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Transistor amplifier

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elec123

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Hi

I'm stuck in the middle of my revision, just wondering if anyone can give me a hand.....

Please see circuit diagram attatched,

I know how to calculate Rc using the gain formula (-Rc=-A.Re) which makes this resistor 1.2k. Could someone expalin the process of calculating the R1 and R2 values which form a potential divider....

Thanks
 

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Hi,

Here is the solution. If Rc is 1.2K then

the current flowing through this resistor is 12V / 1.2K = 10mA

Ic and Ie are approximatly equal therefore the current flowing through the 100 ohm resistor is 10mA. If we know the current we can calculate the voltage across the resistor. 10mA * 100ohms = 1V.

THe voltage drop across the base and emitter is always 0.7V therefore we add this to our 1V

This gives 1.7V across R2. If the sum of the two voltages must equal 12V then 12 - 1.7 = 10.3V

The voltage acrros R1 = 10.3V

the ratio between the two resistors is 10.3 / 1.7 = approximatly 6

therefore if we choose R1 = 6K ohms and R2 = 1K ohms, this keeps the ration as clsoe to 6:1 as possible

I have tried my best to help you here, If anyone has found any mistakes please let me know. Hope it helps you understand.

andy
 
Right, you already know that the collector load has to be 1.2K, to give a gain of 12 - due to the 100 ohm emitter resistor giving negative feedback.

The base bias resistors don't really have anything to do with the gain, you simply choose them to give the greatest available headroom for the stage. So essentially you want the collector at half the HT voltage.

Knowing the collector voltage, and the value of the collector load, allows you to calculate the collector current. The emitter current is (near enough) identical to the collector current, so, knowing the value of the emitter resistor you can calculate the emitter voltage.

All you do now is calculate the base voltage, which is simply the emitter voltage plus 0.7V - calculate your base bias resistors to give that voltage on the base. Bear in mind that the base will draw current from this potential divider, so you can't make the values too large.

You can work out the base current, from the collector current and the Hfe of the transistor, you should pass something like 5 times this current down the potential divider network.
 
andy257 said:
Hi,

Here is the solution. If Rc is 1.2K then

the current flowing through this resistor is 12V / 1.2K = 10mA
<snip>
Actually the current is (12V-Vc)/1.2k, where Vc is the desired collector voltage. As Nigel pointed out, you will probably want Vc to be about 6v, so Ic in this case is about 5ma.
 
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