I have a transformer labeled as providing 9Vac, 3.36 A. It appears to provide 9.4 v ac according to my multimeter.

It runs through a diode bridge (WAB 040 diodes -- already fried the last version), and at the output, I'm getting 11.3 v (which is consistent with 9.4v AC^2 - 2 diodes).

I'd like to bring this down under 9.5 v. Can I just attach two or three more diodes until the voltage is sufficiently low?

Obviously, each one is going to waste energy, but I'm not particularly concerned.

  • \$\begingroup\$ have you measured the voltage when the power supply is running at your full, expected load? \$\endgroup\$ Feb 19, 2013 at 1:28
  • 1
    \$\begingroup\$ I feel that although you want to use diodes, a regulator really is what's called for here. \$\endgroup\$ Feb 19, 2013 at 1:38
  • \$\begingroup\$ #1: I was using a motor, not a silicon chip, so I didn't need great power filtering. #2: Each diode is <Vf1 * 3.3 amps> -- 2-3 watts. These diodes are reasonably big, but they get VERY HOT. \$\endgroup\$
    – gbronner
    Feb 19, 2013 at 5:01
  • \$\begingroup\$ ...and now you find you are concerned with the waste energy--it's heat! (But a linear regulator would get just as hot). \$\endgroup\$
    – gbarry
    Feb 19, 2013 at 6:29

1 Answer 1


If your idea is to bring the output voltage down under 9.5V you can use extra diodes in series.

But the voltage regulation would be really bad even if you use large capacitors. Since the forward bias voltage of diode is a function of forward current, then the effect gets multiplied n times the number of diodes you have added. Not recommended to power semiconductor chips, Please use a proper regulator instead.

Vt    Vf1  Vf2   Vf3 
o----|>|---|>|---|>|---| Vo
                       \ RL
GND        If          |

Vo = Vt - ( Vf1 + Vf2 + Vf3 );

  Vt is the voltage at positive terminal coming out of the birdge rectifier.
  Vf1, Vf2 and Vf3 are the forward biased voltage of Diodes.
  RL is the load resistance.

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