Low Shutoff Voltage Regulator

Thread Starter

axeman22

Joined Jun 8, 2009
54
Hi All,

I'm familiar with using the 7805 reg to some degree, like the one referenced here http://www.datasheetcatalog.org/datasheets/150/44435_DS.pdf

I'm hoping someone could explain..:

1. Dropout Voltage.. I think I get this, if so these would be correct.. no ..? - for the 7805 5V Reg
Vin/Vout
10 / 5
8 / 5
7 / 5
6 / 5
5 / 5
4 / 4
3 / 3
2 / 2 - Dropout Voltage..
1.9 / 0
1.5 / 0
1 / 0

2. What happens to the excess voltage.. is it just dissipated in heat?

3. Slightly outside the scope of the question but what would I use for a situation where my Vin will start at 10V and slowly decline to say 3volts but I want to have a 5V regulated(and no less than 5V) for as long as I can.. i.e. a dropout at 5V.
Curve like this..;
Vin / Vout
10 / 5
8 / 5
6 / 5
5 / 5 - dropout here
4.8 / 0 - below droput
4 / 0
3 / 0

feeling like this is a long winded question.. but thanks in advance, no doubt a simple answer!
 

eblc1388

Joined Nov 28, 2008
1,542
Your interpretations of the dropout voltage is incorrect.

A voltage regulator is meant to regulate its output voltage to a certain fixed voltage level, with changing input voltage level.

Dropout voltage is the extra voltage that is necessary *above* the output voltage such that the regulator can do its job, i.e. regulating.

For 7805 under load, that voltage is typically about 2 to 3V more than the output voltage, so it meant 7V or more input voltage.

If the input voltage falls below 7V, the regulator cannot regulate and so the output voltage falls too, along with the fall of the input voltage. However, it does not drop to zero volt suddenly.

If this 2V dropout voltage in 7805 is too high for your application, then there are "low dropout(LDO)" regulators available that has a very small voltage differential margin between input and output of 0.2V or less. This means the regulator can provide a stable +5V with just 5.2V or more input voltage.
 

gotumal

Joined Mar 24, 2008
99
Slightly outside the scope of the question but what would I use for a situation where my Vin will start at 10V and slowly decline to say 3volts but I want to have a 5V regulated(and no less than 5V) for as long as I can.. i.e. a dropout at 5V.
Curve like this..;
Vin / Vout
10 / 5
8 / 5
6 / 5
5 / 5 - dropout here
4.8 / 0 - below droput
4 / 0
3 / 0
May be a zener of 5.1V inseries with a base of a transistor?
 

jpanhalt

Joined Jan 18, 2008
11,087
You can make a sharp cut-off with another device, as suggested. You might consider using a comparator that turns off the enable pin of your voltage regulator when the supply voltage is too low. A lot of regulators have enable pins.

Or, go to DigiKey and search on voltage regulator >special purpose>voltage regulator/detector

DigiKey carries the Seiko brand. You can set a cut-off voltage such that when the supply falls below that voltage, the output goes to zero.

John
 
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