The Bipolar Transistor: S&S pp. 485-497Short-Circuit Current Gain:Cut-Off Frequency of the Current Amplifier (RL=0?):PSpice Comparison with FormulaComparison of Current Amplifier Current Gain with PSpiceCurrent amplifierSummary:Miller Effect: The Frequency Response of the Transresistance Amplifier (RL=( ?)?PSpice Comparison
Gain-Bandwidth Trade-off: S&S, p.93, pp. 558-559Summary:Appendix 1: PSpice output fileAppendix 2: Using PSpice to obtain fT vs. IEDiscussion of circuitCircuit for the Q2N2907AResultsCheck on the circuitReferenceECE 304: Bipolar Capacitances The Bipolar Transistor: S&S pp. 485-497 Let’s apply the diode capacitance results to the bipolar transistor. There are two junctions in the bipolar transistor. The BC (base-collector) junction is reverse biased in the active mode, and so it has only a junction capacitance contribution to the equivalent circuit, CjC. The EB (emitter-base) junction, on the other hand, is forward biased in the active mode. Therefore, it will exhibit both a junction capacitance CjE and a transit time capacitance CtE. If we look at the simplified version of the bipolar, we can see how to hook these capacitors into the circuit: E B C CjECjCCtE Putting this into the regular hybrid π model we show how to obtain S&S Fig. 5.67, p. 487. The parasitic base resistance rb has been added. This resistor is small (about 10Ω or less). However, r dE CjCCjE + CtEI b β Ib r O + V π -- r b B E in circuits where the source driving the transistor has very low resistance, rb is significant for frequencies high enough that CjE+CdE short-circuits rdE. Then rb is the only impedance in the base lead of the circuit. Also, the presence of rb activates the Miller effect (discussed later) increasing the role of CjC in the frequency response. There are some changes in notation needed to obtain the standard circuit. First, we re-label a few items: CjC → Cµ, (CjE+ CtE) → Cπ, rdE → rπ. Then we notice that the dependent current source depends on only the part of the base current that flows through rdE= rπ (labeled here by phasor amplitude Ib). That is, to use this circuit we have to worry about the current divider made up of (CjE+ CtE) → Cπ and rdE → rπ. This worry is a nuisance, so we prefer to use the voltage Vπ instead. Because the current in rdE = rπ is simply Ib = Vπ/ rπ, the dependent source can be re-expressed as βIb = βVπ/rπ = gmVπ, where the transconductance gm is defined by: gm = [∂ ιC / ∂ υBE]Q = ICQ/VT = βIBQ /VT = β/rπ. Making these changes, the circuit is as shown in Figure 1 below. Unpublished work © 2004 John R Brews Page 1 3/1/2005r b r π Cµ CπgmVπ r O + V π -- FIGURE 1 Hybrid π model with internal capacitances Cπ and Cµ Short-Circuit Current Gain: Read S&S pp. 487-488. We will look at the circuit of Figure 1 with a resistive load in parallel with rO and an ac current source as signal. There are two limiting cases with very different frequency response: the case when the load RL=0Ω (current amplifier) and the case when RL=∞ Ω (transresistance amplifier). Cut-Off Frequency of the Current Amplifier (RL=0Ω): If we hook up our circuit with a short-circuit from the collector to the emitter and neglect rb, we can find the current gain of the circuit A i = (ac collector current)/(ac input current) (S&S hfe, following Eq. 5.158). EQ. 1 πµπµµππµ+ω+ω−=+ω+ω−===r)CC(j1gCj1β)CC(jr1CjghIIAmacmfebci The cut-off frequency fT of the bipolar transistor is defined as the frequency where this current gain drops to unity: that is it is the frequency where there is no gain. The time constant Cµ/gm is very short, so the numerator is well approximated by 1. Setting |A i| = 1, we can solve for the value of fT. Taking βac >> 1 and Cµ/Cπ << 1 (valid if IC is large, but not at low values of IC) we find to lowest order in Cµ/Cπ (S&S Eq. 5.163): EQ. 2 ()()[])CC(2gC2CCr21fm21212acTµπµπππ+π≈+π−β=//. Now let’s put in the current dependence of all the parameters to see how fT depends on the current level. We have gm = ICQ/VT, 1/rπ = IBQ/VT, EQ. 3 Cπ = CjE +IE τF /VT (where we have introduced the forward transit time τF to replace the transit time τT of the pn diode). Approximating IC >> IB we find: EQ. 4 mjEFCQTjEFTgCCIVCCf21)()(µµ++τ=++τ=π That is, if we plot 1/(2πfT) vs. 1/ICQ we should get a straight line with intercept τF. Actually we get a plot like Figure 2 below. Unpublished work © 2004 John R Brews Page 2 3/1/20051/(2πfT) 1/I C τF CjE FIGURE 2 Sketch of EQ. 4 for 1/(2πfT) (dashed line) and more realistic behavior (solid line) In Figure 2, the dashed line is our EQ. 4, the one we will use in hand analysis. The solid line is closer to what is actually observed (and approximated by PSPICE). The departure from our analysis at low 1/IC (large IC) is due to the widening of the electrical base width at large current densities, a subject discussed in ECE 352.1 The point is this: as the collector current varies, the frequency response varies. If the current is low enough (1/IC is big enough) the variation is almost a straight line. But at large currents (small 1/IC) the frequency response begins to get worse (fT drops, 1/fT goes up). PSpice Comparison with Formula If we set the parasitic resistances of the collector rb and of the base rc to zero, we can easily obtain the PSPICE results for Cπ as a function of DC collector current to compare with EQ. 3. For the Q2N2222 the results are shown in Figure 3. 3.07E-111.E-111.E-101.E-091.E-081.E-07 1.E-06 1.E-05 1.E-04 1.E-03 1.E-02 1.E-01 1.E+00DC Collector Current (A)Cπ (F)PSpiceEQ.3Q2N2222 FIGURE 3 Cπ vs. ICQ for Q2N2222 with τF=0.411ns and rb and rc≈0 The results in Figure 3 are obtained by a small-signal AC analysis using the circuit of Figure 4 1 Ask your 352 instructor about this phenomenon Unpublished work © 2004 John R Brews Page 3 3/1/2005PARAMETERS:I_B = 1uAV_CB = 1Cjc = 7.306pRb = 1uRc = 1uACSweepIBac1A0Q2Q2N2222+-CB{V_CB}I2{I_B}I-++-E4GAIN = 1 B2C2FIGURE 4 Circuit for finding Cπ In the circuit of Figure 4 an AC base current is driven into the Q2N2222. The resulting AC base voltage is fed to the collector by VCVS part E, so both base and collector move together, placing zero AC voltage across Cµ so it draws no AC current and does not affect the input impedance of the transistor. The AC base current is given by EQ. 5 Ib = Vπ (1/rπ