Electronic Circuits. Laboratory 1 - Solution

Transcription

1 Institut für Integrierte Systeme Integrated Systems Laboratory Autumn Semester 2016 Electronic Circuits Prof. Dr. Qiuting Huang Laboratory 1 - Solution and Last Update: Task 1 Bipolar Transistor in Common Emitter Stage Task1 Simulation Large Signal Analysis Determine the operating point of the circuit. Try to first calculate the bias voltage V B (assume I B << I B1 ), then calculate V E (assume V BE = 0.7 V) and finally V out (assume I C I E ) V B = 2V CCR B2 R B1 + R B2 = 0.9 V V E = V B V BE = 0.2 V I E = V E R E = 1 ma V out = V CC I C R L V CC I E R L = 3 V How big will the DC offset be? How big is the maximum voltage swing? How could you improve the swing? DC Offset = V out V swing = V CC V out = 2 V Improve swing: increase V CC, i.e. increase R L or decrease R E. Small Signal Analysis Draw the small signal equivalent of the circuit (Note: the small signal equivalent should include C) See Figure 1.

2 Figure 1: Small Signal Solution C, R B1 and R B2 at the input stage represent a high pass filter in the small circuit equivalent. Compute the cut-off frequency f C in Hz of the high pass filter The high-pass has been marked in blue in the small signal equivalent. Its cut-off frequency is: 1 f c = 5.3 khz 2π(R B1 R B2 )C Compute the small signal voltage gain A V at a frequency above f C. This circuit (without the high-pass filter) has been discussed during the first lecture, check the corresponding slides for more information. As given in the hint, the transconductance of the transistor-r E combination can be expressed as G m = i out v in, and consequently A V = v out v in The transconductance g m is defined as = i outr L v in = G m R L g m = I C /V T = 1 ma = 38.5 ms 26 mv using the I C calculated during the large signal analysis. This leads to G m = 4.4 ms and A V = A V = db 2

3 Note that the simplified formula A V = R L R E = 10 leads to an approximation error of about 10 %. In order to set the amplitude for the signal source, calculate the maximum voltage for which the output is not clipped. Task 1 Circuit Simulation V in,max = V swing /A V = 0.34 V Have a look at the multimeter display and check if the analytically calculated V B matches the measured one. V B,measured = 0.9 V Have a look at the two signals on the oscilloscope display. How big is the DC offset? Does the value match with your expectations? If no, why not? DC Offset = 2 V This does not match with the calculated value. This is mostly due to the fact that the assumption V BE 0.7 V is inaccurate in this case. For this transistor model V BE 0.6 V. This leads to the following values: V B = 0.9 V, V E = 0.3 V, I c = 1.5 ma, V out = 2 V, g m = 57.7 ms, A V = db, maximum swing = 3 V, maximum input voltage = 0.32 V. 3

4 Set both channels to AC coupling. How big is the swing of the output voltage? Is this swing reasonable given the calculated A V? Simulated swing is 1.8 V, this is reasonable given the input swing of 0.2 V and A V = 9.2. Have a look at the transfer characteristic on the Bode plotter display. Adapt the vertical settings if you are not able to see the graph for the entire frequency span (10 Hz khz). Compare cut-off frequency calculated analytically to the cut-off frequency in the Bode-plot. Compare the calculated numerical value of A V to the gain you can read out from the Bode plot. Simulated cut-off frequency = khz, simulated A V = 19.2 db. Task 1 Hands On Open the Digital Multimeter from the Instrument Launcher and measure V B and V BE. Do the measurments coincide with the simulation? V B = 0.9 V V BE = 0.6 V Open the Oscilloscope from the Instrument Launcher and connect channel 0 with v in and channel 1 with v out. Connect the clips of the probes to the ground which is located next to the supplies. Set the scaling according to the previous task. What is the DC offset? DC Offset = 1.6 V Change both channels to AC coupling. What is the swing of the output signal? Swing = 3 V 4

5 Raise the amplitude of the input signal. Describe how the output signal changes. Give an idea why this is happening. The gain starts to decrease. As soon as the output signal (incl. DC offset) reaches +5 V the signal gets distorted because the transistor cannot amplify the signal above the supply (+5 V). Figure 2: V in = 0.34 V Figure 3: V in = 1.5 V, output signal is distorted (cannot go above +5 V or below 5 V). Open the Bode Analyzer from the Instrument Launcher. In order to be able to record a Bode plot you need to stop the function generator and the oscilloscope. Record a Bode plot from 100 Hz to 200 khz. Set the peak amplitude to the maximum input voltage and the steps per decade to 10. Compare the Bode plot with the Bode plot of Task 1. What differences do you observe? Where is the jump in phase coming from? 5

6 The simulation and the measurements are almost identical (A V,measured = 19.6 db). The jump is just a matter of displaying the phase. The Bode Analyzer will automatically wrap the phase at ±180. 6