How do we calculate the value of the resistors Rb and Re? The information given are limited:
(β = 600)
Ic = 10mA.
It is always preferred to have a symmetrical output voltage swing.
Answer
Let Ve= Vce/2 = 5V with a ~9Vpp swing possible thus Re=5/10m= 0.5k=500 Ω
As a check, the emitter resistor must always be less than the AC load path R used as an output,
- which for 10mA, satisfies this requirement of 500 < (100+2200)
- This is the same as saying the DC bias current must always be greater than the peak AC load current. .
there are several approaches to solve this if we let Rb=hFe*Re for DC bias this
- when Rb matches the input DC impedance = " Rinput "= hFE*Re , we have a 50% voltage divider, thus Rb=600*500=300 kΩ as a crude estimate.
but for Ib=Ic/hFE = 10mA/600 = 16.7 uA, we expect Vbe=0.65V
- Vbe vs Ib depends on the I max current ratings and thus bulk chip size
- But we know this Vb must be Vbe drop plus Ve=5V so make an assumption
- let Rb = V(Rb)/(Ib=Ic/hFE) = ((10V - 5V - 0.65) / 10mA) * 600
- Rb = 4.35/10mA *600 = 261 kΩ or just slightly below the previous estimate
with test validation, you should expect a 9Vpp swing possible.
- keep in mind the operating point is directly proportional to hFE tolerance
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