0
\$\begingroup\$

I'm building a 5V power supply and I'm hoping to be able to use the 9 VDC, 1 A wall wart that I have lying around (it actually puts out 12VDC). I'm using the following circuit and getting a good response on LTSpice, but I'm wondering if I need to add a bridge rectifier at the front end or if that is only necessary when using a transformer?

Power Supply Circuit

\$\endgroup\$
  • 2
    \$\begingroup\$ Is 16 ohm your worst-case load? If so, you might consider adding a ballast resistor. Your regulator may run hot, approaching its thermal limit. LTspice doesn't simulate smoke. \$\endgroup\$ – glen_geek Aug 17 '17 at 18:35
  • \$\begingroup\$ You might be using that LT3045 regulator just as an exercise. The LT3045 is designed for delivering very low noise and a high line rejection, therefore it might be an expensive component. Why not simply use an LM7805 regulator to do this ? That would be the practical choice unless you have reasons to use the LT3045 of course. \$\endgroup\$ – Bimpelrekkie Aug 17 '17 at 19:25
  • \$\begingroup\$ @glen_geek, where would I add the ballast resistor? The system that I'm powering needs a max of 400 mA in. \$\endgroup\$ – Bex.1233 Aug 17 '17 at 19:51
  • 1
    \$\begingroup\$ @Bimpelrekkie, I have to use the LT3045 (or something like it) because I need lower noise than what the 7805 can give me. The system that it's powering is very sensitive. \$\endgroup\$ – Bex.1233 Aug 17 '17 at 20:00
  • \$\begingroup\$ A series ballast would go between V1 and C1. Its value depends on your 9V supply's minimum valley voltage (guessing 8V). This is always less the average output voltage that a voltmeter reads and depends on how much capacitance sits inside that wall-wart! For 0.4A current, the resistor would be (8-5.3)/0.4=6.75 ohms, 1W \$\endgroup\$ – glen_geek Aug 17 '17 at 22:27
2
\$\begingroup\$

You don't need a bridge rectifier unless you want your circuit to work with either polarity on the input.
Depending on how much output current you need you may want to consider a switching regulator rather than a linear one. With 4V or more across your linear regulator and 500mA out you will be dissipating at least 2W in the regulator. If you need the high PSRR and low noise of this LDO, you could pre-regulate with a switcher so you're not dropping as much voltage across it.

The 2W dissipation is do-able, if you manage PCB layout and thermals correctly.

\$\endgroup\$
  • \$\begingroup\$ A bridge adds 1.4V drop to 0.26 internal drop but efficiency is the same. \$\endgroup\$ – Sunnyskyguy EE75 Aug 17 '17 at 18:18
  • \$\begingroup\$ @TonyStewart.EEsince'75 Of course you are right, I was editing my answer to propose a switcher if the LDO dissipation may be a problem and in that case the bridge would decrease the efficiency. However in a pure linear regulator the efficiency is obviously the same with or without the bridge. (Answer edited to remove the error.) \$\endgroup\$ – John D Aug 17 '17 at 18:21
  • \$\begingroup\$ @JohnD, the system that I'm powering has a max input current of 400 mA. Do you mind explaining more about pre-regulating with a switcher? \$\endgroup\$ – Bex.1233 Aug 17 '17 at 19:52
  • \$\begingroup\$ @Bex.1233 JohnD means that you can use a two-stage regulator system where the first stage is a step-down DC-DC converter that lowers the voltage to a level just a little above what is needed for the second stage, a linear regulator, to work effectively. \$\endgroup\$ – Lorenzo Donati Aug 17 '17 at 20:39
  • \$\begingroup\$ Ok thanks, I think I see what you mean. Could I use a SMD heatsink to help deal with the high power? \$\endgroup\$ – Bex.1233 Aug 17 '17 at 20:46

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.