Power Supply Project

arn't students allowed to use real parts anymore?

using 120v in, a bridge circuit

it is accepted usage that unregulated is rectified and filtered. (but no active elements)

I thought I

right

is the transformer center tapped?

as no ripple is speced it appears not to be relevent to the project.

use ohms law

I feel lost on this project

hint: A REAL transformer has an internal resistance. as you increase the current load the terminal voltage drops. without knowing the R (int) in advance you will have to perform measurements before finalizing the design.

hint: at 2mA a resistor and a Zener will do fine for a +5 supply.

hint: calculate power dissipation in each resistor so you can use sufficient wattage parts.

advise: power supply fundamentals will be important as you progress in electronics. learn it well.

chuckle

When all else works, why not try a few values of C at different loads and find one that gives low ripple. It's the least you can do. Might be worth marks.

Drink some beer, this will help.

Good luck.

OK, I understand my project better, I emailed the teacher. I obviously need a bridge circuit and coming off of that (let's just use the positive side for now) will have a Reistor to ground and a Capacitor to ground which will do the filering. then the output of that will go into a regulator. There is a spec for AC ripple which is less than 10mV, I believe this is the input side of the regulator.

My original thought was that I needed some value(s) of resistors to make the bridge output 15volts, but that's not the case. I guess I was thinking too much like a voltage divider circuit rather than designing a power supply. So I want to make the transformer so the bridge output is around 18 volts with low ripple, then use that into a regulator to produce 15volts out.

Now how do I calculate the resistor that I should use? The regulator output will handle up to 1amp, so do I base the input on around 0.5amps (which is the design I need)?

Here is my guess on the values:

I've been told to use an RC of 5 time constants, so that would make the RC needing to be equal to 41.67msec. So my C value would be:

Now plug those two values in and bang my output should be around 18volts with very little ripple????? Then the negative is easy, that's just the same values as the positive and the regulator is a different number. The 5volt ref. I think I can handle as well.

Thanks

sigh... i feel old

I obviously need

why? resistor to ground? you mean total resistance to ground?

and a Capacitor to ground which will

you need to find out for sure.

just a guess here but since you haven't covered regulators in class yet could it be that that the project involves a simple pi filter? you did originally spec it as an unregulated output.

hint: remember that diodes drop .6 volts

where did the 18 volts come from? what is your AC secondary voltage? its it no load or full load?

I think I'm all set.

I used 18 volts coming off the bridge circuit because I needed 18 to eventually go into a regulator.

The teacher didn't care if we used a regulator or not, he just wanted us to get around 18 volts. So I used a bridge circuit, a resistor and capacitor to ground. The resistor was 40ohms and the cap that I calculated (based on 5 time constants) was about 1100uf, I used a 1500uf.

My ripple wasn't bad at all and I put that into a regulator.

Thanks for everyone's help. I think I understand power supplies a little better now! phew, it seemed easy, but yet it was pretty hard!

Thanks again everyone!!!!