1. Objective:
To design, simulate and analyze a biasing circuit with the current value Ic = 1mA2. Theory:
Fig 1 Voltage divider Transistor biasing
This is a simple circuit of BJT where common emitter is biased using the voltage divider network. The main objective of having resistors is to make this transistor work just by single
voltage source i.e VCC . Here, in the circuit the resistor RB1 and RB2 is connected series which acts as voltage divider thus provides a base voltage VB . A designer can adjust VB by picking suitable value for the resistors RB1 and RB2 .
We set Vcc to 10V. RB1 = RB2 = 2K Using voltage division,
VB = ( RB2*Vcc)/(RB1+RB2)----I
VB = (2K*10V)/(2k+2k)=5V
Also for equivalent resistance,
RTH = ( RB1*RB2)/(RB1+RB2)----II
We calculated the value of RE to be 4.2 K and we picked the value of RC. See figure 6 7 8 for hand calculation and for more details
RTH = (4K*4k)/((4k+4k)=2K
RTH =2K
Now, Rth servs as a base resistor RB and Vth serves as base voltage VBWe calculated the value of RE to be 4.2 K and we picked the value of RC. See figure 6 7 8 for hand calculation and for more details
3. Equipment Used
- Breadboard
- 2N2222 NPN Transistor
- DC power supply
- Wires
- Resistors
- LT Spice IV
4.Schematics and simulation:
5. Results and Discussions
The figure 2 can also be transformed into simple configuration. For this process we have to find the Thevenin voltage which acts as a base voltage which was calculated using equation I. Then we calculated the Rth resistance which is an equivalent resistance using equation II. Now, all the parameters are known we simulated both circuits as shown in the figure 2 and figure 4. Both of these circuits yield same result i.e 1 mA current across the collector terminal for the simulation.On the lab we build figure 2 circuit. We connected a multimeter in series with the resistor RC so we could read current flowing through the resistor R
IC = (Vcc-Vout)/(rC)------III
Table 1 shows the data we have collected though simulation, hand calculation and measured value for figure II on the lab which is shown below.
Table 1: Simulated, Calculated and Measured values
Hand Calculation | Simulation | Lab Result | ||
---|---|---|---|---|
Vin | 5 V | 5 V | 4.7 V | |
Vout | 7 V | 7.06 V | 7.03 V | |
I C | 1 mA | 0.00097 mA | 1.01 mA |
Ic1=(Vcc-Vout)/RC
Ic1=(5-0.658)/4.3k
Iref = Ic1= 1.0097 mA
Ic1=(5-0.658)/4.3k
Iref = Ic1= 1.0097 mA
The table 1 shows that the values are pretty close to each other. The error margin is very small. It is due the fact that instruments and the components that we used were not ideal.