Circuit for reducing DC voltage

Thread Starter

gfvesley

Joined Nov 4, 2017
18
Looking for a way to match speed of HO engines run on DC not DCC. DC v of 15-18 v. I have heard of a diode pair to reduce voltage. I hear resistors are not a good choice. Does anyone have a circuit for a diode pair and type of diode? Obviously I am not an electrical engineer.
 

Reloadron

Joined Jan 15, 2015
5,667
What voltage do you want to go to and from? I want to recall HO as between 0 and about 16 or 17 VDC. Diodes will drop about 0.75 Volt per diode in series with the power. The diode need only have a high enough voltage and current rating. For what you want likely a common 1N4002 would be more than adequate, good to about an amp forward current. Resistive voltage dividers are a poor choice for a number of reasons.

Ron
 

Thread Starter

gfvesley

Joined Nov 4, 2017
18
What voltage do you want to go to and from? I want to recall HO as between 0 and about 16 or 17 VDC. Diodes will drop about 0.75 Volt per diode in series with the power. The diode need only have a high enough voltage and current rating. For what you want likely a common 1N4002 would be more than adequate, good to about an amp forward current. Resistive voltage dividers are a poor choice for a number of reasons.

Ron
Thanks Ron. Voltage 0-12. About 0.5 amp. 0.75 v drop/pair is sufficient. Do I need a resistor?
What voltage do you want to go to and from? I want to recall HO as between 0 and about 16 or 17 VDC. Diodes will drop about 0.75 Volt per diode in series with the power. The diode need only have a high enough voltage and current rating. For what you want likely a common 1N4002 would be more than adequate, good to about an amp forward current. Resistive voltage dividers are a poor choice for a number of reasons.

Ron
 

Reloadron

Joined Jan 15, 2015
5,667
All you need is a diode like a 1N4002 (any 1N4000 series) inline with the power supply. The diode will have a band on one end, that is the cathode side and goes away from your power supply. The diode will drop about 0.7 Volt. Another diode in series with the first will drop an additional 0.7 volt. You do not need or want any resistor(s) in there.

Ron
 

-live wire-

Joined Dec 22, 2017
912
You might want to consider a Buck converter, DIY or commercial. You would need to generate a PWM with a high frequency, and add an inductor with a flyback diode, and some other components to regulate the output if necessary to maximize efficiency and consistency.
 

AnalogKid

Joined Aug 1, 2013
8,706
Voltage 0-12. About 0.5 amp.
What I see so far is this:

Input Voltage: +15 V to +18 V, unregulated
Output Voltage: 0 V to +12 V, regulated, continuously adjustable
Output Current: 0.5 A

If this is correct, then an adjustable linear regulator module on ebay will do what you want. These often are based on the LM317 three-terminal regulator IC from national Semiconductor and others. Note that you will need a heatsink on the IC to keep it from overheating. Note o the LM317 basic circuit adjusts down only to about 1.25 V, not 0 V.

This is the most simple approach both to build and to understand, but the heat could be a problem. If the load draws 0.5 A when the output is adjusted down to 3 V, then 12 W is dissipated in the regulator IC. It is built to do this, but you will have to add a heatsink that might be inconveniently large, something in the 6 to 10 cubic inches range. A switching regulator, in your case a non-isolated buck regulator, is far more efficient. But the output adjustment range probably will be much smaller. Again, ebay.

ak
 
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