LT6015 in Current-Buffering Circuit

I have designed the circuit below to buffer LT6015  output to deliver 4.2V to a 100 ohm load. The load is just a representative of a circuit which sink 44 mA current. I appreciate if you help me find answers for questions listed below,
1) As the output of the OP-Amp going to be 0.7V more than 4.2V, I chose Rc= 50 to provide Vs =2*(4.2+0.7)= (12-50*0.044)= 9.8 for the op-amp power supply. Is my approach correct?
2) I would like to power down or at least decrease the power consumption of my circuit.Therefore, I thought of using Q1 as a switch as shown in the figure. However I do not know if I should place R1 resistor or not and if R1 is required how should I find the proper value? I appreciate i you could give me some hints.



  • 0
    •  Analog Employees 
    on Oct 11, 2019 1:51 AM


    1) I think the computations are correct for the opamp power supply.

    2) R1 could simply be removed and short Q1 collector to the non-inverting pin of LT6015. I think the current developed at 5k (resistor after 2.1V) is small and will not cause Q1 to reach Isat max. I think Ic is typically at mA range. Otherwise if current is large and could reach Isat max of Q1 you can add R1 to limit the current. As you increase R1, this will cause the output voltage on Q2 emitter and current  at 100 ohms to increase. I attached a sample LTspice simulation ckt.

  • Hi Samaneh,

    I think that R1 is not required because in this part of circuit small currents will flow. You can calculated current value through transistor as follows: (2.1 Volts minus collector - emitter saturation voltage) divided by 5 kΩ and compared this value with absolute maximum ratings of transistor.

    After that you will be able to determine the need for R1.

    Also note that if you will use R1, voltage at non-inverting input will be higher than in the case without this resistor and it will cause more output current in "power down" mode.