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[SOLVED] Drop voltage in output of LM317

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H2M

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Hi
I want to made a DC motor driver. the supply of my Driver is 100volt. I use the LM317 to supply my singnal ICs. I have two IC one for generating PWM singal(ne555) and second is a gate dirver to drive n-channel mosfet.
I connect a 1kohm/5w resistor form supply voltage to input pin of LM317 and generate a 14volt output. when i supply NE555 i can generate PWM sinal for gate driver IC, but when i supply gate driver IC with 14volt supply the amplitude of PWM singnal drop. how should i solve this problem?
 

A schematic would help but if I understand what you are doing, the 100V is dropped by the resistor before it reaches the LM317 and when you increase the load on it, the voltage drops.

This suggests the extra current through the resistor is dropping enough voltage that the LM317 no longer has enough at it's input pin to maintain it's 14V output.

Firstly, please be aware that although a resistor dropper will work in that circuit, if the load is removed or becomes too small, the voltage dropped across the resistor will be less and you risk the LM317 maximum input voltage being exceeded.

With that kind of circuit, the only way to allow more load is to drop the resistor value. A better method would be to use a small switch mode supply that can run on a high voltage input but still regulate 14V out, that will allow the load variation without the voltage changing but the circuitry would be a little more complicated.


Brian.
 

I'm with Brian on this one.
Your proposal is fraught with risk, and a perfect opportunity to validate Murphy's Law.

You don't mention the total load current, but dropping 86 volts will be a ton of heat you must get rid of. Heat is the #1 enemy of electronics.

Linear Tech makes some high voltage buck regulators. Search their website.
 

Agree with the above. Drop the voltage 'properly' then feed an appropriate solid state device to control the motor
 

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