DETERMINE COLLECTOR CURRENT For our amplifier to function properly, we need to choose the correct values for Rb and Rc. But first, we need to decide how much current we need to keep flowing between the emitter and collector to keep the transistor operational when there is no input signal present at the base. This is called the quiescent current. We want to choose a quiescent current that's not more than we need if we want to conserve energy. For instance, the PN2222A transistor operates quite satisfactorily with a quiescent current of 1ma. Now that we know how much quiescent current we want, we need to determine how much base current will give us our desired quiescent current. |

The correct values for Rb and Rc can be determined once we know the required base
and collector current. |

DETERMINE BASE CURRENT Because the quiescent current is directly determined by the base current, we need to calculate what base current will give us 1 milliamp of quiescent current. Since we know that the collector current in this particular transistor will always be 200 times greater than the base current, we can see that the base current needs to be 5 microamps. Ic / Hfe = Ib .001A / 200 = .000005A |

DETERMINE BIAS RESISTOR With 9 volts supplied at the base through the bias resistor, we need to know what value this resistor needs to be in order to draw 5 microamps into the base. Actually, the base voltage is not quite 9 volts. Since the base-emitter junction drops .70 volts out the 9 volts applied at the base through the bias resistor, the actual base voltage becomes 8.35 volts. Vcc - Vbe = Vb 9V - .70V = 8.30V I would like to mention that it does not matter what voltage is applied at the base of the transistor as long as the correct value of resistor is chosen to provide the correct base current. The higher the base voltage, the higher the value the bias resistor needs to be. Since the voltage dropped across the bias resistor is 8.30V, we can see that the value of the resistor (from here on, all formulas are derived from Ohm's Law) needs to be 1.67 Megaohms in order to pass 5 microamps across the base-emitter junction. Vb / Ib = Rbias 8.30V / .000005A = 1,660,000 Ohms |

DETERMINE LOAD RESISTOR With no signal applied at the base, the collector current will always be 1 milliamp as established by the base current no matter what resistance we choose for the load resistor*. So changing the load resistance will only change the collector voltage - not the collector current (as long as the load resistance isn't too high to sustain the desired collector current). According to Ohm's Law, the voltage dropped across the load resistor is equal to resistance times collector current. So if we just happen to pick a load resistance of 6000 ohms, the voltage across the load resistor will be: R x Ic = V 6000 x .001 = 6 Volts Whatever voltage is dropped across the load resistor, the rest of the supply voltage will be dropped across the transistor (the collector voltage). So for the example given above, 6 volts dropped across the load resistor will produce a collector votage of 3 volts. Obviously, if you need a specific collector voltage, you need to subtract the required collector voltage from the supply voltage to determine what the voltage drop should be across the load resistance. Once we know what the voltage drop should be across the load resistor, we can solve for the load resistance. For example, if we want a collector voltage of 7 volts, we know that the load resistor has to drop 2 volts. Since resistance is equal to voltage divided by current, a 2000 ohm load resistor will do the job nicely. (Vcc - Vc) / Ic = Rc (9 - 7) / .001 = 2,000 Ohms However, a collector voltage of 7 volts may not do us much good. By applying a signal to the base, causing the collector voltage to vary, it would only be possible for the collector voltage to rise by 2 volts before "bumping" into the supply voltage. The least distortion will be produced if the collector voltage is half the supply voltage. So with a supply voltage of 9 volts, the collector voltage should be at 4.5 volts. That way, the collector voltage can vary equally about the 4.5 volt value to a higher or lower voltage. 1/2Vcc / Ic = Rc 4.5 / .001A = 4,500 Ohms |

*With a 9 volt supply voltage, our transistor will always try to maintain a total
combined resistance of 9000 ohms between the supply voltage and ground in order
to provide 1milliamp of current as determined by our chosen base current (9 /
.001 = 9000). In other words, the transistor can vary it's resistance according
to the the load resistance. For example, (assuming we have a collector current
of 1ma) if the load happens to be 6000 ohms, the transistor will act like a 3000
ohm resistor. |

Now that we have a functional small signal amplifier, it would be nice if the output
signal could vary about a 0 volt baseline like the input signal does. Then
the output would consist of a.c. voltages instead of just variations in direct
current (most audio equipment or devices require an ac signal input). A simple
capacitor tied between the collector and output terminal will do the trick nicely.
The capacitor blocks unchanging d.c voltages, but does respond to voltages
that are continuously changing. Since C2 blocks the collector's baseline voltage
of 4.5 volts, the output's baseline is set to 0 volts - changing the highest
and lowest possible output voltages to 4.5V and -4.5V respectively. We should also
include a capacitor tied between the input terminal and the base to block any
unwanted d.c. voltages that would otherwise upset the collector's baseline voltage.
A 1 microfarad capacitor for C1 and C2 should be adequate for general audio
applications. |

With the inclusion of C2, the output now follows only continuous changes in collector
voltage - not the collector voltage itself. For instance, if the collector
drops from 4.5 to 3.5 volts, the output will drop from 0 to -1 volt. Let's say
that the collector voltage stays at 3.5 volts. The output will then instantly
return to 0 volts. As already stated, the output can only follow changes in voltage. With C1 able to block any d.c. voltages other than the audio
signal, the collector voltage will be unable to stay at any voltage other than
it's baseline of 4.5 volts anyway.continouos |

© 2007 Dave Cline. All Rights Reserved. |

NextAmplifier with gain control |