Hello friends, I hope you all are doing great. In today’s tutorial, we will have a look at Transistor BJT Voltage Divider Bias. The most common and normally used method for biasing a transistor is a voltage divider bias circuit. It consists of some resistances for division or voltages and distribution among resistance at a proper level.
Voltage divider bias also is known as emitter current bias. In today’s post we will have a detailed look at its circuit, working and other related factors. You will also learn as biasing linear transistor operation through using a single-source resistance voltage divider. So, let’s get started with Transistor BJT Voltage Divider Bias
Transistor BJT Voltage Divider Bias
- In below figure you can see the circuit which using VCC as a single bias source.
- To make simple representation for VCC instead of battery symbol a line with a circle is shown.
- The dc bias voltage at the base terminal of the transistor can be provided through resistive voltage divider circuit which has two resistances denoted as R1 and R2.
- VCC is a dc collector supply voltage source. There are 2 paths for current between point A and ground one from the resistance R2 and second from the base-emitter junction of transistor and resistance RE.
- Usually, voltage divider bias circuitry is created as IB is less than the current flowing through the resistance R2 that is I2.
- After finding the value of VB you can determine the voltage and current in the circuitry as.
VE =VB -VBE
VC =VCC -ICRC
- After calculating VC and VE you can find VCE.
VCE = VC – VE
Loading Effects of Voltage-Divider Bias
DC Input Resistance at the Transistor Base
- The value of dc input resistance of the transistor is directly proportionate to the βDC so its value will be different for a different transistor.
- When the transistor functions in its linear region the emitter current will be βDCIB.
- When we observed the emitter resistance from the base circuitry the resistance looks greater than its real value due to dc current gain of a transistor.
RIN(BASE)= VB/IB = VB /(IE/βDC).
- It is effective to load on the voltage divider circuit as shown in the figure.
- You can rapidly guess the loading effect by relating RIN(BASE) to the resistor R2 in the voltage divider.
- If RIN(BASE) is at least 10 times greater than R2, the loading effect will be ten percent or less and the voltage divider is stiff.
- If RIN(BASE) is less than 10 times R2, it must be connected in parallel with R2.
Thevenin’s Theorem Applied to Voltage-Divider Bias
- Now we will apply Thevenin theorem to for analyzation of voltage divider biased transistor circuit for loading effect of base current.
- First take an equivalent base-emitter circuitry shown in figure denoted as (a)by applying Thevenin theorem.
- By observing from the base terminal bias circuitry can reconstruct as denoted in figure as (b).
- Apply thevenin theorem on left side of point A with VCC substituted with a short to ground and removed the transistor from the circuitry.
- The value of the voltage at point A with respect to ground terminal is given as.
the value of resistance is
- The Thevenin correspondent of the bias circuitry, associated to the transistor base, is shown in the beige box denoted as (c).
- By applying KVL we get.
VTH – VRTH – VBE – VRE = 0
- By applying ohm law we have.
VTH = IBRTH + VBE + IERE
- By putting IE/βDC for IB
VTH = IE(RE + RTH/βDC) + VBE
- By solving for IE we have.
- If the value of RTH/βDC is less than the RE the output is like to the voltage divider without load.
- Voltage-divider bias is extensively used due to fine bias stability is obtained with a single supply voltage.
Voltage-Divider Biased PNP Transistor
- We know that for PNP transistors there is a need for opposite biasing polarities. It can be obtained through the negative collector supply voltage as denoted in figure by (a).
- It can also obtained through providing positive voltage to the emitter as shown in the figure by (b).
- The analyzing process is alike to the NPN transistor circuit applying Thevenin and KVL as described in these steps according to given figure.
- After applying KVL about the base-emitter circuit you have.
VTH + IBRTH – VBE + IERE = 0
- Applying Thevenin theorem.
Base current is given as.
- The equation for IE is given as.
- For circuit denoted as (b) the analysis is given as.
-VTH + IBRTH – VBE + IERE – VEE = 0
VTH =(R1/R1 + R2)VEE
RTH=R1R2/R1 + R2
IB = IE/βDC
- The equation for emitter current will be.
IE=VTH +VBE-VEE/(RE +RTH/βDC)
So, friends, it is a detailed post about Transistor BJT Voltage Divider Bias if you have any questions ask in comments. Thanks for reading. See you in next post have a good day. thanks for reading.
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