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I've got a 1.2 V DC battery operated display LED light (three low wattage bulbs) that seems to drain my rechargeable 1.2 V batteries. I want to replace the batteries with an AC to DC battery replacement source rated at 1.5 V DC.

What I don't know is whether the 1.5 V DC will hurt the 1.2 V DC display. If not, no problem; if it would do harm, how do I reduce 1.5 V DC to 1.2 V DC? A simple inline resistor, perhaps? The current draw is very low as to three LCDs.

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    \$\begingroup\$ No visible LED can run directly on 1.2V, so the device must have a voltage booster in it. So the question is can the booster handle 1.5V? (it probably can). Is the battery removable? What size is it? \$\endgroup\$
    – Bruce Abbott
    Commented Feb 11, 2022 at 4:16

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You can run the LEDs on 1.5 V no issue. I run them on 5 V quite often. what is more important is to make sure the current (amps) is not too high for the LEDs. 20 mA is plenty for a low power LED. To get 20 mA you would want a 75 Ω resistor from Ohms law (V = IR > R = V/I, R = 1.5/0.02 = 75 Ω).

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  • \$\begingroup\$ It seems awfully unlikely that a battery-powered lamp would require a 75 Ω series resistor. After all, even a AAA battery doesn't have 75 Ω of internal resistance, and there's no reason why replacing the battery with a battery eliminator would somehow cause the lamp to start requiring a resistance. \$\endgroup\$
    – Sophie Swett
    Commented Feb 11, 2022 at 4:51
  • \$\begingroup\$ When calculating the resitance required with an LED, you use the voltage across the resistor, not the full supply voltage - for a 1.5 V supply, 1.2 V LED and 20 mA, you want 0.3 V/0.020 A = 15 Ohms. But a 1.2 V LED would be infra-red, not visible, so the poster's lamp must have a voltage boost circuit to drive a visible LED. \$\endgroup\$
    – Peter Bennett
    Commented Feb 11, 2022 at 7:19

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