Hello!

When you look this topic http://www.allaboutcircuits.com/vol_3/chpt_4/8.html

and this circuit there is voltage divider of R3 and R2 to keep the transistor open all the time. But why it is so when I simulate this circuit without R3 and there is only R2, then circuit still will work normally(will amplify all AC)

When you look my attachment then there is circuit with amplification of 10x and this don't have this one voltage divider resistor. So why I should put there one, when the circuit will work anyway fine? And why does it work?

 

Originally posted by napos@Mar 3 2006, 07:14 PM
Hello!

When you look this topic http://www.allaboutcircuits.com/vol_3/chpt_4/8.html

and this circuit there is voltage divider of R3 and R2 to keep the transistor open all the time. But why it is so when I simulate this circuit without R3 and there is only R2, then circuit still will work normally(will amplify all AC)

When you look my attachment then there is circuit with amplification of 10x and this don't have this one voltage divider resistor. So why I should put there one, when the circuit will work anyway fine? And why does it work?

[post=14602]Quoted post[/post]​

Your circuit will work, but the Q-point will not be stable. Voltage Divider Bias is a way to make the amplifier's Q-point stable over variations in supply voltage, temperature, transistor beta and so forth.

The idea behind a quiescent point or Q-point is to put the transistor in such a condition that it behaves as a LINEAR system for small changes around the Q-point. If the input signal or other circuit condition causes the Q-point to move out of the linear range then what you get is non-linear distortion going from input to output.

 

Originally posted by napos@Mar 3 2006, 06:14 PM
Hello!

When you look this topic http://www.allaboutcircuits.com/vol_3/chpt_4/8.html

and this circuit there is voltage divider of R3 and R2 to keep the transistor open all the time. But why it is so when I simulate this circuit without R3 and there is only R2, then circuit still will work normally(will amplify all AC)

When you look my attachment then there is circuit with amplification of 10x and this don't have this one voltage divider resistor. So why I should put there one, when the circuit will work anyway fine? And why does it work?

[post=14602]Quoted post[/post]​

The resistor between the base and ground is there to set the output quiescent operating voltage at the collector. Without this resistor in place, the collector quiescent voltage ends up being a strong function of the beta of the transistor being used. Since beta varies from transistor to transistor as well as changes in temperature then this could result in the transistor output clipping due to excessive output dc offset at the collector. The cap coupling between the collector and the load will mask some drift in the collector offset but clipping can still take place if the collectror voltage drifts too close to the positive power rail or the saturation point of the transistor.

hgmjr