Voltage Regulator

Guest
Quick question:

Using a 7812 voltage regulator to maintain a constant voltage to a
circuit (@ about 100ma) where the input can be up to 15 volts, what
will happen if the supply voltage drops to 11 volts? Will the
regulator pass the supply voltage through to the circuit being
powered?
 
In article <4ed96930-45d9-48b0-a237-ade4b7722405
@p69g2000hsa.googlegroups.com>, nancydow26@hotmail.com says...
Quick question:

Using a 7812 voltage regulator to maintain a constant voltage to a
circuit (@ about 100ma) where the input can be up to 15 volts, what
will happen if the supply voltage drops to 11 volts? Will the
regulator pass the supply voltage through to the circuit being
powered?


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 :
http://www.tranzistoare.ro/datasheets/150/44435_DS.pdf

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.
 
nancydow26@hotmail.com wrote:
Quick question:

Using a 7812 voltage regulator to maintain a constant voltage to a
circuit (@ about 100ma) where the input can be up to 15 volts, what
will happen if the supply voltage drops to 11 volts? Will the
regulator pass the supply voltage through to the circuit being
powered?

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 <= 12V + Vdrop,
you'll get Vin - Vdrop, more or less (less when Vin < Vdrop!).

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
 
nancydow26@hotmail.com wrote:

Quick question:

Using a 7812 voltage regulator to maintain a constant voltage to a
circuit (@ about 100ma) where the input can be up to 15 volts, what
will happen if the supply voltage drops to 11 volts? Will the
regulator pass the supply voltage through to the circuit being
powered?

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
 
<nancydow26@hotmail.com>
Using a 7812 voltage regulator to maintain a constant voltage to a
circuit (@ about 100ma) where the input can be up to 15 volts, what
will happen if the supply voltage drops to 11 volts?

** The output will be 9.5 volts ( ie 11 - 1.5 ) .

Will the regulator pass the supply voltage through to the circuit being
powered?

** 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
 
On Mon, 19 Nov 2007 14:01:35 -0800 (PST), nancydow26@hotmail.com
wrote:

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

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


Here is what I am seeking:

15v in = 12v out 12
14v in = ??v out 12
13v in = ??v out 11.5
12v in = ??v out 10.5
11v in = ??v out 9.5
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
 
On Mon, 19 Nov 2007 12:43:02 -0800 (PST), nancydow26@hotmail.com
wrote:

Quick question:

Using a 7812 voltage regulator to maintain a constant voltage to a
circuit (@ about 100ma) where the input can be up to 15 volts, what
will happen if the supply voltage drops to 11 volts? Will the
regulator pass the supply voltage through to the circuit being
powered?
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.

http://www.zetex.com/3.0/pdf/ztx604.pdf

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
 
On Wed, 21 Nov 2007 00:44:53 +1100, "Phil Allison"
<philallison@tpg.com.au> wrote:

"Hammy"



** PISS OFF YOU FUCKING WANKER !!






..... Phil


Well if it isn't the dipshit who's afraid of anything over 50Vrms.

I see you're off your meds. I hear medication for bipolar disorders
can cause bad side effects. Is this why you didn't take them today
Asshole? Tell your mommy to slip it in with the peanut butter
sandwiches she makes you for lunch.

"Wanker " Gee that's original you dumb Aussie fuck.

I bet you don't talk to people like that in person. You snivelling
little pussy.

Usually spineless loud mouth twits are so vocal because they are
compensating for other shortcomings. Quit falling for those penis
enlargement spams asshole you just encourage them.
 
"Hammy"


** PISS THE FUCK OFF


YOU VILE AUTISTIC PIG WANKER !!





........ Phil
 
On Tue, 20 Nov 2007 07:54:18 -0800 (PST), nancydow26@hotmail.com
wrote:

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 guess I need another way to regulate the input voltage. Would a
circuit with a zener diode (maybe 1N4742) do the trick?
 
On Mon, 19 Nov 2007 21:40:05 -0800, John Larkin wrote:
On Mon, 19 Nov 2007 14:01:35 -0800 (PST), nancydow26@hotmail.com wrote:

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

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

Here is what I am seeking:

15v in = 12v out 12
14v in = ??v out 12
13v in = ??v out 11.5
12v in = ??v out 10.5
11v in = ??v out 9.5

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.
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
 

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