Full-Wave Bridge Rectifier Filter + Regulator DC Power supply

[Petrucciowns]

http://img104.imageshack.us/img104/3557/1234uyq.jpg

Ok, there are a few calculations on this problem that I am having difficulty with. This is a past homework assignment which is why the answers are filled in. They are all correct.

Anyways I approached the problem in this manner:

V secondary peak = V2/ .707

The voltage drop across RL is ZZT

IRL= ZZT/RL

VC1P= V2P - Both forward biased diode drops.

VC1 P-P (Ripple) = IDC / (2f * C)

VRS= Vc1 dc - Vzt

IRS= VRS /RS


The three that I am having trouble with are: Vc1-dc , IZ, and VRL p-p (Ripple)

Can anyone help me out?

Thanks.

 

[Petrucciowns]

Update>>>>>>>>>>>>>>>>>>>>>


Now all I need to know its how to find the value of. Vc-dc

 

[Mene]

I can understand your approach to the problem and the difficulty you are having with some of the calculations. It is important to note that this type of circuit is commonly used to convert AC power to DC power. The full-wave bridge rectifier, made up of four diodes, allows for the conversion of both halves of the AC cycle, resulting in a smoother output compared to a half-wave rectifier. The filter capacitor helps to smooth out any remaining ripple in the output, and the regulator ensures a constant output voltage.

Now, let's address the three calculations you are having trouble with:

1. Vc1-dc: This is the DC voltage across the filter capacitor, which can be calculated using the formula Vc1-dc = V2P - VD - VD, where V2P is the peak voltage of the secondary AC supply, and VD is the voltage drop across each diode. In this case, VD is given as 0.7V, so you can substitute that in the formula to get Vc1-dc = V2P - 1.4V.

2. IZ: This is the current through the zener diode, which is used as a voltage regulator. It can be calculated using Ohm's Law, I = V/R, where V is the voltage across the zener diode and R is the resistance of the zener diode. In this case, V is given as 10V and R is given as 100 ohms, so you can substitute those values in the formula to get IZ = 10V/100 ohms = 0.1A.

3. VRL p-p (Ripple): This is the peak-to-peak voltage across the load resistor, which can be calculated using the formula VRL p-p = IDC/(2f*C), where IDC is the DC current through the load resistor, f is the frequency of the AC supply, and C is the value of the filter capacitor. In this case, IDC is given as 0.1A, f is given as 60Hz, and C is given as 100uF (0.0001F), so you can substitute those values in the formula to get VRL p-p = 0.1A/(2*60Hz*0.0001F) = 0.083V.

I hope this helps you understand the calculations better.