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I am dealing with the following circuit:

Universal Voltage Divider Bias Circuit with both BJT and JFET I am trying to understand if I can just assume that, much like in the JFET voltage divider circuit, V_G = 10/(40+10) * 16V , or if the presence of the BJT after the JFET will affect the voltage at this point.

Any help is appreciated. Thank you so much!!

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The current into the base of the BJT is about \$\beta\$ (100) times smaller than the current through the emitter/collector. The upshot of this is that the effect of the 1k2 resistor as seen at the base can be regarded as the same as a grounded base resistor that is 100 times higher or 120kohm. This will make Vg a few percent lower.

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