7805 GND Ref.

INCREASING THE OUTPUT VOLTAGE
The output voltage of a supply can be increased by "jacking up" the voltage produced by the 7805. The way the 7805 works is this: It maintains a voltage of 5v between output and common terminal. If the voltage on the common terminal is increased (jacked up), the output voltage will be 5v higher. The 7805 always maintains 5v between output and common. The circuit below produces an output of 12v.

1 amp 12v Power Supply Circuit

Almost any voltage between 5v and 30v can be obtained by this method. This saves stocking the complete range of regulators.
The output voltage is determined by two resistors in VOLTAGE DIVIDER MODE. Five volts is always present across the 120R resistor and if another resistor is placed in series, it will have a proportional voltage across it. In the circuit above, 7v is developed across the 180R resistor, making a total of 12v on the output.
To increase or decrease the voltage, only one resistor has to be changed in the circuit above. The 120R is retained and the 180R is changed. If it is increased to 220R, the output voltage will be 14v, for a 330R, the output voltage will be 18v. The resistor in the common line can be a potentiometer. This will produce an adjustable output voltage. The dropper resistor for the LED will also have to be increased so that the LED is not over-driven on the higher voltages.
A meter can be placed on the output to monitor the voltage and current taken by the load.
There is only one problem with an adjustable supply. The regulator must be heatsinked so it is capable of dissipating the heat for the worst condition. In addition, the input voltage must be sufficient to cater for the maximum output voltage.

http://www.talkingelectronics.com/pay/BEC-3/Page57.html

Hang on, if this is true .... "It maintains a voltage of 5v between output and common terminal"

"five volts is always present across the 120R resistor and if another resistor is placed in series, it will have a proportional voltage across it. In the circuit above, 7v is developed across the 180R resistor, making a total of 12v on the output. "

if it's always 5v between output and common..

How the hell is 7v developed across the 180R resistor????

http://gecieeeunit.x10.mx/project/Variable%20Power%20Supply%20Using%20A%20Fixed%20Voltage%20Regulator%20IC.pdf

Much better, R1 - 470, R2 - 510ohms = 2.602v + 5v = 7.6v! Eureka!

...... Is the first "reference" i found "wrong" or am i just missunderstanding?

if it's always 5v between output and common..

How the hell is 7v developed across the 180R resistor????

Because it is a potential divider.
5V will cause a current to flow through the 120R and that will flow to ground through the 180R, nothing wrong with the explanation as I see it.

For the chip to produce 5 volts across the 120ohm (load) resistor it has to output 5 volts between its output terminal and its common terminal. But the common terminal is NOT the negative supply line; it's a point floating somewhere determined by the potential divider created by the 120ohm and the 180ohm resistors. So a voltage (X volts) must be developed between the 0v supply line and the common terminal of the two resistors. To simplify the explanation ignore any current demand placed on the series circuit by the regulator chip. With 5 volts across the 120ohm the current flowing through the chain must be 5/120 amps. Since this current flows through both resistors in series, the voltage across the 180ohm resistor must be X = 180 x (5/120) = 7.5volts

7.5 + 5 = 12.5 Hence 12.5 volts will appear at the output terminal of the regulator relative to the 0v supply rail.

Or in other words the 7805 with a 120ohm connected between its legs (output and gnd) works as a current source, where the current is

I = 5V/120ohm + I_7805, where I_7805 is about 2-3mA (that is the 7805's internal current, and may vary).

That is 43.6mA. This current flows through the 180ohm down to gnd and creates a voltage on the 180ohm resistor:

V=43.6mA*180ohm=7.86V

This voltage adds to 5Volt = 12.86V against the ground at the 7805 output pin.

The issue with 780x in this wiring is the I_7805 current which is quite high. Therefore you may rather use an LM317, which works in the very same manner, except its I_317 current is 50 microampers, and the voltage between its output and gnd legs is 1.25V.

I'm having issues with "out" is it ALWAYS 5v? when referenced to common? or does this value change as well? (can't get 12v out without the voltage divider!)...

5v (out?)
|
|
470ohms
|
| ---- 2.6v
|
510ohms
|
|
Common (GND)

5 + 2.6v = 12 volts?

with a 12v resistor divider

12v (out?)
|
|
470ohms
|
| ---- 6.2v
|
510ohms
|
|
Common (GND)

5 + 6.2 = 11.2

i'm sorry for being so dense right now, either i'm over looking something very basic or this does not act like a voltage divider?

I Created this "12v" output circuit... and the simulated results show 10v not 12v!

(4 new replies have been posted!, i'll read them first)

7805 ALWAYS keeps 5V between its output leg and gnd leg (mind the "leg" or "pin"). That is because it is a "fixed 5V voltage regulator" thus it does its best to keep 5 V between its output leg and its gnd leg.

Your simulation - the 7805 model is imperfect - so just try to increase the battery to 35V for example and you get better results. My simulation shows the 7805 internal current is 5.27mA, therefore the output voltage is higher. But it copies what we had written before:

I=5/120+5.27mA = 46.9mA
V(180)=46.9mA*180=8.43V
Vout=5+8.43=13.43V

Again - you must consider the internal current of the 7805 as well (ie 4mA) to get precise results..

7805stab.jpg

pito:
Your simulation - the 7805 model is imperfect - so just try to increase the battery to 35V for example and you get better results. My simulation shows the 7805 internal current is 5.27mA, therefore the output voltage is higher. But it copies what we had written before:

I=5/120+5.27mA = 46.9mA

V(180)=46.9mA*180=8.43V
Vout=5+8.43=13.43V



Again - you must consider the internal current of the 7805 as well (ie 4mA) to get precise results..

I'm with it now, cheers.

7805 needs at least 2.5V higher input voltage than its output. So with 12V input voltage you cannot get 12V output in that wiring. You need at least 14.5V input voltage in your wiring to make the 7805 happy.. :slight_smile: