How to calculate Load Regulation

Thread Starter

jegues

Joined Sep 13, 2010
733
How would I go about calculating the Load Regulation in this circuit? Note that we don't know the value of RL.

After doing some wiki'ing, I have to find the Voltage(Full Load), where IL = 20mA, and Voltage(Min. Load), where IL = 0mA.

I have done this correctly?

Also if I calculate a load regulation of 84% does that mean that the output changes by 84% or that it regulates 84% of the output?

See figure attached for my attempt and question statement.

EDIT: Also, for part e), how do I calculate the maxmimum power dissipation if I've solved for the maximum current?
 

Attachments

Last edited:

Thread Starter

jegues

Joined Sep 13, 2010
733
I'm pretty what I've got in my OP is incorrect.

Here's my 2nd shot at it. Am I on the right track?

What would IL be for no load? Is it modeled as a short circuit or an open circuit? If I know this I can find IL.

Also, what do they mean by full load condition?

Thanks again!
 

Attachments

Jony130

Joined Feb 17, 2009
5,599
What would IL be for no load?
No load = 0A for lad current.
All current provides provided by R flow through Zener diode
Also, what do they mean by full load condition?
load current is equal 20mA.

R = (15V - 9.825V)/ ( 25mA ) - 7Ω = 200Ω

ΔVo = rz/(rz+R) * ΔVs = ± 169mV

Output impedance is equal

Rout = R||rz = 6.7632Ω ≈ 7Ω

Load regulation
ΔVo = Rout*ΔIL = -135mV
---> ΔVo/ΔIL = - 6.7632mV per 1mA of load current.

Max diode Zener current
Iz_max = (25V - 9.825V) / (200Ω + 7Ω) = 73.3mA
 

Thread Starter

jegues

Joined Sep 13, 2010
733
No load = 0A for lad current.
All current provides provided by R flow through Zener diode

load current is equal 20mA.

R = (15V - 9.825V)/ ( 25mA ) - 7Ω = 200Ω

ΔVo = rz/(rz+R) * ΔVs = ± 169mV

Output impedance is equal

Rout = R||rz = 6.7632Ω ≈ 7Ω

Load regulation
ΔVo = Rout*ΔIL = -135mV
---> ΔVo/ΔIL = - 6.7632mV per 1mA of load current.

Max diode Zener current
Iz_max = (25V - 9.825V) / (200Ω + 7Ω) = 73.3mA
Thank you this is much more clear, however I don't think we are expected to recalculate the value of R each time, we just kept it at 205ohms as assumed.
 
Top