Lecture 6: Transistors Amplifiers. K.K. Gan Lecture 6: Transistors Amplifiers

Transcription

1 Lecture 6: Transistors Amplifiers ommon mitter Amplifier ( Simplified ): What's common (ground) a common emitter amp? The emitter! The emitter is connected (tied) to ground usually by a capacitor To an A signal this looks like the emitter is connected to ground. What use is a ommon mitter Amp? Amplifies the put voltage (the voltage at the base of the transistor). The output voltage has the opposite polarity as the put voltage. We want to calculate the followg for the common emitter amp: oltage Ga out / Input Impedance Output Impedance

2 D oltage Ga: The voltage ga we are about to derive is for small signals only. A small signal is defed here to be the range of a few m. As all of what follows we assume that the transistor is biased on at its D operatg pot. out = cc I Sce cc is fixed (its a D power supply) we have for a change output voltage out Δ out = ΔI Δ stands for a small change either the voltage or current. The put voltage is related the emitter voltage by a diode drop: = B = Δ = Δ We want to relate the emitter voltage to the emitter current (I ): = I Δ = ΔI We can relate the emitter and collector currents by rememberg that for a transistor: I I ΔI ΔI Δ = ΔI = ΔI Δ = Δ = ΔI = ( Δ out / )

3 D voltage ga (G) for a common emitter amp: Ga = Δ out Δ = What happens if = 0??? Do we have fite ga? No, we get a new model for the transistor. The base-emitter junction is a diode. Describe the behavior of the junction usg the bers-moll equation: I = I s [e q/kt ] = B kt/q = 5 m at 0 o Neglectg the - term: The mus sign the ga means that the output is the opposite polarity as the put (80 O out of phase). B = kt q [lni lni s ] alculate the dynamic resistance of the base-emitter junction, r B = d B di = kt qi = / I r B = 5 Ω for current of ma Ga = r B + X 3

4 We can now write the ga for the case = 0 (neglectg X too): Ga = / r B = (I / 5) with I measured ma. Simpson (page 7) writes an equivalent formula for the ga usg the transistor parameter β and a slightly different temperature, T = 300 O K. In terms of the hybrid parameter model (we will see this model soon) r B = h ie / h fe Usg r B to design a circuit is a dangerous practice as it depends on temperature varies from transistor to transistor even for same type of transistor. Input impedance Input impedance of the common emitter amp can be calculated from the equivalent circuit: = + + t t Δ B ΔI B = Δ ΔI /β = ΔI ΔI /β t = β For A case, we usually have and > t t = β = βr B = 500 Ω for ma of collector current and β = 00. 4

5 Output impedance Harder to calculate than the put impedance and only a hand wavg argument will be given. The output impedance of the amp is the parallel impedance of and the output impedance of the transistor lookg to the collector junction. The collector junction is reversed biased and hence looks like a huge resistor compared to. The output impedance is simply assume that the load impedance (the thg the amp is hooked up to) is less than. ommon ollector Amplifier: Sometimes this amp is called an emitter follower. What's common (ground) a common collector amp? The collector! The collector is connected (tied) to a D power supply. To an A signal this looks like the collector is connected to ground. We want to calculate: voltage and current ga, and put and output impedance. oltage Ga: The put is the base and the output is taken at the emitter cc = B 0.6 Δ = Δ B Δ out = Δ The amp has unity ga! s Q 0 out 5

6 urrent Ga: As always we can use Kirchhoff's current rule. cc I = I B + I = I B (β + ) ΔI ΔI B = β + ΔI out ΔI = β + Sce a typical value for β is 00, there is lots of current ga. Input impedance: By defition the put impedance is s Q 0 out = Δ ΔI = Δ B ΔI B = Δ ΔI /(β + ) = ΔI ΔI /(β + ) = (β + ) Sce is usually a few kω and β is typically 00 the put impedance of the common collector amp is large. 6

7 Output impedance: This is trickier to calculate than the put impedance. In the figure below we are lookg to the amp: s s is the put impedance of the transistor and t is the voltage drop across it. t = + s cc Q 0 out β β + s If we look from the other (output) side of the amp with out the output impedance of the transistor The voltage drop at A is the same as the voltage at the base ( B ) sce the amp has unity ga. We can rewrite the equation to a voltage divider equation to fd out. A = out A + out out = s β = t = β β + s = + s /β out is small sce β is typically 00. 7

8 What good is the common collector amp? xample: In stereo systems very often loud speakers have 8 Ω put impedance. Assume that you want to drive the speakers with a 5 olt voltage source with 9 Ω of serious resistance. Lets look at ways of drivg the speakers and the power each method delivers to the speaker. a. Hook the speakers directly to the voltage source: 9 Ω (5 rms) 8 Ω speaker The voltage delivered to the speaker is (8/00). The power delivered is: P = / = (5 8 /00) /8 = 0.0 Watts not much power! 8

9 b. Use a common collector (emitter follower): cc (5 rms) 9 Ω 50 K 5 K Q 9 Ω 5 K 50 K K β β 8 Ω 0 K 8 Ω speakers out An A signal at the put sees. β sp = β 8 Ω = 800 Ω From the speakers pot of view the amp impedance looks like 9 Ω/β ~ Ω The power delivered to the speaker can now be calculated: sp = (β8 Ω )/(β8 Ω+ 9 Ω) = 0.9 P sp = sp / sp = (0.9 5) /8 =.5 Watts (rms) over a hundred times more power delivered to the speaker. mitter Followers (common collectors) are used to match high impedances to low impedances 9