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Need Help Understanding Hi Current Add On For LM78XX

Rory Starkweather

Nov 13, 2014
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Nov 13, 2014
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I understand how the LM78XX works, but I'm having trouble with the current booster. The data sheet says that even at half rated current the LMXX needs a heat sink. That's OK, but I would rather avoid it.

So they have a diagram for a High Current Add On. (Question 1. Will this keep the LM78XX from getting hot if it turns on at, say, .25 max I?)

I have a good picture, but can't figure out how to post it. Anyway, there are only two components in the High Current part of the circuit.

1. A resistor which is connected to the input on one end, and to the junction between the input to the LM78XX and the filter capacitor which is connected to the LN78XX input and ground.

2. A transistor connected between the input to the circuit and the output from the LM78XX. This transistor has the Emitter connected to the circuit input, Collector connected to the output of the circuit, and the Base connected to the input pin of the LM78XX.

Two formula are given.
1. R1 = Vbeq1 / (Ireg - Iq1) / Bq1.
What is Vbeq1? Base to Emitter voltage of Q1?
2, I assume Iq1 = current out of the transistor.
3. Is Bq1 = voltage on the base of Q1?

Other formula; Io = Ireg + Bq1 * (Ireg - Vbeq1 / R1).

I would like to build one of these, but am not sure enough of what values are being used.
 

Arouse1973

Adam
Dec 18, 2013
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Hi Rory
This sounds like the common approach used to shunt current from the regulator to increase the total current that the circuit can supply. Your right with most of your assumptions but BQ1 I think is the gain of the transistor. I have added the picture to help others. If you work it out so that each component has a current of 200mA and a gain of 100 with Vbe of 0.7V then this gives a resistor value of 3.5R. Remember you still might need a heat sink for both components dependant on the amount of current you are drawing from the circuit and the split ratio.
Adam

7805.PNG
 

Rory Starkweather

Nov 13, 2014
77
Joined
Nov 13, 2014
Messages
77
Hi Rory
This sounds like the common approach used to shunt current from the regulator to increase the total current that the circuit can supply. Your right with most of your assumptions but BQ1 I think is the gain of the transistor. I have added the picture to help others. If you work it out so that each component has a current of 200mA and a gain of 100 with Vbe of 0.7V then this gives a resistor value of 3.5R. Remember you still might need a heat sink for both components dependant on the amount of current you are drawing from the circuit and the split ratio.
Adam

View attachment 17787

Thank you very much. That is exactly the drawing I was talking about.

I guess that since I'm not planning to use large currents now, the smart thing to do would be to buy a few more heat sinks.

Another question, I used to see circuits with Diodes to protect the LM78XXs. A simple diode loop from input to output. I believe that the Anode was connected to the input terminal. Does this sound reasonable?

And 1 more. This would be an input over-current protector. I can't remember the name of the component, though. The description of a thermistor just doesn't sound quite right. I have some of these. Mine are in thin square packages with two leads coming out of one side. They are thin enough that you can see wires inside.The ones I have are about 3/4 inch on a side but 1/4 inch or less thick. Any idea?
 

Arouse1973

Adam
Dec 18, 2013
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Yes the diode discharges any output capacitance when the power is switched off. Some people do some people don't. I don't personally as most of the time I don't have that larger capacitance on the output.

Are you thinking of PTC or NTC. PTC is used for over current protection and NTC is used for reducing large in rush currents. You would typically see these in switch mode power supplies.

Adam
 
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