I need 6volts to supply about 0.700 watts.

My supply is 11.4 to 14.2 volts, I need to drive a small DC motor that operates on 6 volts and draws about 0.112 amps. Unless I am forgetting something, a voltage divider won't provide enough amps to drive the motor unless I use some 2 or 3 watt resistors, which I would rather not do.

Why is there not a single unit such as the 5volt 7805?
Seems that converting 12volts to 6 volts would be just as easy as going to 5 volts.

So whats the easiest way to get 6volts?

 

Hello,

There is also an 7806 available.
Otherwise you can use a LM317 with some resistors to adjust it.
See attached drawing how to calculate the resistors.

Greetings,
Bertus

PS this is my post 500

 

Thanks! I don't know how I over looked the 7806, all my searching I never came across that one. That should suit me just fine.

 

The lower voltage 78xx series are more versatile; you can add resistance between the GROUND terminal and ground to raise the output voltage.

Simply measure the current from the GROUND terminal to ground. Then, calculate the resistor you'll need.
Example:
You have a 7805 that you want to get 6v out of.
You measure 5.5mA current from the GROUND terminal to ground.
6V-5V=1 volt to add
Since R=E/I, and you want to add 1 volt, 1/5.5mA = 1/0.0055 = 181.818... Ohms.
180 Ohms is the closest standard value.
Since E = I x R, and I = 0.0055 and R = 180, E = 0.99V. That's close enough for me. If you wanted closer, you could use a 500 Ohm potentiometer to "tweak it in".

 

The lower voltage 78xx series are more versatile; you can add resistance between the GROUND terminal and ground to raise the output voltage.

Simply measure the current from the GROUND terminal to ground. Then, calculate the resistor you'll need.
Example:
You have a 7805 that you want to get 6v out of.
You measure 5.5mA current from the GROUND terminal to ground.
6V-5V=1 volt to add
Since R=E/I, and you want to add 1 volt, 1/5.5mA = 1/0.0055 = 181.818... Ohms.
180 Ohms is the closest standard value.
Since E = I x R, and I = 0.0055 and R = 180, E = 0.99V. That's close enough for me. If you wanted closer, you could use a 500 Ohm potentiometer to "tweak it in".

Thanks for the tip, Never thought about using a pull up on the 7805.
I'm an electrical engineer, mainly 240/120 AC and 24/12 DC, I just dabble in the electronics side.

I'm curios how that works though, I assume the output is in reference to ground so the 1 volt across the resistor adds to the 5v output? That in turn then lowers the output current capacity of the 7805?

 

Thanks for the tip, Never thought about using a pull up on the 7805.

Ahh, I wouldn't exactly call it a pull-up resistor.

I'm an electrical engineer, mainly 240/120 AC and 24/12 DC, I just dabble in the electronics side.

I'm curios how that works though, I assume the output is in reference to ground so the 1 volt across the resistor adds to the 5v output? That in turn then lowers the output current capacity of the 7805?

If you look at the schematics of the 78xx series of regulators, you'll see that internally there's a pair of resistors near the output. Those internal resistors perform the same function as the external resistors do in an LM117/LM317 regulator. Adding resistance between the GROUND terminal and ground basically increases the value of that lower resistor.

It doesn't change the output current capacity; just the output voltage. The current from the GROUND terminal will be constant, thus you will get a constant voltage across a constant resistance.

The current from the GROUND pin will vary somewhat from regulator to regulator, due to how it's trimmed internally at the factory.

If you wanted, you could use forward-biased 1N4148 or 1N914 or 1N400x diodes from the GROUND pin to ground to boost the output voltage; each diode would add roughly 0.65V to 0.7V. Just another way to skin the proverbial cat

 

Cool, thanks.