Voltage Regulator

Sorta, maybe, but I wouldn't count on it. According to the datasheet it takes at least 14.6V for regulation and then the unit will give you a linear in to out V once you drop under that 14.6V.

See :

They show the drop out data for the 7805 only, but I'm sure the 7812 acts similar.

Jim

Thanks for the data sheet, but I still can't find the answer. Maybe it's there and I just don't see it.

Here is what I am seeking:

15v in = 12v out 14v in = ??v out 13v in = ??v out 12v in = ??v out 11v in = ??v out

If I had a variable power supply, I could find out on my own, but I don't have one available.

You need more than that, the 7812 series require around 1.5 volts on the input side above the mark to provide a voltage of spec. anything below that most likely will be 1.5 volts less than the input. This would mean that if you have 11 volts input, your output most likely will be more more than 10 volts or less. What you need is a LDO type (Low drop out).. Something like a LM2940CT-12 LM2990T-12

That's just the tip of the ice burg.

--
"I\'m never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

Linear voltage regulators have a characteristic known as the "drop out voltage". It'll be on the data sheet somewhere, trust me.

For Vin > 12V + Vdrop, you'll get 12V out. For Vin

** The output will be 9.5 volts ( ie 11 - 1.5 ) . ** Yep - but with 1.5 volts taken off the top by itself.

So, with 1.5 going in, the output will finally be zero.

....... Phil

Depends on the current. Assuming a light load, 100 mA maybe, expect roughly...

In other words, the output is about 1.5 volts below the input, but doesn't go above 12.

The lost 1.5 volts is the "dropout voltage" for this regulator. At higher currents, the dropout voltage goes up a bit, 2 volts maybe at 1 amp.

"LDO" regulators have lower dropout voltages than the ancient 7812.

John

The L7812 uses a Darlington NPN as its pass device this accounts for some of the voltage drop and it will vary depending on temp, loading and manufacturing tolerances. A good experiment to see this is get yourself a power NPN in a TO-3 package and a thermal couple and vary Ic and measure Vbe note Tc .Tc is the case temperature of the TO-3 NPN.

The junction temperature can be calculated by Tj = Tc + (theta (jc) * Pd)

Tc is the case temp.

Theta jc junction to case thermal resistance see the data sheet.

Pd is Vce*Ic

Vbe will increase under load

Don't exceed Tjmax or you will fry the transistor.

Heres a data sheet on a NPN TO-92 Darlington showing Vbe in relation to Ic and temp.

The 7800 series also has thermal protection. R11 in the schematic which is a TDR temperature dependent resistor. When VR11 hits about

0.7V q15 turns on cutting drive to the pass device. So r11 also contributes to the dropout voltage.

http://www.ortodoxism.ro/datasheets2/8/0ishsf7y9sp31h690e60g8gclc3y.pdf

ldo's use typically p-channel mosfets to obtain drop out voltages in the mv range.The dropuot voltage is determined by Rdson of the FET.

To get a higher output voltage then input voltage you will need either a SMPS or charge pump.

"Hammy"

** PISS OFF YOU FUCKING WANKER !!

..... Phil

"Hammy"

** PISS THE FUCK OFF

YOU VILE AUTISTIC PIG WANKER !!

....... Phil

I guess I need another way to regulate the input voltage. Would a circuit with a zener diode (maybe 1N4742) do the trick?

Depends on the numbers... load current range, expected input range. But zener regulators are inefficient and can get very nasty.

What's your specific problem?

John

I've also seen a reverse-biased diode from output to input, in case the input drops abruptly, and some capacitance on the load side wants to discharge itself into the supply - the diode protects the regulator.

Cheers! Rich