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I made a power supply using two 47uf electrolytic (aluminum, I believe) caps and a L7805CV 5v power regulator. I also added a 100ohm resistor and an led to vout. I'm using a battery pack of 4 NiMH AA batteries. The pack reads 5.43 on the multimeter with no load. When I hook up a motor to vout and ground, I get 3.85v instead of the expected 5v.

I'm wired up like so:

  • battery positive -> vin
  • battery negative -> ground
  • vout -> 100ohm resistor -> led -> ground
  • vin -> 47uf cap -> gnd
  • vout -> 47uf cap -> gnd

Here's a pic: http://sphotos.ak.fbcdn.net/hphotos-ak-snc4/hs1169.snc4/154180_170665839624424_100000430264071_471279_1472098_n.jpg">

Any ideas?

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  • \$\begingroup\$ I added the picture to automatically display. \$\endgroup\$
    – Kortuk
    Commented Nov 19, 2010 at 5:04

3 Answers 3

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The L7805C has a dropout voltage of 2v typical.

So with 5.43 volts input you can expect an output of 3.43 volts typical.

(Although dropout is not really spec'd when the output is below regulation voltage.) alt text

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  • \$\begingroup\$ Yeah, this is it. I get 4.95v when I use a 9v alkaline as the power source. Thanks! \$\endgroup\$
    – Matt Williamson
    Commented Nov 19, 2010 at 5:28
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    \$\begingroup\$ You could use a fresh 4-cell NiMH battery pack with a low-dropout (LDO) regulator and get 5 volts out. But note that battery pack voltage will be about 4.5 volts at the end of discharge. \$\endgroup\$
    – markrages
    Commented Nov 19, 2010 at 5:37
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Did you measure the voltage of your battery pack under load?

Battery packs can have a small dip in voltage due to charge level, but often the way a battery fails is best simulated as an ideal voltage source with an increasing resistance in series.

This means they measure almost full charge for most of their life if there is no load. If you then place a load their charge can change significantly. The lithium battery I use at work measures 2.7V under no load but under a 1mA load drops below 1.6V. These batteries are effectively dead, but still show quite a bit of voltage if you do not need current.

Chances are your battery pack is between 4.3 and 4.9V under load. this depends on your regulator.

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  • \$\begingroup\$ I get 5.37v when plugged directly into a motor \$\endgroup\$
    – Matt Williamson
    Commented Nov 19, 2010 at 4:59
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    \$\begingroup\$ hahahahahaha. Okay, point taken. I will leave the answer and write another, this is honestly what it will be in 99% of people that have this problem. \$\endgroup\$
    – Kortuk
    Commented Nov 19, 2010 at 5:00
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This may seem obvious, but did you wire it correctly. linear regulators cause a rather large drop if you mix up ground and power. There are a number of other ways dependent on your package that can cause odd problems.

It just takes mixing up power and ground to cause funny things to happen.

Secondly, try removing the caps to see if it helps. They are required for transients, not for a constant DC(although still suggested). You may have a bad one.

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  • \$\begingroup\$ double checking... \$\endgroup\$
    – Matt Williamson
    Commented Nov 19, 2010 at 5:04
  • \$\begingroup\$ Also check the caps as I just noted. I have to head to sleep, got an early morning, I will check back in the morning. \$\endgroup\$
    – Kortuk
    Commented Nov 19, 2010 at 5:05
  • \$\begingroup\$ Hmmm, I get 4.3v if I reverse vout and vin.... \$\endgroup\$
    – Matt Williamson
    Commented Nov 19, 2010 at 5:10
  • \$\begingroup\$ Thanks a lot for the help. Im pretty sure the issue is the 2v dropout. +1 for hanging in there with me :) \$\endgroup\$
    – Matt Williamson
    Commented Nov 19, 2010 at 5:30
  • \$\begingroup\$ @MattWilliamson, I have been doing this so long that I forget to check the dropout voltage. If it was still a problem today I was going to open the datasheet. Last night I just dropped a couple things that it often is. \$\endgroup\$
    – Kortuk
    Commented Nov 19, 2010 at 14:54

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