1
\$\begingroup\$

I've been researching power supply design techniques and information for the last few days and feel like I have a good understanding of how this is to work, but I can't seem to get my simulation to fruition.

The PS needs to have maximum of 5% error and .5% ripple. Current output isn't an issue, just something usable. SO basically the idea is I need something very smooth at 10V with basically no ripple. The other trick is that I can't use a solid state regulator, I have to design my own circuit.

So What I know is that I need a transformer to step down the ac voltage. Which then goes through a rectifier to convert to DC which steps down ~1.4 V if I use a full wave bridge rectifier (which I understand is the best for noise reduction)? Then I need a filter capacitor, which goes into a regulator circuit.

I have a design that simulates correctly on SPICE with a 12v rms transformer, but I don't actually have one. I think this would be cooler if I could use my 5V rms cell phone charger to get the 10V DC final result. Is this possible in a simple sense? This project is supposed to be fairly simple and I have most of the quirks figured out, I just can't produce the voltage I want with the 5v transformer.

BTW the regulator circuit I created is a shunt regulator just using a reference zener and some resistors to create feedback. I will post my circuit, any help is greatly appreciated!

Also the circuit is a complete rough draft which works for a 17v output from a transformer, but I would like to do it with a 7v if possible. Really Im just going for simplicity but need to meet specs.!

Power Supply

\$\endgroup\$
  • \$\begingroup\$ You need to accept that any linear regulator will have LESS voltage at the output than at the input. If you want to build a switched mode power supply from 5V to 10V it is possible but requires an occilator with a tapped inductor at least, you can get an idea at the link electroschematics.com/7032/12v-to-120v-voltage-inverter on what is about the minimum to get a higher output voltage, you will still need your regulator circuit as well. \$\endgroup\$ – KalleMP Jul 14 '15 at 16:44
1
\$\begingroup\$

No, you don't have a shunt regulator. Q1 is the main pass element. It is working like a linear regulator.

Q2 is the wrong way around. You have C and E flipped. It will still probably have some gain like that, but performance will be better if you flip it the right way around. The concept is that when the output of the R3-R5 voltage divider gets high enough to be the zener voltage plus the B-E drop of Q2, Q2 will turn on, which will steal current from the base of Q1, thereby turning it off more.

I have no idea what you think R2 is doing. It makes no sense.

\$\endgroup\$
  • \$\begingroup\$ That makes sense about Q2, I will switch it, but do you have any suggestions for me or do you get more pleasure just tearing my ideas down? \$\endgroup\$ – StefKat Jan 23 '13 at 0:43
  • 1
    \$\begingroup\$ @StefKat - I think you got some good feedback. When you put out a design and ask for feedback about it you really need to be prepared for and accepting of any review that you get. \$\endgroup\$ – Michael Karas Jan 23 '13 at 1:30
  • \$\begingroup\$ Alright that's true. My apologies Olin. I guess I was looking for something to steer me in the right direction because my knowledge of this topic is very limited. \$\endgroup\$ – StefKat Jan 23 '13 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.