How to Reduce Voltage but not reducing Current???

[jimmy2447]

HELLO GUYSS...
I have a dc supply 5v 600mA and I want to convert it to 2.5V 500mA.
I tried Voltage Divider Circuit but it doesn't work because I think it reduces the Current very much.
Please Help Guyzz..

[Psi]

2.5V voltage regulator

[w3amd]

Hi,

As PSI said voltage regulator. You might be interested in LM317 adjustable voltage regulator and checking out a circuit named emitter follower.

John

[Psi]

I don't think you will be able to get 2.5V from a LM317,  min is 3V.
Will need to be a different regulator

[jimmy2447]

Thankss 4 reply.. Yes I know LM317 can give the output of 2.3 or 3V will be good but does it effect in Current Value???

[Whales]

If you connect a load to a constant voltage source (eg a 5V power supply) then current will flow through the load.  The load can be anything: a resistor, motor, circuit, etc.

The amount of current that flows through depends on the load and the supplied voltage.  If I have a 100 ohm resistor as my load hooked up to a constant 5V power supply, I can calculate the amount of current drawn through the resistor using ohm's law (V=IR).  For nonsimple loads it gets a little messier, but you can measure it instead of calculating it.

Your '5v 600mA' power supply is probably a constant voltage power supply.  The amount of current it provides will change -- 600mA is likely just a maximum value. 

If you have a 2.5V constant-voltage power supply, it will supply any amount of current up to its limit (where it will hopefully shut itself off rather than damage itself).  If your load will only draw 500mA at 2.5V then all you need to do is get/make a constant 2.5V power supply that can supply at least 500mA without blowing up.  On the other hand your load may want to draw more than 500mA when given 2.5V, so you would have to drop the voltage to compensate.

[jimmy2447]

thank u 4 information.. actually i had a trimmer with unknown dc motor and attached with some sort of charging circuit and battery is of 2.4V 600mA where the charger is of '5V 85mA' written on it. So I just removed the circuit and Battery and now I dont know how to provide the motor with that same supply like Battery.. I managed to provide the Motor 2V using Voltage Divider Circuit but the motor was not working.. So now I will try to use LM317 with the adapter have output voltage: 5V 350mA, maybe it works???

[bitshift]

I don't think you will be able to get 2.5V from a LM317,  min is 3V.
Will need to be a different regulator

You can get as low as 1.25V. Reference

[aaronwiz]

Voltage dividers are used for really low current situations.

Either use a linear regulator, such as the LM317, which will reduce the voltage by burning it off as heat. Or if efficiency is important, you can buy a cheap adjustable DC-DC switchmode regulator for like $3 on ebay.

[w3amd]

thank u 4 information.. actually i had a trimmer with unknown dc motor and attached with some sort of charging circuit and battery is of 2.4V 600mA where the charger is of '5V 85mA' written on it. So I just removed the circuit and Battery and now I dont know how to provide the motor with that same supply like Battery.. I managed to provide the Motor 2V using Voltage Divider Circuit but the motor was not working.. So now I will try to use LM317 with the adapter have output voltage: 5V 350mA, maybe it works???

If it's for a motor you might need some protection in case the motor locks. You might get a voltage spike in this case?

Also the charging rate for the battery will likely be different than the current drawn from the motor. You can measure the starting and running current with the battery to know how much it takes.

John

[Joule Thief]

http://www.reuk.co.uk/LM317-Voltage-Calculator.htm

use a 470 ohm resistor for R1 and R2 to regulate to a 2.5 volt output.

[pmbrunelle]

If you try to set an LM317 for 2.5 V output from a 5.0 V input, this is only a 2.5 V difference, so that seems to be borderline for the dropout voltage...

I would try 1N400x diodes in series with the supply. To change the motor speed, change the number of series diodes.

Use another diode for freewheeling.

[mikerj]

If you try to set an LM317 for 2.5 V output from a 5.0 V input, this is only a 2.5 V difference, so that seems to be borderline for the dropout voltage...

LM317T dropout at room temperature and 500mA is around 1.75volts, so provided the 5v supply is regulated it looks like there is sufficient overhead.  Dropout reduces as temperature increases.

I would try 1N400x diodes in series with the supply. To change the motor speed, change the number of series diodes.
Use another diode for freewheeling.

That will give very poor regulation performance, the voltage drop will increase as the current demand increases.

[AlxDroidDev]

Use a MAX603/MAX604

Although they have a preset voltage output of 5V/3.3V, respectively, they feature dual-mode operation, and can be eaily adjusted for any voltage you want, up to the preset voltage.

The circuit and formula are similar to the LM317, but Vset = 1.20V, instead of 1.25V on the LM317.

Dropout voltage is less than than 320mV @ 500mA.  They allow for up to 500mA and the input voltage can be between 2.7V and 11.5V.

[Monkeh]

Use a MAX603/MAX604

Although they have a preset voltage output of 5V/3.3V, respectively, they feature dual-mode operation, and can be eaily adjusted for any voltage you want, up to the preset voltage.

The circuit and formula are similar to the LM317, but Vset = 1.20V, instead of 1.25V on the LM317.

Dropout voltage is less than than 320mV @ 500mA.  They allow for up to 500mA and the input voltage can be between 2.7V and 11.5V.

.. they're $5 a piece!

[Seekonk]

Seems like a waste for just a motor.  Just PWM it to get the speed you want.

[jimmy2447]

Thank you guys for helping me.. i will try LM317 with 470ohm to get 2.5V and  see if motor runs proper or not.. thanks again