1. Calculate all voltages and currents in this circuit. hfe = 200

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1 BJTs Note: the symbol for transistor I used in this document is wrong. It actually belongs to a IGBT (insulated- gate bipolar transistor, BJT with isolated gate). I did not have a BJT symbol in the software I use to draw schematics. 1. Calculate all voltages and currents in this circuit. hfe = 200 If you do not know where to start find: VE, IE, IB, IC, VC in this order. Remember that B E behaves as a diode with fixed voltage drop. Check if VC > VE. VE = VB VBE = = 3.3V IE = VE/Re = 3.3/165 = 20mA What we know about the transistor: IE = IB + IC IC = IB hfe Combining these two we get: IE = IB + IB hfe IE = IB(1 + hfe) IB = IE/(1 + hfe) = 0.02/(1+200) = 99.5μA IC = IB hfe = 19.9mA Voltage drop on RC VRc = IC RC = 4.38V So VC is VC = 10 VRc = 5.62V Check if VC > VE. This is true. Transistor is operating normally.

2 What happens at RC = 500Ω? Recalculate all currents and voltages for this value. Check if VC > VE. For RC = 500Ω we need to redo final equations. Voltage drop on RC VRc = IC RC = 9.95V So VC is VC = 10 VRc = 0.05V Check if VC > VE. This is not true. Transistor is saturated. This means that equation IC = IB hfe is not true any more and all our calculations are wrong. Transistor tries to pull 20mA through the collector but RC does not permit it by limiting the current. If we slowly change the RC value at around 326Ω the VC will reach 3.5V. This means VCE is now 0.2V. It cannot be lower because we operate in saturation now. Use a simulator so find the maximum RC value. Sweep RC resistance and see what happens with IC. problem- 1/

3 As you can see up to 326Ω the collector current is 20mA. Above this resistance the collector current drop as transistor goes into saturation. 2. Each of the transistors in this circuit has hfe = 20. The circuit behaves as a single transistor. What is the hfe of this equivalent transistor? Imagine we inject 1mA into base of the first transistor. IC of this transistor is 20mA. Emitter current is 21mA. This current flows into the base of the second transistor. Collector current of this second transistor is = 420mA. Therefore a 1mA base current yields IC1 + IC2 = = 440mA current. The gain is equal to hfe1 hfe2 + hfe1 + hfe2 This is a Darlington configuration. 3. Analyse this circuit. It looks like an amplifier but is hat two outputs. How do these outputs look when you connect a sinusoidal input signal? What is the purpose of this circuit? It is really necessary that RC = RE. Why? phase- splitter/ Use time domain simulation to see how this circuit works. IC is approximately equal to IE. Because RC = RE are equal it means the voltage drops on these resistors are equal. If voltage drop on these resistors increase VE increases and VC decreases. Therefore each signal you put on the input is going to be split into two outputs with opposite phases. Try changing the input voltage amplitude. See how the output voltages will look like. If you increase the peak input voltage above 5V what will happen?

4 How can we use this? Therefore you can use a differential line in high- speed communication in noisy environment.

5 4. Calculate all voltages and currents in this circuit. hfe of both transistors is 100. Solution to this problem is in the book. I put it in the Moodle. 5. How does this circuit work? Amplifier tries to equalize potentials on its inputs. Voltage on the resistor is then equal to Vin. Therefore resistor current is Vin/100Ω. What is the relationship between Vin and load current? If gain is very high then Iload = Vin/100Ω (ignogint gate current) Does the operation of this circuit depend on hfe? What happens if we use a different transistor with bigger hfe? It does not depend on it. It also does not depend on VBE variation. Will this circuit operate with Vin = 5V? No, because VBE is then reverse biased.

6 6. How does this circuit work? How does it differ from push- pull? Normal push- pull will have problems when Vin is close to 0V because we need 0.7V to bias the diodes. In this circuit diodes D1 and D2 are always forward bias because of R1 and R2. Therefore when input voltage is 0V then base voltages are actually 0.7V and 0.7V, enough to bias the base emitter barrier. So transistors are always ready to conduct.