Calculating Voltage Ripple in a Full Wave Rectifier Circuit

In summary, the conversation discusses a power supply with a voltage regulator LM317T at the end. The question is to find the voltage ripple across the capacitor, and the peak voltage is found to be 17 volts. Due to the diode bridge rectifier, 1.5 volts are lost, leaving 15.5 volts across the capacitor and LM317. The conversation then moves on to solving for the voltage ripple, with equations Vr = I / (2fC) and Vr = Vp / (R_load * C) mentioned. The problem is discussed further, with a sketch provided as a guide. The question of whether the equations are correct is also brought up.
  • #1
foobag
75
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?
 
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  • #2
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:
  • #3
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?
 

Related to Calculating Voltage Ripple in a Full Wave Rectifier Circuit

1. What is power supply voltage ripple?

Power supply voltage ripple refers to the small fluctuations or variations in the output voltage of a power supply. These variations can occur due to internal factors such as the design and components of the power supply, or external factors like changes in the input voltage.

2. Why is power supply voltage ripple important?

Power supply voltage ripple can affect the performance and reliability of electronic devices. High levels of ripple can cause voltage spikes or drops, which can lead to malfunctions, overheating, or even damage to components. Therefore, it is important to keep the voltage ripple within acceptable limits for the proper functioning of electronic devices.

3. How is power supply voltage ripple measured?

Power supply voltage ripple is measured in terms of peak-to-peak (p-p) or root mean square (RMS) voltage. P-p voltage is the difference between the highest and lowest voltage points in the ripple waveform, while RMS voltage is the average of the absolute values of the voltage variations.

4. What causes power supply voltage ripple?

Power supply voltage ripple can be caused by a variety of factors, including the design and quality of the power supply, variations in the input voltage, and the load on the power supply. Other factors like temperature, aging of components, and electromagnetic interference can also contribute to voltage ripple.

5. How can power supply voltage ripple be reduced?

There are several methods for reducing power supply voltage ripple, such as using high-quality components, proper design and layout techniques, and implementing filtering and regulation circuits. It is also important to ensure that the input voltage and load are within the recommended limits for the power supply. Additionally, using a linear or switching regulator can help reduce voltage ripple in a power supply.

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