how do I use resistors of certain values to get 19vDC down to 18vDC?

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the device can put out around 2 amps exactly on the mark and no higher.
I need to drop it one volt to run my laptop because the original laptop is gone and I don't want to buy a new charger for it.
the charger I have runs fine and I wish to use it to run my newer laptop. but the laptop requires 18 volts DC and 1.5 amps and since I don't have a charger for it I wish to use the one I already have. if I can use some simple resistors to drop it by one volt that would be great.
 
sofaspud is right, many have 3 conductors so they will not work without the correct brands powersupply. The plug has a central spike and the plus and minus are the inner and outer of the barrel. Dell are infamous for this - the actual charger could be standard but for the lockout from the data line.
dell-laptops-reject-third-party-batteries-and-ac-adapterschargers-hardware-vendor-lock-in/

If this is not one of the trapped brands then a couple of 5Amp diodes in line will drop 1-1.2 v or so.
 
the power cord chargher going into the laptop only has 2 wires.
one negative one positive.
Are you sure? 3 of the 4 laptops around here have 3 conductors. There's probably an embedded controller at each end, brick and battery.
The 4th uses NiCads.
Otherwise, using diodes or an LDO regulator would be the way to go. A resistor's voltage drop will vary with the current through it so makes a poor regulator.
 
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