How to bias a transistor into cutoff

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To bias a transistor into cutoff, the user is analyzing a circuit with a collector cutoff current of 100nA. They performed DC analysis to derive the threshold voltage (Vth) and Thevenin resistance (Rth) equations. The user questions whether to assume the collector current (Ic) is slightly above 100nA to ensure the transistor operates in the active region before adjusting resistance values to achieve cutoff. The circuit is intended for RF applications. The user successfully resolved their query regarding the biasing process.
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hi
i would like to bias a transistor into cutoff. i have the schematic shown below and for the transistor i am using it has a collector cutoff current at 100nA. so my approach was frist to perform Dc analysis:

Vth=5*(R10/(R10+R4)) Rth=(R4*R10)/(R10+R4)

so my input equation is:
Vth=Rth*Ib+0.9+Ie*R12
and output equation is:
5=Vce+Ie*R12

i would ussually say Ie=Ib+Ic=(B+1)Ib but i don't think this equation holds at cutoff so should i assume Ic>100nA like 105nA so the amplifier will be in the active region(so i can use Ic=B*Ib) and then increase resistance values so Ic<100nA and the transistor will be in cutoff please help!
 
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zak8000 said:
hi
i would like to bias a transistor into cutoff. i have the schematic shown below and for the transistor i am using it has a collector cutoff current at 100nA. so my approach was frist to perform Dc analysis:

Vth=5*(R10/(R10+R4)) Rth=(R4*R10)/(R10+R4)

so my input equation is:
Vth=Rth*Ib+0.9+Ie*R12
and output equation is:
5=Vce+Ie*R12

i would ussually say Ie=Ib+Ic=(B+1)Ib but i don't think this equation holds at cutoff so should i assume Ic>100nA like 105nA so the amplifier will be in the active region(so i can use Ic=B*Ib) and then increase resistance values so Ic<100nA and the transistor will be in cutoff please help!

Looks like the schematic did not post. Can you try again? What is the application for the circuit?
 
ops i taught i attached it. anway i figured it out, it is used for RF applications
 
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