Calculating Voltage Ripple in a Full Wave Rectifier Circuit

Click For Summary
SUMMARY

The discussion focuses on calculating voltage ripple in a full wave rectifier circuit using the LM317T voltage regulator. The peak voltage is determined to be 17 volts, accounting for a 1.5-volt drop across the diode bridge, resulting in 15.5 volts across the capacitor. The user seeks guidance on applying the ripple voltage formulas, specifically Vr = I / (2fC) and Vr = Vp / (R_load * C), emphasizing the importance of using 120 Hz for full wave rectification. The conversation highlights the relationship between output current demand and voltage droop in the capacitor.

PREREQUISITES
  • Understanding of full wave rectifier circuits
  • Familiarity with the LM317T voltage regulator
  • Basic knowledge of capacitor behavior in power supply circuits
  • Proficiency in using ripple voltage equations
NEXT STEPS
  • Study the impact of load resistance on voltage ripple in capacitors
  • Learn about the effects of diode forward voltage drop in rectifier circuits
  • Explore advanced capacitor filtering techniques for power supplies
  • Investigate the use of simulation tools for analyzing rectifier circuits
USEFUL FOR

Electronics engineers, hobbyists designing power supplies, and students studying rectifier circuits and voltage regulation techniques.

foobag
Messages
75
Reaction score
0
http://img683.imageshack.us/img683/5329/unledtw.png

So its basically a power supply with a voltage regulator LM317T at the end.

The question asked to find the voltage ripple across the capacitor.

So I basically solved the peak voltage found which would be 12V * SQRT(2) = 17 volts.

and because of the diode bridge rectifier, 1.5 volt would be lost, so like 15.5 volts would go across the capacitor and eventually the LM317.

Now my question is how do I solve for the voltage ripple. I understand since this is a full wave rectifier, I should use 120 Hz instead of the 60Hz.

I found an equation as Vr = I / (2fC), where I is the current going into the capacitor? and another equation Vr = Vp / (R_load * C), where Vp is the peak voltage.

Could someone help me out and guide me through the problem?
 
Last edited by a moderator:
Physics news on Phys.org
foobag said:
http://img683.imageshack.us/img683/5329/unledtw.png

So its basically a power supply with a voltage regulator LM317T at the end.

The question asked to find the voltage ripple across the capacitor.

So I basically solved the peak voltage found which would be 12V * SQRT(2) = 17 volts.

and because of the diode bridge rectifier, 1.5 volt would be lost, so like 15.5 volts would go across the capacitor and eventually the LM317.

Now my question is how do I solve for the voltage ripple. I understand since this is a full wave rectifier, I should use 120 Hz instead of the 60Hz.

I found an equation as Vr = I / (2fC), where I is the current going into the capacitor? and another equation Vr = Vp / (R_load * C), where Vp is the peak voltage.

Could someone help me out and guide me through the problem?

The higher the output current demand, the more droop you will get in the capacitor storage voltage each 120Hz cycle. Start with a sketch like the following, and figure out the voltages and currents. There will be no current into the capacitor from the rectifier when the cap voltage is higher than the input voltage (and account for the diode drop in there too).

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

.
 
Last edited by a moderator:
well so were my equations to solve for ripple close?

my understanding is I need to find the current running through the capacitor am I correct?
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 52 ·
2
Replies
52
Views
3K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
23K
  • · Replies 15 ·
Replies
15
Views
18K
  • · Replies 1 ·
Replies
1
Views
3K