Re: Transistors
- From: Phil Hobbs <pcdh@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Wed, 14 Nov 2007 17:07:02 -0500
Bruce Varley wrote:
"john" <conphiloso@xxxxxxxxxxx> wrote in message news:1194902332.623320.152710@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxHey! This is SED--no fair being helpful!Hi,
I am new to electronics and working with transistors for the first
time. I chose 2N4401transistor from Fairchild to begin with. You can
find the data *** at http://www.fairchildsemi.com/ds/2N/2N4401.pdf
I have following questions
1. The base current Ib controls the collector current Ic. But I am
unable to find the base current from the data ***.
2. I am also unable to find the maximum and minimum values of Ib too.
3. Is it ok to determine Ic and then calculate Ib using Ic = beta x
Ic.
4. What should I consider to do regarding DC biasing of the transistor
before applying the AC signal to the transistor?
Please advice!
John
Here's something simple you can do to get a feel for the situation. Assuming you have a 9v or 12V DC supply, and your transistor is NPN small case:
Emitter to V- through a 1K resistor.
Collector to V+ through a 1K resistor.
Base to V- through a 3K3 resistor.
Base to V+ through a 10K resistor.
Attach the battery and measure some voltages.
The key voltage is between the base and emitter. Base will be roughly 0.6V positive above the emitter.
*Rule 1: Whenever the transistor is 'working' (let's not worry about exactly what this means) this voltage, Vbe, will be around 0.6 volts.
*Rule 2. Whenever the transistor is 'working' collector current Ic = base current * 'beta', where beta will be somewhere between 50 and a few hundred, and will vary a lot from device to device.
Now measuring voltages from V-.
Emitter will be at about +2V.
Collector will be at about 2 volts less than the supply (+7V?).
How should you view this?
You've fixed the base voltage at about 2.5V with the 10K/3K3 divider (1/4 supply).
So by Rule 1, the emitter voltage Ve will be about 0.6 volts less, or about 2 volts. Or IOW, the voltage across the emitter resistor is 2 volts.
So by Ohms law, the current through the emitter resistor is V/R = 2volts / 1K = 2mA.
Now the collector current Ic = emitter current - base current. BUT by rule 2, since beta is 'big', we can say that the collector current roughly equals the emitter current within a percent or two.
So the voltage across the collector resistor, by Ohms law, will be Ic * 1K = 2 volts.
Note that nowhere here did you consider the base current at all. Rather you set up a system where the transistor took the base current that it needs to satisfy the above rules itself. A great deal of bipolar circuit analysis uses these principles. It's worth taking some time to get the concepts clear.
Good luck
Cheers,
Phil Hobbs
.
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