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RF Choke in Amplifier Voltage Divider Bias

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fuseless

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Referring to the attached schematic diagram. The idea is to keep all RF out of the bias circuitry. The 5230 Ohm resistor in the voltage divider bias is automatically bypassed by a .1 micro Farad capacitor in order to put the shunt 33nH impedance matching coil at ground while allowing DC to be available to the base of the transistor. This in effect removes the 5230 Ohm resistor from the AC or RF circuit. Ok, I understand that part. The same holds true for the output impedance matching network and the 100 Ohm collector resistor. Ok, I understand that part to. The emitter resistor has the typical RF bypass. I got that part along time ago.

But when it comes to the 3240 Ohm resistor from the base of the transistor to ground in the voltage divider bias network, it's a little different here. At first, when the design was completed, there was only the 3240 Ohm resistor without the .9 micro Henry choke to keep RF out of that part of the circuit. I decided to add the .9 micro Henry choke in to completely isolate the voltage divider network from the RF signal.

However, can anyone tell me if the .1 micro Farad capacitor in series with the .9 micro Henry choke to ground that bypasses the 3240 Ohm resistor is necessary?
 

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The impedance of the 0.9uH choke is 915 ohms at 162 MHz. Because you have the 0.1 cap to ground, you have the +j915 ohms shunting your base input. If you don't want this shunt circuit involved in your AC circuit, you would be better off with a higher shunt impedance, would you not? In that case, it makes more sense to eliminate both the 0.9uH inductor and the 0.1 uF cap and have only the 3.24K resistor to ground. In that case, the shunt impedance would be much higher, at 3240 ohms rather than 915 ohms, and this very high resistance would have negligable effect on the AC behavior of your input matching circuit.
 
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