The problem of Bjt Biasing techniques.

Thread Starter

Rhýthm Khanna

Joined Feb 8, 2016
1
Mods Edit:
This thread was split from -- Bjt Biasing techniques.
Please don't hijack other member's thread, now you have your own.



Actually, capacitors have very little to do with DC biasing, which is where you start. They have everything to do with AC frequency response however.

There are more than one model for describing how a transistor works. The one I use starts from the base assumetion, Ic = β X Ib . From there you can derive almost every biasing formula used.

Given the 3 basic configurations, Common Emitter, Common Collector, and Common Base, and their subclassifications you have to figure each set of equations on a case by case bases, using the root one I described. If you want to understand a specific example you have to pick one. My personal favorite is this one.



It is a general purpose amplifier.
This given circuit looks like a combo. R1 and R2 are used as a voltage divider for providing biasing for class A operation. The Re resistor is used for emitter biasing.
Am I going right with this? Please let me know.
Also, I do not understand the usage of 2 input sources, one with the Rinput and the other one with input pin. Plus the use of the equation Rc=R3||RL is somewhat confusing as they do not seem to be forming a parallel network of resistors.
Will be eagerly waiting for a reply.
 

Wendy

Joined Mar 24, 2008
21,930
Look closely at the quote in post 1, a schematic is embedded within.

All signals and batteries have a resistance included, I was showing them for demonstration purposes. It is a model after all. The input and output shown is where you would look with your instrumentation, be it an Oscope or a meter.

The transistor is a dynamic resistance, while R3 is fixed. It helps sets the output impedance of the amplifier, and interacts with the load. R3 is still part of the AC signal path.
 

dl324

Joined Mar 30, 2015
9,620
Welcome to AAC! And getting your first thread hijack out of the way...
Also, I do not understand the usage of 2 input sources, one with the Rinput and the other one with input pin.
Same here; insufficient context.
Plus the use of the equation Rc=R3||RL is somewhat confusing as they do not seem to be forming a parallel network of resistors.
This is the Homework Help forum; you need to show your work or ask questions if you're stuck.
 

Russmax

Joined Sep 3, 2015
82
Mods Edit:
This given circuit looks like a combo. R1 and R2 are used as a voltage divider for providing biasing for class A operation. The Re resistor is used for emitter biasing. Am I going right with this? Please let me know.
That is correct, or at least the designer hopes it is correct. There is a voltage loop formed by Vcc, R1, R2, Vbe, and Re. This loop is used to calculate Vb, Ve, and Ib. Remember, if Ic= β*Ib, then Ie=(β+1)*Ib.
Also, I do not understand the usage of 2 input sources, one with the Rinput and the other one with input pin.
Only the drawn source, with Rinput, is used. The input pin is test point.
Plus the use of the equation Rc=R3||RL is somewhat confusing as they do not seem to be forming a parallel network of resistors.
Think of a circuit like this as 2 different circuits, a DC circuit with the capacitors open and an AC circuit with capacitors shorted and supply voltages = GND. The DC circuit is for calculating bias voltages and currents. Vcc, β, Vbe, Ic, Ib are used to analyze the DC circuit. The AC circuit is used for calculating voltage gain, Rin and Rout at AC frequencies (in the middle of the circuit's bandwidth). Small signal parameters βac, Rπ, and Ro apply to this circuit, which can be redrawn using the small-signal model.

In the AC circuit, C2 is shorted and Vcc=GND, making R3 parallel with RL.
ALWAYS keep these 2 circuits, DC and AC, separate in your mind. You must do the DC analysis 1st, because the small signal parameters are calculated from the DC bias current, and also to verify the transistor is not in saturation or cutoff.

Hope this helps.
Will be eagerly waiting for a reply.
 
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