Unlocking the Mystery: Calculating Values for a Full Wave Bridge Circuit

[Shahil]

Help! Full Wave Bridge!

Hi guys!

Okay - I know that this is a straight-forward intermediate level circuit theory problem but uh...I've forgotten how to do it so I need help!



I have a full-wave rectifying bridge connected to a rectifying cap and a regulating zener diode. I do have values for the cap/zener/resistor (worked out through trial and error) but: What equations do you use to work out values for this circuit? I've looked all over the net and couple textbook but can't seem to find any good information on it! If it does help,
Please help people - I really need to figure this out!



Note on pic:
Zener = 3.9V
R = 1100 ohms
C = 1000uF

Only the non-inverted coloured part of the circuit is the part that is concerend in my question.

 

[juming]

I'm having the same problem with my powersupply circuit. Some people have suggested to use 4700uF in the rectifying caps... but I have no idea what the theory behind it is... I've just worked mine by trial & error in some simulation software called "Protel 99".

 

[thepatientmental]

Hi there,

No need to panic, let's break it down step by step. The full-wave bridge circuit is used to convert alternating current (AC) to direct current (DC) by utilizing four diodes in a bridge configuration. This allows for a more efficient conversion compared to a half-wave rectifier.

To calculate the values for this circuit, we need to use the following equations:

1. Peak AC voltage (Vp) = RMS AC voltage (Vrms) x √2
2. Peak DC voltage (Vdc) = Vp - Vd (where Vd is the voltage drop across the diode, typically around 0.7V)
3. Capacitor charging time (t) = 2 x R x C (where R is the resistor value in ohms and C is the capacitor value in farads)
4. Ripple voltage (Vr) = Vdc / (2 x f x C) (where f is the frequency of the AC input)

Using these equations, we can calculate the values for the circuit in your picture.

1. Vp = Vrms x √2 = 120V x √2 = 169.7V
2. Vdc = Vp - Vd = 169.7V - 0.7V = 169V
3. t = 2 x R x C = 2 x 1100 ohms x 1000uF = 2.2 seconds
4. Vr = Vdc / (2 x f x C) = 169V / (2 x 60Hz x 1000uF) = 1.41V

Therefore, the values for your circuit are as follows:
- Peak AC voltage = 169.7V
- Peak DC voltage = 169V
- Capacitor charging time = 2.2 seconds
- Ripple voltage = 1.41V

I hope this helps! If you have any further questions, please don't hesitate to ask. Good luck!