# The peak-to-peak ripple voltage

1. Apr 7, 2016

### gazp1988

1. The problem statement, all variables and given/known data
A 9 V power supply delivers 2 A to a resistive load. The a.c. supply is 230 V, 50 Hz and a bridge rectifier is used in conjunction with a 0.047 farad reservoir capacitor.

2. Relevant equations
Vpp = 1/(2FCRL) * VC

3. The attempt at a solution
1/(2*50*0.047*2) * 9
=0.957v
after doing research it says that the frequency will be double due to it being a bridge rectifier because its a full wave. if this is the case then is the 2 in the equation taking this into consideration. am i on the right track with my answer

cheers

Last edited by a moderator: Apr 7, 2016
2. Apr 7, 2016

### Staff: Mentor

Could you clarify this equation? Where did you get it? Are the parenthesis correct? The units don't seem consistent to me -- for example, the units of the LHS are Volts, but the units of the RHS don't seem to come out as Volts (unless I'm misinterpreting the parenthesis.

BTW, it helps if you can post equations in LaTeX format -- that usually removes the ambiguity. A tutorial on LaTeX can be found here: https://www.physicsforums.com/help/latexhelp/

Finally, it would help if we could have a figure to talk this problem through. I'll post one that I found via a Google Images search:

http://macao.communications.museum/images/exhibits/2_16_0_12_eng.png

3. Apr 7, 2016

### gazp1988

Vpp=/ frac{1} {2FCRl} {Vc}

This equation came from my notes that were attached to the assignment equation

4. Apr 7, 2016

### Staff: Mentor

Is this intended to be this?
$$Vpp = \frac{1}{2FCRI} Vc$$

5. Apr 7, 2016

6. Apr 7, 2016

### Staff: Mentor

The common approximation for the ripple voltage for a full-wave rectifier is

$V_{pp} = \frac{V_o}{2 f C R_L}$

where $V_o$ is the DC output voltage, $R_L$ is the load resistance.

Since $V_o / R_L$ is the DC load current, it can also be written as:

$V_{pp} = \frac{I}{2 f C}$

This approximation makes certain assumptions about the form of the discharge curve (straight line assuming constant load current) and that the discharge occurs over the full time interval between the waveform peaks. In reality the curve will be an exponential decay and the discharge is "caught" by the rising sinewave voltage just prior to the next peak. As a result the approximation formula will produce a value that is slightly high.

7. Apr 10, 2016

### gazp1988

yes berkeman, its suppose to be like that.

i have used both above formula and gave me the same answer of 0.957

Vpp = 9/(2 x 50 x 0.047 x 2) = 0.957
Vpp = 4.5/(2 x 50 x 0.047) = 0.957

8. Apr 10, 2016

### Staff: Mentor

Can you explain each of the quantities in the formulae? What value are you giving the load resistance?

9. Apr 10, 2016

### gazp1988

9 = dc output
2 = double the frequency
50 = frequency
0.047 = reservoir capacitor
2 = load resistance given in the question, is this information you wanted

10. Apr 10, 2016

### Staff: Mentor

All is good except for the load resistance. How did you obtain its value? It was not given directly in the problem statement.

11. Apr 10, 2016

### gazp1988

i was assuming that 2A stated in the question was the load resistance or as quoted in the question resistive load

12. Apr 10, 2016

### Staff: Mentor

2 A has unit A for Amperes. It's the load current. Let me re-write the opening statement of the problem:

"A 9 Volt power supply delivers 2 Amps to a resistive load."

13. Apr 10, 2016

### gazp1988

i see i have made a mistake now.
using the second formula above:
Vpp= 4.5/(2 x 50 x 0.047)

14. Apr 10, 2016

### Staff: Mentor

I think you're plugging things into the wrong places in the formulas. The formula I believe that you're trying to use now is:

$V_{pp} = \frac{I}{2 f C}$

where $I$ is the load current. There is no load resistance value in this version.

15. Apr 10, 2016

### gazp1988

i see what i have done, sorry.
I = 2...
Vpp = 2 / (2 x 50 x 0.047) = 0.426

16. Apr 10, 2016

### Staff: Mentor

Yes. Much better