EECE 312L Notes Experiment 11 Notes, Fall 2013

Experiment 11 Bipolar Junction Transistor BJT Characteristics:  We first start by connecting the circuit of figure 1 which will be used to determine the characteristics of the 2N222 BJT, and its different regions of operations.  We start with VBB = 1 V, and we increase VBB to 8 V in steps of 0.2V. While measuring the collector current IC, the Base current IB, the collector-to- emitter voltage VCE, and the base-to-emitter-voltage VBE. In the active or the saturation region: 0.7< VBE <0.8 In the saturation region: VCE= 0.2 V In the saturation region: VCE> 0.2 V How do we find β?  Vary IB with a certain range and measure VCE, IC and VBE  Divide VCE by VBE  Find β= IC/ IB  BJT as a Switch:  We first start by determining the values of RB and R of figure 2 so to meet the following specifications: When VIN = 5V, the transistor is fully saturated with IC/IB = β/10 and with a LED current of 10 mA (LED drop voltage is 2V).  Then we start varying VIN from 0 to 2 V in steps of 0.2 V and we measure Vout Then we increase VIN from 2 to 5 V in steps of 0.5V and we measure Vout. WE also note if the LED is glowing and for what values of VIN. Switch:  ON → SATURATION  OFF→ CUTOFF The drop across the BJT is very small. if it increases then there would be too much heat and thus the transistor would burn out. The drop across the transistor could be calculated using the formula: (VCC- o.2). If we divide this value by the current I we calculate the resistor’s value Rc. General Remarks:  A transistor is a current controlled device.  In a transistor the base current defines which region we would be working in.  A MOSFET is a voltage controlled device.  We usually take a constant value of β throughout the experiment. How do we measure IB? At the edge of saturation or in the active region only IC= β IB If we use a value of β equal to 100 we might burn the transistor. Thus we have to increase β by a factor of 10 for example (usually) and thus we would have β ‘ = β /10 The β force is the factor that forces the transistor to go to saturation by β ‘ = β /10. Thus the current IB would increase and IC would remain fixed with IC= (β /10) IB the RB could be calculated by RB= (VCC - 0.7)/ IB calculated. Connect the circuit in figure 2. Vary VIN from 0 V to 2 V in steps of 0.2 V and measure VOUT (by the oscilloscope). Increase VIN from 2 V to 5 V in steps of 0.5 V and measure VOUT. We also note the value of VIN at which the LED starts glowing.  BJT as an Amplifier: Why is the DC analysis important?  Point of operation Q (we need to get Q)  Place in small signal model and analyze DC measurements: gm is needed to place the transistor in the active region that is to check if the transistor is in the active region. VCE>0.3 VCC = 1/3 VCC Connect the circuit of figure 3- c The voltage source is a 100 mV peak-to-peak, 10 KHz sine signal (by the function generator) Measure off-circuit the values of the resistors RS and RL. 1. Measure the output voltage across the load resistor RL. Calculate the voltage gain vO/vS. Note: Even thought the ratio vO/vS is less than 1, we still refer to it as gain. When we connect the BJT amplifier block, the ratio will become much larger than 1. Vo is (Vs – Vi)/VCC - 0.3 And Vi = Vo/RL Find Av= Vo / Vi  As we can see in the first 2 steps the gain does not change though Vin is much bigger in step 2 then in 1 . This is because the gain does don’t depend here on the amplitude of Vin . In fact we have that Vout = (RL / (Rs + RL)) Vin, so the ratio Vout / Vin is equal to RL / (Rs + RL) which is a constant (here it is equal to 0.41). As we can also see for low frequencies the gain is approximately equal to RL/(Rs + RL) and for very high frequencies the gain is much more less then RL / (Rs + RL). This is due to coupling capacitance in the circuit.  BJT as a Current Source: Why do we use a Zener diode? We use a Zener diode to maintain VBE (VB- ground) at a certain voltage. That is to hold a constant voltage between the base and the ground. We could use a Zener diode or a battery. IB should be constant then should be constant as well  Connect the circuit shown in Fig. 4-a. Start with RC = 0 (short circuit). We measure the base voltage and the emitter voltage. Also measure off-circuit the value of the 4.7 KΩ emitter resistor. Increase the value of RC from 0 to 12 KΩ using 1 KΩ, 1.5 KΩ, 2.2 KΩ, 2.7 KΩ, 3.3 KΩ, 4.7 KΩ, 5.6 KΩ, 6.8 KΩ, 8.2 KΩ, 10 KΩ, and 12 KΩ resistors, and measure the collector current of the transistor and the collector-to-emitter voltage. Note: RC should be measured off-circuit. General Shape of the BJT