I have a converter that outputs 5v and 1 amp. The input voltage can be 2.5-6 volts. Does it matter as long as it is in the range? Also, will the current draw from the batteries differ if I change the input voltage? Here is the link to the usb converter: http://www.ebay.com/itm/261097668334?euid=cbb7b0f1c6c84804827dc94d84ad2d39&cp=1.

  • \$\begingroup\$ Please post a link to the converters datasheet or specification. I don't think I understand the comment "will the current draw from the batteries differ if I change the voltage?"; does it have a fixed output voltage of 5V, or not? \$\endgroup\$ – gbulmer May 7 '16 at 18:51
  • \$\begingroup\$ @gbulmer Good question, in my answer below I assumed the OP was talking about runnning at different INPUT voltages but it's not 100% clear. \$\endgroup\$ – John D May 7 '16 at 18:54
  • \$\begingroup\$ @JohnD - I think the OPs comment after your answer shows that it is the input voltage, and the link is to a fixed 5V output DC-DC converter. Nick Solonko, it would help us if you would clarify the question, please change "will the current draw from the batteries differ if I change the voltage?" to "will the current draw from the batteries differ if I change the input voltage?", I assume that is what you mean. \$\endgroup\$ – gbulmer May 7 '16 at 19:00
  • \$\begingroup\$ If you use a boost configuration (what you most likely mean by "step-up converter"), and the input goes up to 6 V, the output won't stay at 5 V. You'll need a buck-boost configuration to allow that input voltage range. \$\endgroup\$ – The Photon May 7 '16 at 19:11

If your source (batteries?) can deliver the required power, it doesn't matter what the input voltage is as long as it's in the range. The current draw from the batteries will be higher as the voltage gets lower for a constant output load.

A DC-DC converter with a constant output load is a constant power device. In your case your converter puts out 5W (5V*1A). There are some conversion losses, so the input power will be higher, for example maybe 6W (the output power divided by the efficiency.)

So your input must supply 6 watts. At 6 volts that will be 1A. At 3V that will be 2A, etc.

  • \$\begingroup\$ So if I use 6 volts, the current drawn will be less than an amp? And if I use 3 volts, it will be more? Thank you \$\endgroup\$ – shurup May 7 '16 at 18:52
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    \$\begingroup\$ Well in the example I gave with 6V the input current would be 1A. In your case depending on the efficiency of your converter it would be somewhere around 1A. At 3V it would be somewhere around 2A. \$\endgroup\$ – John D May 7 '16 at 18:55
  • \$\begingroup\$ Although now that I see your link, it's advertised as a boost, not a buck-boost converter meaning that the output may rise above 5V for inputs above 5V. (The eBay listing isn't really clear.) But the constant power operation would hold for Vin<=5V. \$\endgroup\$ – John D May 7 '16 at 18:58
  • \$\begingroup\$ How does the converter output 1 amp? Doesn't the current that is outputted depend on the total resistance of the circuit? Or do all usb charging devices have somewhat the same resistance? \$\endgroup\$ – shurup May 7 '16 at 19:02
  • \$\begingroup\$ The output current is dependent on the load resistance like you said, which is why i said "A DC-DC converter with a constant output load." But whatever it puts out the constant power relationship still holds: Power in = power out/efficiency. \$\endgroup\$ – John D May 7 '16 at 19:09

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