Changing capacitance effect on ripple in full rectification circuit

  • #1
kaiiliana
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[Mentor Note -- thread moved to the schoolwork forums]

Hi. So I'm a Physics HL student in IB and I'm designing a circuit for my Physics IA. (think of it as a lab/project if you're not familiar with IB)

I'm required to have an independent variable of 7 dif. values and a dependant variable. For my IA, I'm trying to build a diode bridge full rectification circuit. What I will do is I will connect different capacitors of different capacitance values, and connect the circuit to an oscilloscope to read the ripple values generated. According to formulas I've seen online, my hypothesis is that when I increase the capacitance, the ripple will decrease. By showing the relationship between capacitance and ripple, I plan to conclude that since different mobile phone chargers have different capacitors within with capacitances, the ones with higher capacitance values will be more qualified.

I have a few questions regarding this circuit I'm planning to build. I have the oscilloscope and cables at school so that's fine. I know I need an AC power generator, a transformer, 4 diodes, 7 capacitors and one resistor for the circuit. Though, I'm not sure for what values I should buy these. I'm going to an electronics store tomorrow, but how many ohms should my resistor be, or what interval of capacitances should I buy my capacitors in, how many volts should the ac generator have etc. ? Also I know the basic set up of a diode bridge rectifier circuit, sth like this:

xwp7x05qj6k81.png

Though I'm not sure where I would connect the oscillator. Also, the value that I'll read from the oscillator, is it the ripple voltage or the ripple current, and in that case how do they differ? Like, I know the diff. between voltage and current obv but doesn't the ripple current/voltage give the same thing technically? I had a hard time trying to grasp that. So yeah, I would appreciate if you could answer the questions I had, or any advice regarding this circuit or IA is appreciated. If you advice me on changing something, or are confident about your knowledge in this topic, feel free to dm please. Thanks a lot.
 
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  • #3
Welcome to PF.

kaiiliana said:
By showing the relationship between capacitance and ripple, I plan to conclude that since different mobile phone chargers have different capacitors within with capacitances, the ones with higher capacitance values will be more qualified.
No, that's not how phone chargers work. They use switching power supplies, not bridge rectifiers to generate the 5Vdc output. Their output ripple is at the switching frequency (100kHz-1MHz typically), and depends more on the Equivalent Series Resistance (ESR) of the output capacitors.

kaiiliana said:
Though I'm not sure where I would connect the oscillator.
So do you have a signal generator at school that you will use? What is its output resistance (look on its datasheet)? You will need to include that series output resistance in your circuit diagram in order to make it more accurate.

kaiiliana said:
Also, the value that I'll read from the oscillator, is it the ripple voltage or the ripple current, and in that case how do they differ?
To see the ripple in this type of circuit, you will want the RC time constant of that output circuit to be about the same as the frequency of your oscillator.
 
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  • #4
Thank you. My school does not have a signal generator, though I won't be needing that as I'll use the AC power supply and just buy a transformer. Thanks for the correction about the phone.

I tried creating a simulation of the circuit I'm planning to build, though the ripple graph (I mean output voltage - time) only shows the way I want it to when the resistance of the resistor is extremely high for some reason. What would you suggest changing in the circuit to fix that, as I probably won't be able to find such a high resistance resistor.

https://www.multisim.com/content/XY...-of-ac-to-dc-power-supply-unregulated-1/open/
 
  • #5
kaiiliana said:
the ripple graph (I mean output voltage - time)
That doesn't look like the ripple graph, but like the 'voltage over the resistor' graph ...

It does try to show you the ripple (i.e. the AC component of the output) , but that ripple is so small you don't see it.

Start your simulation with a fully charged capacitor, or change the vertical axis of the graph you have, and you'll be happy :smile:

 
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