Tutorial #4: Bias Point Analysis in Multisim


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1 SCHOOL OF ENGINEERING AND APPLIED SCIENCE DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING ECE 2115: ENGINEERING ELECTRONICS LABORATORY Tutorial #4: Bias Point Analysis in Multisim INTRODUCTION When BJTs are used in actual circuits, there is typically a single voltage supply. This is due to where the transistor will be used. For example, in a cell phone there is just one battery, perhaps 3.3V, in a car a 12V battery, in an MP3 player 1.5V, etc. We would like to use two separate voltage sources to bias our transistor, as shown in the example circuit in Figure 1, a 5V source for the base, and a 15V source for the collector and emitter. However, due to the fact that we have only one voltage source, say it is 15V, we need to apply the proper voltage to the base and the collector from only one single voltage supply, as shown in Figure 2. This tutorial will show that Figure 1 and Figure 2 apply the same voltage to the base, collector, and emitter. Most importantly, all of the currents (, I C, and I E ) will be identical for each circuit. RTH 33. VTH 5 V Figure 1 Circuit to Bias Transistor Q 1 Using Two Voltage Sources Figure 2 Circuit to Bias Transistor Q 1 Using One Voltage Source The circuit in Figure 2 is sometimes called a Beta Stabilizer because the value for β is fixed, since and I C will be fixed at a particular value. The example used in this tutorial follows Example 6.10 on p. 390 in the Sedra/Smith Microelectronic Circuits textbook, except that no steps are skipped. 1
2 INSTRUCTIONS Part I Hand Calculations I C Given Information β = 100 Equations true for any BJT: I E β = I C I E = + I C (DC Current Gain) (KCL) Figure 1.1 Typical Bias Circuit for a BJT In the circuit in Figure 1.1, we are typically asked to find the voltage and current at each node, specifically:,,, I C,, and I E. 1. We do not know the voltage or the current, so we start by finding the Thévenin equivalent for C, R B1, and R B2. a. Recall that Thévenin s Theorem states that any combination of voltage sources, current sources, and resistors with two terminals is electrically equivalent to a single voltage source and a single series resistor. We can see from Figure 1.1 that R B1 and R B2 have 15V across them. So, we redraw the first part of the circuit as follows: + VBB  Figure 1.2 Simple Voltage Divider Where B = 5V R B2 B = ( ) V R B1 + R CC = ( ) 15V = 5V B2 + 2
3 b. From Thévenin s Theorem we know that to find the Thévenin resistance, we short the voltage source and find the equivalent resistance R BB. R BB is simply two resistors in parallel: R BB Figure 1.3 Finding R TH R BB = ( R B1R B2 R B1 + R B2 ) = ( ()( ) + ) = 33. c. The next step is to put this all together and replace R B1 and R B2 with the Thévenin equivalent: I C RBB VBE  VTH 5 V L 1 I E Figure 1.4 Simplified with Thévenin Equivalent 2. The next step is to use the Thévenin equivalent circuit to determine, I E, and I C. a. In Figure 1.4, we can use KVL on Loop L 1 to determine and I E : Note: We must assume that the BJT is in its active region, meaning that the device is ON. Because the BaseEmitter junction is essentially a diode, when it is turned ON, it has roughly a 0.7V drop across it. In Lab 5, you determined the actual value for E. Looking back, you will see it was approximately 0.6V0.7V, so this is a valid assumption. This assumption makes it so that E = 0.7V in the equation we derived from KVL. B = R BB + E + I E R E 5V = ( (33.)) + 0.7V + (I E ()) 3
4 b. We do not yet know or I E, but we remember that for a BJT, KCL states that the current going into the BJT must equal the current coming out of the BJT: + I C = I E c. We also remember that DC Current Gain for a BJT is called β (beta): β = I C d. Using these two facts, we can plug one equation into the other and put emitter current (I E ) in terms of base current ( ). I E = (1 + β) e. Now we can substitute this expression for I E back into the KVL equation for Loop L 1. 5V = ( (33.)) + 0.7V + ((1 + β) ()) f. Since β is given to be 100 for this BJT (in the active region of operation), we can solve for in the equation above: 5V = ( (33.)) + 0.7V + ((1 + β) ()) 5V = ( (33.)) + 0.7V + ((101) ()) 5V 0.7V = ( (33.)) + ( (30)) 4.3V = ( ) = 12. 8μA g. Once we have one current, we can easily find the other two: I E = (1 + β) I E = (101)(12.8μA) I E = 1. 29mA I C = β I C = (100)12.7μA I C = 1. 28mA 3. Using the currents found above (, I E, and I C ) and the size of the resistor s R BB, R E, R C, the last step is to use Ohm s law to determine,, and. a. From Figure 1.4, find,, and using Ohm s Law: = B R BB = 5V (12.8μA)(33.) = 4. 57V = I E R E = (1.29mA)() = 3. 87V = C I C R C = 15V (1.28mA)() = 8. 61V b. From,, and we can find E, E, and B, and we can see that E = 0.7V, which matches the assumption that we made earlier. E = E = 4.57V 3.87V E = 0. 7V 4
5 Part II Bias Point Analysis in Multisim To verify that our hand calculations are correct, we perform what is called a bias point analysis in Multisim. This will allow us to view the node voltages and currents in our Beta Stabilizing Network. Figure 2.1 Beta Stabilizing Network 1. Build the circuit in Figure 2.1 in Multisim. a. Rename the wire nets at,, and to B, C, and E, respectively by doubleclicking on the wires and entering the preferred net names. This will make it easier to select and view the desired outputs. 2. Select a DC Operating Point analysis and output, I C, I E,,, and as shown below. Figure 2.2 DC Operating Point Analysis Settings 5
6 3. Run the simulation and the following table should appear listing the desired DC values. Figure 2.3 Bias Point Analysis Output Note: Notice that our currents and voltage do NOT match our hand calculations! This is because the BJT has a β equal to about 150 for the values of I C and for this circuit s configuration. The β used in the Sedra example was only 100. In our labs we will only use the, so in our hand calculations for the, we will use β = 150, and our calculations will then match the simulated results. 6
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