BJT: Should we bias BJT for Base Voltage or Base Current ?

Thread Starter

Vignesh D

Joined Feb 18, 2019
6
Hello, I am trying to design my own CE Amplifier using BJT. I am having issues with fixing the Q-Point.

Biasing can be performed in two ways:
(1)
Base Bias.PNG
Here, we are biasing for a fixed Base Current i.e
Ib = (12 - 0.7) / 1k
Assuming, Vbe as 0.7V.

(2)
2. Voltage Divider.PNG
Here, we are fixing base voltage using voltage divider. (This voltage divider can be made equalent to (1) using thevinin transformation).​

So, Which of the above two approach is correct ? Based on what parameter should we bias the base ? Base Current or Base Voltage?
Also, Kindly point me for any resource, that explains selection of base , Collector and emitter resistances.
 

Thread Starter

Vignesh D

Joined Feb 18, 2019
6
hi VD,
Welcome to AAC.
Is this a college or homework assignment.?
E
Neither of them. I am trying to make my own function generator for which I am trying to bias BC107 with certain input and output resistances. I can't simply go for values used in Internet. So I am trying to design my own CE amplifier.
 

Alec_t

Joined Sep 17, 2013
12,009
Which of the above two approach is correct ?
Both are 'correct', but method 1 requires an additional voltage source so is not used as frequently as method 2.
The resistor values would depend on the intended use of the CE stage. Do you want it to work as a class A, class B or class C amplifier stage? Is it required for medium power, low power, high frequency or low frequency amplification? What impedance load will it drive?
 

Thread Starter

Vignesh D

Joined Feb 18, 2019
6
Both are 'correct', but method 1 requires an additional voltage source so is not used as frequently as method 2.
The resistor values would depend on the intended use of the CE stage. Do you want it to work as a class A, class B or class C amplifier stage? Is it required for medium power, low power, high frequency or low frequency amplification? What impedance load will it drive?
I require Class A Operation, with high input resistance (In Range of MOhms) with Max Frequency of operation at 1MHz and High Power.
 

MrChips

Joined Oct 2, 2009
23,268
You need to calculate for both base current and voltage.

In simple terms, here are the steps to designing a simple CE BJT amplifier.



1) Choose your supply voltage Vcc (e.g. Vcc = 5V)
2) Select your collector load resistor (e.g. Rc = 4700Ω).
3) Choose your voltage gain, (e.g. Av = 10).
4) Calculate emitter resistor RE = Rc / Av (e.g. RE = 4700 / 10 = 470Ω)
5) Calculate max collector current = Vcc / ( Rc + RE) = 5 / 5170 = 1mA
6) Calculate max voltage across RE, (e.g. 1mA x 470Ω) = 0.47V
7) Take half of this voltage (e.g. 0.47 / 2 = 0.23V = VBE)
8) Add one diode voltage drop to this (e.g. 0.23 + 0.65 = 0.88V). This is your base bias voltage = VB.

9) Take your collector current (e.g. 1mA) and divide this by 10, (e.g. 1mA / 10 = 0.1mA)
10) Calculate the total voltage divider resistance (R1 + R2) = 5V / 0.1mA = 50kΩ
11) Use this as a starting point for base current drive R1 = 47kΩ
12) Calculate R2 to provide the required base bias voltage (e.g. 0.88V / 5V = R2 /( R1 + R2). This gives R2 = 10kΩ)

13) Insert input DC blocking capacitor C1 (about 10μF) on input to base.
14) Apply AC signal to input capacitor C1.
15) Adjust R1 and R2 for minimum distortion at collector.

btw, for consistency, I would have preferred to relabel the resistors as R1 and R2 on the base, R3 on the collector and R4 on the emitter.

Edit: I have relabeled the resistors to R1, R2, Rc and RE.
 

Jony130

Joined Feb 17, 2009
5,230
Use the secend circuit with Re resistor and a voltage divider.
Why BJT amplifier ? Why not to use a opamp instead?
 

Thread Starter

Vignesh D

Joined Feb 18, 2019
6
Use the secend circuit with Re resistor and a voltage divider.
Why BJT amplifier ? Why not to use a opamp instead?
I have too many BJT's lying simply in my collection :D. Instead of buying Opamp, I am trying to use whatever I have in hand.
 

Thread Starter

Vignesh D

Joined Feb 18, 2019
6
You need to calculate for both base current and voltage.

In simple terms, here are the steps to designing a simple CE BJT amplifier.

1) Choose your supply voltage Vcc (e.g. Vcc = 5V)
2) Select your collector load resistor (e.g. R4 = 4700Ω).
3) Choose your voltage gain, (e.g. Av = 10).
4) Calculate emitter resistor R1 = R4 / Av (e.g. R1 = 4700 / 10 = 470Ω)
5) Calculate max collector current = Vcc / ( R1 + R4) = 5 / 5170 = 1mA
6) Calculate max voltage across R1, (e.g. 1mA x 470Ω) = 0.47V
7) Take half of this voltage (e.g. 0.47 / 2 = 0.23V)
8) Add one diode voltage drop to this (e.g. 0.23 + 0.65 = 0.88V). This is your base bias voltage

9) Take your collector current (e.g. 1mA) and divide this by 10, (e.g. 1mA / 10 = 0.1mA)
10) Calculate the total voltage divider resistance (R3 + R2) = 5V / 0.1mA = 50kΩ
11) Use this as a starting point for base current drive R3 = 47kΩ
12) Calculate R2 to provide the required base bias voltage (e.g. 0.88V / 5V = R2 /( R2 + R3). This gives R2 = 10kΩ)

13) Insert input DC blocking capacitor (about 10μF) on input to base.
14) Apply AC signal to input capacitor.
15) Adjust R2 and R3 for minimum distortion at collector.

btw, for consistency, I would have preferred to relabel the resistors as R1 and R2 on the base, R3 on the collector and R4 on the emitter.
Thank you. I will try this and get back to you. The image I posted is simply for demonstration. I did not select any name or value for it.
 

crutschow

Joined Mar 14, 2008
27,017
I require Class A Operation, with high input resistance (In Range of MOhms) with Max Frequency of operation at 1MHz and High Power.
That's difficult to do with a one transistor Class A stage.

Why such a high input impedance? BJT amps usually don't go that high.

Define "High Power".
 

Audioguru

Joined Dec 20, 2007
11,249
I was taught to NEVER bias the base of a transistor with a current from a single resistor because transistors have a wide range of current gain (hFE or beta). A transistor with a high hFE would be saturated and a transistor with a low hFE would be cutoff.
The voltage divider at the base and an emitter resistor which might have a bypass capacitor will work with all transistors of one part number.

For a high input resistance use an emitter-follower at the input. For high power use NPN and PNP complimentary emitter-followers at the output.
the class-A circuit you showed is in between.
 
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