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Hi Murugesh ,, thanks for replying .. Its clear that reducing the RL i can achieve more currents but in reality i have a fixed load resistance of 120 ohms ..
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How to overcome this problem .. can i choose another transistor .. My base current is not absolute ,, i have already considered around 30 % more than what i computed with IC*hfe ...
You can get less voltage drop if you use a MOSFET instead of a BJT. With a MOSFET you also don't have to worry about base current, just apply +12V to the gate to switch it on.
For proper switching of a bipolar transistor a beta value of 10 is typically used, not 100. The beta value of 100 is only applied when using the transistor in linear or small signal applications, not when using it as a switch.
Hi John.stm,
One thing to be noted here is that you are using a 12V supply and want to have a 100mA through 120 ohm load. Hence the voltage drop of 120*100mA = 12V should happen across the load. Hence there is no voltage available for Vce. This is not a practical case.
The very low value of Vce say 0.05V could be achievable if you require lower collector current. Since you require more current of the order 100mA, the Vce drop should not be reduced to zero and it will stay around 0.2V and hence the current of around 98 - 99 mA is flowing through it. If you use the supply voltage of around 12.3V(min) then what you want will be achieved.
An LTspice simulation of a 2N3904, 200mA transistor with 12V supply, 120 ohm collector resistor, and 10mA of base current (forced Beta value of 10), gave a Vce value of .062V and a collector current of 99.48mA. Is that good enough? :smile:
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