There is a 10 ohm resistor in series before a regulator

This 10 ohm resistor takes up half of the wattage of the regulator doesn't get to hot?

The resistor is rated at 1 watt

Without the 10 ohm resistor before the regulators input pin, the regulator would have the full wattage and heat across the regulator?

When there is a SHORT on the output of the regulators output pin, the current is increased, why does the 10 ohm resistor open?

When the current increases

 

when the current increases very high on the regulators output pin , the 10 ohm resistor takes half of the Heat and wattage? to prevent the regulator from overheating and getting damaged?

The 10 ohm resistor is a voltage divided with the Regulators internally components to ground

A Regulator has a Ground Pin , so the 10 ohm resistor is a voltage divided with the internally components to the regulators ground pin?

 

here is the schematic

 

The resistor does take half the power dissipation (heat), it's more complicated than that.

The current drawn from the regulator output determines the voltage across the resistor and the regulator, thus, the power dissipated by each. Consider the following two examples:

Output current = 200mA
Voltage across the 10 ohm resistor is 2V
Voltage across the regulator is 27-2-15 = 10V
Thus...
Power dissipated by the resistor is 0.4W
Power dissipated by the regulator is 2W

Output current is 500mA
Voltage across resistor is 5V
Voltage across the regulator is 27-5-15 = 7V
Thus
Power dissipated by resistor is 2.5W (your 1W resistor isn't big enough!)
Power dissipated by regulator is 3.5W

 

The current drawn from the regulator output determines the voltage across the resistor and the regulator

I don't understand or picture this

How can the output of the regulator draw current? I can understand if the regulator VCC pin can draw current

But also how does the regulators output current determine the resistors voltage across it?

When there is a SHORT on the output of the regulator, current increases HIGH, so the 10 ohm resistor OPENS

But I thought a regulators input and output pins were isolated , the current on the output pin is HIGH and input pin should be normal

I would think the 10 ohm resistor should be on the output pin of the regulator
not the input pin of the regulator

 

How can the output of the regulator draw current?

It doesn't. It delivers current.

But I thought a regulators input and output pins were isolated , the current on the output pin is HIGH and input pin should be normal

Not for most regulators. Input current would be mostly the same as output current.

 

I don't understand or picture this

How can the output of the regulator draw current? I can understand if the regulator VCC pin can draw current

That's NOT what I said. Whatever you are powering up with the 15V output of the regulator will draw current from the regulator's output pin. In other words, the regulator supplies the current that is demanded by the circuit/load connected to its output.

Since the input and output current of a linear regulator are nearly the same, I simply used the same values when running my examples for you.

 

Not for most regulators. Input current would be mostly the same as output current.

If there is a SHORT on the output pin of the regulator, I would think all the current would get grounded through the internal ground inside the regulator to the ground pin of the regulator

The SHORTED currents path would be entering in the output pin? going internally to the ground pin of the regulator

 

Since the input and output current of a linear regulator are nearly the same,

So they the input and output are not isolated in linear regulators?

Are switching regulators input and outputs isolated?

 

The current flows thru the regulator - just like the resistor. The ground on the regulator is only there so it can measure the output voltage (to ground) and try to keep it constant.
You don't say what kind of regulator you have so we can't tell you how high the short circuit current will be without the resistor, but with the resistor it will be less than 2.7 amps.

 

79 series for the positive regulator
78 series for the negative regulator

but with the resistor it will be less than 2.7 amps.

So the 10 ohm resistor is used as a voltage divided with the regulator?

The 10 ohm resistor takes half of the heat, amps, wattage when there is a SHORT on the output of the regulator?

The ground on the regulator is only there so it can measure the output voltage (to ground) and try to keep it constant.

If the regulator is floating without a ground, the output voltage of the regulator will not be constant? it would drift , why?

 

79 series for the positive regulator
78 series for the negative regulator

I think you mean the other way around.

So the 10 ohm resistor is used as a voltage divided with the regulator?

Kind of.

The 10 ohm resistor takes half of the heat, amps, wattage when there is a SHORT on the output of the regulator?

No. If you look at the data sheet you will see the drop out voltage is 2 volts. So the regulator will turn on as hard as it can trying to make the output 15 volts - but alas it can't. so there will only be 2 volts across the regulator and 25 volts across the 10 ohm resistor. So 2.5 amps.

If the regulator is floating without a ground, the output voltage of the regulator will not be constant? it would drift , why?

The same thing happens to it as happens to you when you don't put the ground of your volt meter on ground. It thinks there is no voltage so it turns all the way on.

 

the data sheet you will see the drop out voltage is 2 volts

The drop out voltage on the date sheet, is the voltage across the regulator from input to output? its 2 volts from input to output?

The same thing happens to it as happens to you when you don't put the ground of your volt meter on ground. It thinks there is no voltage so it turns all the way on.

So if the regulator turns all the way on, the output voltage on the regulator will be the same as grounded right? what's the difference

It seems that the regulator is a voltage divided internally, because it changes 27 VDC input to a + 15 VDC output

A regulator is like a variable voltage divider referenced to ground
The Regulator is a variable voltage divider that compares the input to the output voltage

Variable voltage sensing voltage divider

 

If there is a SHORT on the output pin of the regulator, I would think all the current would get grounded through the internal ground inside the regulator to the ground pin of the regulator

The SHORTED currents path would be entering in the output pin? going internally to the ground pin of the regulator

What I told you is correct, no matter what you think happens when the output is shorted.

 

So when the output pin is shorted , the regulator is fully turned On and open

The input and output voltage should be the same with max. current rating

A switching regulator with a shorted output would not be the same, it's input and output voltage will be different because the input and output is isolated?

 

It might be worth getting hold of a few books to help start you off, I find books a great help especially when trying to learn the basic's.
Try some of these

http://www.amazon.co.uk/Magnets-Batteries-Ladybird-Junior-Science/dp/072140118X
http://www.amazon.co.uk/Simple-Elec...1397141082&sr=1-3&keywords=simple+electronics

http://www.amazon.co.uk/Electronics...1397141120&sr=1-1&keywords=simple+electronics

Mods please note I am not responsible for the title of the one above!
http://www.amazon.co.uk/Electronics...397141233&sr=1-16&keywords=simple+electronics

 

So when the output pin is shorted , the regulator is fully turned On and open

The input and output voltage should be the same with max. current rating

That's not really how it works. A voltage regulator would not have the same input and output voltages under any operational conditions, even if the output circuit is shorted. The very definition of a voltage regulator implies different input and output voltages, of at least the minimum dropout voltage. In most cases, if the output circuit is shorted, the voltage regulator would be in current limit mode, and the output voltage would decrease to a value to maintain the limit current.