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schematic

simulate this circuit – Schematic created using CircuitLab

Its just the basic diagram (the actual circuit is a low noise amplifier at 5.5GHz).It does work very correctly.In this circuit the BJT operates in active region with Ve=1.61V and Vc=-0.3mV but if I detach JFET,the BJT starts operating in saturation.I don't find this very strange but would like to get a good explanation.Secondly my instructor insisted that I use PNP here.I could use NPN.Why would he insist for me to use PNP?

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  • \$\begingroup\$ What is your proposed NPN circuit? \$\endgroup\$
    – Spehro Pefhany
    Jul 30, 2015 at 16:12

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Since the output of the BJT has to be negative with respect to ground and the input to it is required to be in the range of 0 to +5V. I don't see how you could use an NPN in that circuit without making it much more complex.

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  • \$\begingroup\$ I am using dual supply here.I definitely can use NPN with emitter (connected to -5V through a resistor) providing me negative voltage and collector giving positive. \$\endgroup\$
    – muhammad muheeb
    Jul 30, 2015 at 16:35
  • \$\begingroup\$ Yes, but then how are you going to sense the drain voltage of the JFET (the node you are trying to control). You will need to shift the voltage of the drain down to the base of the NPN which will be near the -5V rail. You could use a zener but not very flexible. If you use resistors you will attenuate as well as shifting. I think it will be much more complex. \$\endgroup\$
    – Kevin White
    Jul 30, 2015 at 16:40
  • \$\begingroup\$ But why does the BJT operation mode changes when I detach the JFET? \$\endgroup\$
    – muhammad muheeb
    Jul 30, 2015 at 16:48
  • \$\begingroup\$ That is because you have a feedback loop - the BJT is attempting to keep the drain of the JFET at 1.6V by adjusting the gate voltage of the JFET. If you remove the JFET the BJT will keep applying a more positive voltage to the gate of the (nonexistent) JFET to correct the imbalance. The drain voltage will go towards +5V and the BJT will go into saturation with around 2-3v on the gate pin. \$\endgroup\$
    – Kevin White
    Jul 30, 2015 at 16:54
  • \$\begingroup\$ The BJT is actually applying a negative voltage to the gate here and why would the emitter voltage go toward +5V?The BJT is not going to be in cut off,it is going to be saturated. \$\endgroup\$
    – muhammad muheeb
    Jul 30, 2015 at 21:54

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