# design a transistor done, just need a little help

Discussion in 'General Electronics Chat' started by Tahmod, Feb 25, 2013.

Feb 23, 2013
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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You can change the collector current by changing R1 resistor or R2 resistor.

$Ie = \frac{Vcc - Vbe}{\frac{R1}{(\beta +1)} + R2}$

$R2 = \frac{Vcc - Vbe}{Ie} - \frac{R1}{(\beta+1)} = \frac{6V - 0.66V}{150mA} - \frac{188}{110} = 35{\Omega} - 1.7{\Omega} = 33{\Omega}$

3. ### #12 Expert

Nov 30, 2010
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Graphs of gain, current, and voltage are usually on the datasheet for the transistor.

Feb 23, 2013
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5. ### WBahn Moderator

Mar 31, 2012
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Transistors intended for switching applications frequently don't have very thorough characterizations in the data sheets because most applications don't care. You are mote interested in curves of hFE and saturation voltage in those apps.

I'm a little concerned when you say that you want to change the collector current from 300mA to 150mA because you want to use it to control a motor. What is it that you are hoping to accomplish by this change? Where is this motor going to go in the circuit?

6. ### #12 Expert

Nov 30, 2010
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This is the datasheet I have.

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7. ### Tahmod Thread Starter New Member

Feb 23, 2013
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-By having the current at 150ma one of my motor will reduce speed and the other will remain on. This will able the motor to move more smoothly.
If I have only a current of 300ma and 0ma. This make one motor stop while one off...

But If I provide a voltage to the base of 6 volts for having a collector current of 300ma, a voltage of 3 volts to the base will provide 150ma. This is correct?

8. ### WBahn Moderator

Mar 31, 2012
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What kind of motors are these? Most simple permanent magnet motors produce a torque that is proportional to the current in the motor. There is also a voltage across the motor that is proportional to the speed of the motor.

If you are putting your motor in the collector circuit, you are only establishing a maximum current. The actual current will reflect the interaction of the speed and the load.

Putting that aside, your design is not good because it relies on the beta of the transistor being a fixed value. A much better approach is to use a voltage divider to establish a base voltage and size the voltage divider resistors such that the base current is a small fraction of the current in the resistors. Then you can reasonably assume that the voltage at the emitter of the transistor will be one diode drop below the now-fixed base voltage which let's you control the emitter current with the emitter resistor and the collector current will essemtially be equal to this (until the transistor saturates).

Last edited: Mar 2, 2013