdc

But (of course?) I do have some comments/questions:
First: did you draw lots of diagrams ? of the circuit layout, of the 'effective circuit' during one half period,
of the secondary voltage as a function of time, of the voltage of the capacitor as a function of time ?
Especially the last two (in one picture): can you post it ?

I can understand 1) and 2). Could you explain a bit more what you do in 3) ? What's Vr = IL/fC stand for ? (I recognize the expression, but you want to look at it in detail to understand). And yes, the ripple will be considerable when you have such a heavy load. Just consider discharging 25 V on 1 mF over 25 $\Omega$ and see how steep that goes.

4. Apr 5, 2017

### Staff: Mentor

Hi sr0019,

Welcome to Physics Forums!

Looks like you're doing okay for (1) and (2).

When the simple approximation method for ripple (based on constant load current) yields a value that's an appreciable chunk of the total output voltage you know that it will be an overestimate as the discharge current will have dropped significantly between the start and end of the capacitor discharge.

You can get a better value by looking at the exponential RC discharge curve. Again there are different levels of accuracy you can obtain based on what approximations you're willing to make. Assuming that the capacitor discharges from one voltage peak until the next voltage peak is one level of approximation. Finding the actual intersection of the capacitor discharge curve and the rectified AC curve as it returns to peak is better, but when you're dealing with curves that are a mix of exponentials and sinusoids you need to be ready to solve graphically or via numerical methods.

5. Apr 5, 2017

### sr0019

Hello tech99, Having read some technical literature online, I do agree with what you are saying. Unfortunately that is the value I have been given to work with.

6. Apr 5, 2017

### sr0019

Good Afternoon gneill,

Many thanks for the warm welcome!

The literature I have been given for this particular set of questions seems to use the approximation method to work out the ripple voltage. There is a question further on which requests that I use simulation software to produce a graphical output of the circuit & then compare/comment the results I worked out in the previous questions. I presume that when I produce the graphical format the ripple voltage will be smaller (different) than what I have approximated, which will be one aspect of the results I can comment on.

However, saying this, it would be more than useful to actually be able to work out the ripple voltage rather than an approximation. I will look into this more now and get back to you (if that is ok!) with any queries I have.

Thanks Again!

7. Apr 5, 2017

### sr0019

Good Afternoon BvU,

Thank you for the warm welcome & for the advice regarding additional information etc.

I do have a diagram - when I can get access to scanner, I will provide this.

My understanding of the circuit is as follows:

The transformer steps down the AC voltage at a 10:1 ratio. The frequency remains the same as the mains side
The bridge rectifier converts the AC voltage to DC. This is achieved by opposite diodes in the bridge conducting either side of the load at each half cycle. This also results in the frequency doubling
The capacitor will charge up near to the peak of each rectified cycle and will discharge into the load (resistor) for the rest of the cycle. This is what causes the smoothing effect.

Vr = IL/fC is an approximation for the ripple voltage. It takes an approximation for the load current & divides it by the product of the frequency and the capacitance.

8. Apr 5, 2017

### CWatters

I think you are doing everything correctly.

Looks like the problem is designed to have large ripple to highlight that the approximation method is only accurate when the ripple is small.

Another approximation method you can use is to calculate dv/dt = I/C (assumes constant current)
Then plot the dv/dt line over the rectified sine wave.
Read off where it crosses the next peak (about 22V in this case).
So the ripple is 31-22 = 9V pk to pk by this method.

9. Apr 5, 2017

### CWatters

In case it's not obvious where I got dv/dt=I/C from...

Q=VC ("Queen = Victoria Cross")
dQ/dt = Cdv/dt
dQ/dt = I
I = Cdv/dt
dv/dt = I/C

10. Apr 5, 2017

### sr0019

Good Afternoon Everyone,

Many thanks for all of your input so far - it has been extremely helpful. I have been stuck in a bit of rut with a couple of these problems recently and this has given me a bit of confidence going forward.

Based on the use of the approximation Vr = IL/fC, I end up with a result of 12.4V pk to pk. To find the rms value, I assume that I carry out the following:

0.5 x 12.4V = 6.2V pk . Then 6.2V/sqrt2 = 4.38V rms . Which gives me the two answers required for Q3.

For Q4. - Load d.c. voltage and current

Load d.c. voltage = V pk - Vr pk

Vdc = 31.1V - 6.2V . Therefore Load d.c. voltage = 24.9V

Idc = 24.9/25. Therefore load d.c. current = 996mA

I hope this is on the right path?

Thanks again!

P.S. I will get round to posting the sketches, as requested above - I just need to open a photobucket account.

11. Apr 5, 2017

### Staff: Mentor

Unless the curve is a sinusoidal the rms value won't be given by the peak/√2 . The ripple has more the shape of a sawtooth. Check out the Wikipedia article on rms.

12. Apr 5, 2017

### CWatters

+1

An RMS value is essentially an average. You might decide the ripple is like a triangular wave of height 12.4V. The average height of a triangle is it's area divided by the base.

13. Apr 6, 2017

### sr0019

Hello,
Apologies for the delayed response.

Based on the information above, would it be reasonable to assume then that the discharge period is approximately equal to the total period, therefore creating the sawtooth shape?

If so, this then gives the rms value as: peak/√3 ?

14. Apr 6, 2017

### BvU

Repeat CWatters' drawing using the values from your exercise

15. Apr 6, 2017

### Staff: Mentor

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