T1 is a current limiter. When the current thru T2 reaches 4 amps, you want T1 to turn on. The base-emitter voltage of T1 is 0.6 volts, so by Ohm's Law: R2 = 0.6/4 = 0.15 ohms.

The 7805 needs about 7 volts at the input to work, so the drop across R1 is 3 volts max. R = E/I = 3 ohms.

Sorry Russ,
When T2 is conducting 4A, its base-emitter voltage will be about 1V. So R1 should be 1 ohm plus the voltage across R2.
EDIT: R1's calculation.

The power rating of the parts depends on their current that we know, and on the voltage across them, so Pr1 = 4W (use 5W or more) and Pr2 = 2.4W (use 3W or more). We don't know the voltage across the transistors so we can't calculate their power.

Your school's examination request is not realistic because you can't buy transistors with a beta of 100 at 4A. Their beta will be within the range of from 20 to about 80. Also, transistors have a range of Vbe that is wide. Therefore you would be crazy to design the circuit with the regulator's current as high as 1A which is so close to its limit.

In National Semi's datasheet, they have the same circuit and use the same 3 ohm current sensing resistor as Russ proposed. But their regulator is operating at a current much less than 1A. They have a detailed formula involving beta so use it. :lol:

Sorry, with a regulator current of 1A, R1 should be 1.875 ohms, if you could find a transistor with a beta of 100 at 4A, and with a Vbe of 1V.

Look what happens if you build it with resistors calculated for the transistors having a beta of 100 and a Vbe of 1.0V, but end-up with low spec transistors with a beta of only 20 and a Vbe of 1.2V. The regulator will try to pass 1.25A but won't do a good job and might even shutdown if it gets too hot.

R1 is not essential, its value could be zero and the circuit will still work. R1 reduces the power dissapated in the LM7805, so with 10 volt supply and max load, the dissapation in the 7805 is minimized when R1 = 3 ohms.

Originally Posted by Russlk

R1 is not essential, its value could be zero and the circuit will still work.

If R1 is zero ohms, then Q1 could never turn ON (as it's base and emitter would be shorted together), so the 7805 would provide all the current, limiting it to the 1A maximum of the 7805.

So without R1 the circuit wouldn't work!.

Originally Posted by Russlk

R1 is not essential, its value could be zero and the circuit will still work. R1 reduces the power dissapated in the LM7805, so with 10 volt supply and max load, the dissapation in the 7805 is minimized when R1 = 3 ohms.

The power dissipation in the 7805 is reduced when it has a low current, so R1 must have a high value. Allowing for only 10mA through R1 at full output to bypass leakage current in the boosting transistor, R1 = 169.8 Ohms.

Nigel is correct, as usual. I still think R1=3 ohms is the best value.

Originally Posted by Morgen

"When T2 is conducting 4A, its base-emitter voltage will be about 1V."
(How was this 1 volt Vbe determined?)

Any power transistor has a high Vbe when it has 4A of collector current. I couldn't find a datasheet for the ones in my example, but a common 15A-rated 2N3055 power transistor has a max Vbe of 1.5V at 4A.

"The power rating of the parts depends on..."
(The problem is to solve for the resistor values.)

Patrt of the excercise was to calculate power dissipation of its parts. I missed seeing a 10V input voltage in my 1st reply.

"If R1 is zero ohms, then Q1 could never turn ON (as it's base and emitter would be shorted together)"
(Has R1 in the original circuit has been confused in the NS circuit?)

The circuits are identical.

"The power dissipation in the 7805 is reduced when it has a low current, so R1 must have a high value."
(The current through the 7805 is 1A.)

No it isn't. The current through the 7805 depends on the value of R1 and on the Vbe of the booster transistor.

If R1 is 3 ohms and the Vbe is 1V, then the 7805 passes only 644mA and you lost your bet. :lol:

2N3055 datasheet: