Yes, me again.
I have worked through another of Harry Lythall's biasing examples. **broken link removed**
This time it promises to improve the control of Transistor gain using an emitter resistor and potential divider to bias the transistor.
Here is my working out:-
Assumption 1. The collector current is 10mA
Assumption 2. The transistor base is fed from a potential divider of 2V
Assumption 3. The gain (for sake of argument) is 200.
Emitter Resistor
Re = (Vb - 0.7) / Ic
Re = (2 - 0.7) / 10mA
Re = (1.3) / 0.010
Re = 130 ohms
Collector Resistor
Rc = (Vcc - Mean Level between 10V and 2V) / Ic
Rc = (10 - (10+2/2)) / 10mA
Rc = (10 - 6) / 10mA
Rc = (4) / 0.010
Rc = 400 ohm
Potential Divider
Total Resistance = Vcc / (Ic /200)
" " = 10 / (10mA / 200)
" " = 10 / (0.010 /200)
" " = 10 / (0.00005 or 50uA)
" " = 200,000 (200k)
Use values 40k\160k for 2V supply to base.
As you can see from the attachment I have plugged in the values, but as you can see the meter readings arent immediately what I expected.
For instance I was expecting to see 6V and 10mA at the collector. Also a PD of 2V at the potential divider junction feeding the base.
I have been thinking this through and I just wanted confirmation.
1. The reason the base voltage is registering 1.298v is because it has dropped 0.7v through the "be" junction of the transistor?
2. Because the simulation has done this, the Collector Voltage has dropped to 8.068v. The only way I can explain this is by Vcc - (Rc x 4.83mA) = 8.068?
Do you agree with this?
I can't explain Ic though?
Thanks in advance.
Mark