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LT3080 Higher Current Output: How does it work?


Stephen Hill:
So I saw this circuit schematic on the LT3080 datasheet and I would please like help understanding how the MJ4502 PNP, 50? Resistor and the LT3080 work together to control the output voltage.

My current guess (from playing with the circuit in LTSpice) is that the 50? Resistor create a small voltage drop between Vctrl and Vin which creates a voltage potential between the emitter and base. This then allows the current drawn by the LT3080 to adjust the output voltage of the PNP by pulling current out of the base.

Am I correct or completely wrong?

Many Thanks

Hi Stephen,

you got it correct.
This arrangement is called complementary Darlington or Sziklai pair. The advantage is that the base  turn-on voltage is only 0.6V instead of 1.2V with a classical Darlingtonpair.

BTW. Sometimes it helps to simply redraw the circuit in another arrangement of the components (at least it helps me) to understand what's going on.

Stephen Hill:
Hi Fox,

Thank you :) I didn't think to redraw and thus simplify the circuit - something to remember for next time.

Many Thanks

Well, Darlington pair does not explain why the 50 Ohms is there;

"Reduced turn on voltage" is not the key either since to have the MJ4502 do anything useful, one needs to have  (VIN-VOUT)  > 1.2 plus V at least at larger load level; then, min (VIN-VOUT) > 1.2V plus V needs to be fixed from design for all load levels. In other words, "reduced turn on voltage" did not help much.

The OP's statement "the 50? Resistor create a small voltage drop between Vctrl and Vin which creates a voltage potential between the emitter and base"  is also a wrong way to approach the answer. MJ4502 does not need the resistor to bias it; LT3080 is the one bias MJ4502; In fact, the 50 ohms is to remove the bias effect of Lt3080 for small load level. It is not required to have voltage drop btw Vin and Vctrl either.

The main effect of the the 50Ohms is to reduce the the required minimum current load of the design to that of LT3080(0.5mA) by effectively shutting off MJ4502 at small current load. If w/o this resistor, the 0.5mA required for LT3080 to work properly will go through EB of MJ4502, assuming a gain of 40, the minimum current load of the design will be 40x0.5=20 mA now. In addition, the design has to accomendate variations in MJ4502's gain. To have easier understanding, a dummy resistor that provade the min current load passage can be draw btw Vout and ground.


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