Hello,

Can someone please tell me how to calculate the ratio of the output ripple voltage to the input ripple voltage of a simple Zener regulator circuit when:

V(in)=25.5V, Series Resistor Rs=500Ω, Zener Voltage Vz=11.8V, Zener Resistance Rz=8Ω, Voltage Load VL=11.86V and Load Current IL=19.77mA.

I think that I may be missing something but I am not sure what so any help will be fully appreciated.

Kind Regards,

Sparkyian.

 

The input voltage Vin equals a DC voltage Vdc plus a ripple voltage Vr

thus Vin=Vdc+Vr

The output voltage Vo equals a DC voltage Vodc plus a ripple voltage Vor

thus Vo=Vodc+Vor

The equivalent circuit of a zener diode is a battery in series with its internal resistance Rz.

Now, if you reduce all the DC voltages (reduce the battery which represents the zener voltage too) to zero you will be left with only the ripple voltage of Vin and Vo which are Vr and Vor respectively. Thus, you will have a circuit with only an AC input voltage source, which represents Vr and a voltage divider network connected to it composed of Rs in series with Rz in parallel with RL:

Rs+Rz//RL

Thus the output ripple voltage Vor is the voltage across the Rz//RL parallel combination and its value equals

Vor=[Vr*(Rz//RL)]/(Rs+Rz//RL)

thus the output to input ripple voltage ratio is Vor/Vr=(Rz//RL)/(Rs+Rz//RL)

 

Thank you mik3, you have been a big help and I can now see where I was going wrong.