Hi guys. Im trying to design a simple power supply using a half wave rectifier and a voltage regulator. The aim is to produce a output dc voltage of 2.8 V and maximum current of 430mA. I got the voltage but the current never reaches 430mA. if it did the voltage will be higher than 2.8 V. ive attached my circuit to see if i made any mistakes. the power supply is low AC with 50 Hz so i take it to be 5. The D2 diode has a breakdown voltage of 2.8 so it's a must. the values of other components are just my guessing to produce the required current and voltage. So any advice would be appreciated.

 

Hello,

Take a look at this page from our eBook:
http://www.allaboutcircuits.com/vol_3/chpt_3/11.html

This will explain the voltage drop at higher currents.

Bertus

 

Hi guys. Im trying to design a simple power supply using a half wave rectifier and a voltage regulator. The aim is to produce a output dc voltage of 2.8 V and maximum current of 430mA. I got the voltage but the current never reaches 430mA. if it did the voltage will be higher than 2.8 V. ive attached my circuit to see if i made any mistakes. the power supply is low AC with 50 Hz so i take it to be 5. The D2 diode has a breakdown voltage of 2.8 so it's a must. the values of other components are just my guessing to produce the required current and voltage. So any advice would be appreciated.

Your diagram seems to label D1 and D2 the other way around. As shown, D2 is a 1n4004, which has a 400V breakdown rating. http://www.fairchildsemi.com/ds/1N/1N4004.pdf

D1, on the other hand, does have 2.8V typical breakdown voltage. http://www.fairchildsemi.com/ds/1N/1N5224B.pdf

You cannot get 430mA out of this arrangement with a 500Ω resistor in series with the Zener. You would need a far lower resistor: if the rectifier produces about 6V, and the output is 2.8V, by Ohm's law you would need (6-2.8)/0.43 = 7.4 ohms. (Actually the resistor would need to be a bit less, to give the Zener some current and allow for its voltage rising a bit, but let's not worry about that just now.)

Unfortunately, if you did not connect an external load, the Zener would overheat, because it would dissipate roughly 2.8V*0.43A = 1.2W and the 1N5224 appears to be only a 0.5W rated device.

You could solve this problem by getting a bigger Zener, say 1.5W rated or more, but this is a clumsy way of doing this job. Why not use a voltage regulator - something like this? http://www.national.com/ds/LM/LM117.pdf

 

Thanks for the advice. I ve just started to learn about diodes so i want to keep my design as simple as possible. Also if i connect an external load to the zener diode , will it be able to keep the power dissipation of the diode under 0.5 W?

 

Also if i connect an external load to the zener diode , will it be able to keep the power dissipation of the diode under 0.5 W?

Yes,but the current for the load should be atleast(min) (430mA - 178mA) 252mA or more , the max current for the zener diode is 178mA (2.8V * 0.178A = 0.4984W i.e.. approx 0.5W),if you pass any more current through the diode it will dissipate more then 0.5W.

For example your load may be of 11ohms,so the current for your load will be (2.8V/11ohms = 255mA),so the current for the zener diode will be (430mA - 255mA = 175mA) and the zener will dissipate (2.8V * 0.175A = 0.49W )

And check out the stuff posted by Bertus, it will help you more on zener diodes.

Good Luck