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I got a full drawer of second hand wall warts which I'm planning to use for powering arduino projects (mostly 5V "pro mini" or "nano").

I'm using a voltmeter to check that they are still working, however most of them seem to output a far greater voltage than what is printed on the case.

For example the Arduino pro mini accept up to 12V so I've been looking at 9V wall warts, but they often give more than 12V.

Example (but typical reading):

  • written specification: 9 V 300 mA
  • Measured voltage: 12.37 V
  • Measured voltage while powering a led (through a 300 ohms resistor): 11.2 V

I tried to add a led because I've read that measuring a voltage with no load may give a wrong result, but perhaps that's not enough to get a proper reading...

My questions are:

  • Is there a way to measure the correct voltage that will be issued in "normal usage conditions"?
  • Can I safely use a "9V wall wart" to power a "12V max" device if I measure more than 12V?

PS: So far I've use 6V wall warts which are giving about 10V, but I don't have any left so I'm trying to use 9V ones

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marked as duplicate by Ricardo, Daniel Grillo, PeterJ, W5VO Dec 23 '15 at 13:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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It sounds as though you've got some unregulated power supplies. Generally these have a transformer, rectifier and smoothing capacitor. With no load the capacitor will charge up to peak voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

There is always some internal resistance in the transformer, represented by R1, which will cause a voltage drop which increases with current drawn. In addition, C1, which maintains output voltage during troughs in the rectified AC (i.e., between pulses from the rectifier) will droop between cycles causing the average DC value to droop with increasing current.

To test you should load them with something simulating the intended load. e.g., for 100 mA load on a 9 V PSU you would use a 9 / 0.1 = 90 Ω test load. Nearest standard values are 82 Ω or 100 Ω. Connect the test load across the terminals / connector and measure the voltage.

An alternative is a 12 V, 6 W car tail-light lamp. At 12 V this should pull about 0.5 A. Be aware that the resistance value will change dramatically with temperature which is voltage dependent. To get a good feel for what's happening you would need to measure both the voltage and the current.

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  • \$\begingroup\$ Thanks for this complete answer! I didn't realize I should draw the total rated current to get the expected voltage. It does make sense though. \$\endgroup\$ – LeFauve Jan 9 '16 at 11:37
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If they're transformer supplies (rather than switched mode) then yes, you're going to get higher than the stated voltage.

This is likely to continue until you draw the maximum stated current, or within a certain percentage of it.

Your best option is to use a pre-regulator. With something small like a Nano or Pro Mini you are better off dropping as much voltage as you can externally to save heat being dissipated on the Arduino board. A 7805 is beefy enough to handle the currents typically required by an Arduino without a heatsink, even with a 12V input.

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  • \$\begingroup\$ Thanks! I will probably add a 7805 since this is for a 24/7 application and I wouldn't want to fry the Arduino. \$\endgroup\$ – LeFauve Jan 9 '16 at 11:39
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It's normal for unregulated wall-warts to show higher voltage than the nominal one when tested without (or with small) loads. It will show voltages closer to nominal as you increase the load, reaching nominal voltage when the current approaches the specified limit.

In your example, the LED isn't enough of a load to bring the voltage to nominal value, but it already shows that the wall-wart voltage decreases with higher loads. You'll probably measure 9V if you apply a load that draws close to 300mA.

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