Other Transistor Circuits CHAPTER 6 Other Transistor Circuits

OBJECTIVES Describe and Analyze: Common Collector Amplifiers Common Base Amplifiers Darlington Pairs Current Sources Differential Amplifiers Troubleshooting

Common Collector Amplifiers The common-collector amplifier, more commonly called an emitter follower, is used as a “buffer”

Buffers Amps The ideal buffer amplifier has unity voltage gain (Av = 1), infinite input impedance (Zin = ), and zero output impedance (Zout = 0). The power gain would also be infinite (Ap = ) The “job” of a buffer amp is to prevent loading of a signal source. If a high-impedance signal source is connected to a low-impedance point in a circuit, most of the signal will be lost in the source’s internal resistance. The buffer goes in between the source and the rest of the circuit.

The Emitter Follower Buffer As we shall see, the emitter follower has a voltage gain slightly less than one (Av  1), a high input impedance (Zin  Re), and low output impedance (Zout  Re || Rb / ). It has a reasonably high power gain. Emitter followers are used very often in linear circuits, even in linear ICs. They are simple, yet effective.

Biasing the Emitter Follower Emitter followers typically use resistor divider biasing, just like the common-emitter amplifier. Usually, the collector is tied directly to Vcc, so the collector to emitter voltage is Vce = Vcc – Ve. If Vcc is too high, then the transistor can get hot since the power dissipated is PD = Vce  Ic. Remember that Ic is basically equal to Ie = Ve / Re = (Vb – 0.7) / Re. Emitter followers sometimes use a collector resistor to lower the Vce drop.

A Biasing Example The Specifications: Suppose we have a circuit like that of figure 6-1. Vcc is 12 V, and we want Ve to be 6 V  5%. Find Rb1 and Rb2 so that there is 2 mA of current through Rb2 (Ib2 = 2mA). The minimum beta is 70. The emitter resistor is 600 Ohms (Re = 600). Also, find the power dissipation in the transistor.

Biasing Example (cont.) Find Vb: Vb = Ve + 0.7V = 6.0V + 0.7V = 6.7V Find Rb2: Rb2 = Vb / Ib2 = 6.7V / 2mA = 3.35k Choose a standard resistor value: Let Rb2 = 3.3k Find Rb1: [Rb2 / (Rb1 + Rb2)]  Vcc = Vb So, Rb1 = Rb2  [(Vc – Vb) / Vb] Rb1 = 3.3k  [(12V – 6.7V) / 6.7V] = 2.61k Choose a standard resistor value: Let Rb2 = 2.7k

Biasing Example Now let’s check to see if we got it right: Vb = [Rb2 / (Rb1 + Rb2)]  Vcc Vb = [3.3k / (3.3k + 2.7k)]  12V Vb = (3.3 / 6.0)  12 = 6.6V Maximum base current is 10mA / 70 = 0.14mA The Thevenin’s equivalent of Rb1 & Rb2 is RTH = (Rb1  Rb2) / (Rb1 + Rb2) = 1.5k and 0.14mA  1.5k =0.2V, so Vb = 6.6 - .2 = 6.4V But 5% of 6.7V is 0.34V. The minimum Vb is 6.36V, so Vb = 6.4V seems OK.

Biasing Example Let’s find the power dissipation of the transistor: PD = Vce  Ic = (12V – 6V)  0.30A = 1.8 Watts Most likely, this guy needs a heat-sink.

Input Impedance Let’s find Zin for the emitter follower that we just biased: Zin is the parallel combination of the biasing resistors together with the impedance “looking into” the base: Zin = Rb1 || Rb2 || Re But Rb1 || Rb2 = RTH, which we calculated to be RTH = 1.5k and Re = 70  600 = 42k Since Re is so much bigger than RTH, we can say: Zin  RTH = 1.5k

Output Impedance Zout is the parallel combination of Re and the equivalent base resistance divided by beta

Output Impedance Now let’s find Zout for the same emitter follower we biased: Zout = (RTH / ) || Re = (1.5k / 70) || 600 But, RTH /  = 1.5k / 70 = 21 Ohms So, for all practical purposes, Zout  20 Ohms Let’s just check that: Actual Zout = (21  600) / (21 + 600) = 20.3 Ohms

We should find that Av is close to 1 Voltage Gain We should find that Av is close to 1

Voltage Gain The equation for the voltage gain of an emitter follower is: Av = re / (re + r’e) where r’e = 25mV / Ie, and re is Re in parallel with the load being driven by the emitter follower. Let’s find Av for the circuit we biased: R’e = 25mV / 10mA = 2.5 Ohms Re = Re = 600 Ohms Av = 600 / (600 +2.5) = 600 / 602.5 = 0.996 0.996 is close enough to 1 for most purposes.

Power Gain Power gain (Ap) is output power divided by input power: Ap = Pout / Pin Since P = V2 / R, Pout = (Vout)2 / Rout and Pin = (Vin)2 / Rin Some algebra, and: Ap = (Vout / Vin)2  (Rin / Rout) For a buffer, Vout = Vin, so Ap = Rin / Rout For the emitter follower we’ve been using, Rin = Zin and Rout = Zout, so its power gain is: Ap = Rin / Rout = 1.5k / 20 = 75

For a Darlington pair, Ic / IB = 1  2  2 Darlington Pairs For a Darlington pair, Ic / IB = 1  2  2

Darlington Pairs Some things to know about Darlington pairs: Since the collectors are tied together, the transistors can not saturate. When used as a switch, Vce  Ic can generate a lot of heat when Ic is big. A transistor’s beta is often lower for very low values of Ic. So the beta of Q1 may be a lot less than the beta of Q2. The Vbe of a Darlington pair is 2  0.7 = 1.4V The equivalent fT of the pair is lower than the fT of either transistor.

Common Base Amplifiers Work at higher frequencies than a common emitter can

Common Base Amplifier The Miller Effect Common emitter amplifiers lose gain at higher frequencies because of what’s known as the Miller Effect. The capacitance of the collector-base junction (CCB) looks bigger at the base than it really is. It looks like CM = Av  CCB where CM is the Miller Capacitance. It’s caused by negative feedback from the output (collector) back to the input (base). An RC low-pass filter is formed by CM and the resistance of the signal source driving the base.

Common Base Amplifier The base is at signal ground in a common base amplifier (but not necessarily at DC ground). So CCB can only shunt some signal to ground, not back to the input. That eliminates the Miller Effect. The trade-off is that Zin is very low, on the order of 50 Ohms. But in a high-frequency amplifier, it’s usually required that Zin and Zout be around 50 Ohms.

Comparison of Configurations All three configurations have their place in circuits the base is at signal ground

Differential Amplifiers The “diff amp” is commonly used in linear ICs

Differential vs. Single-Ended All the amplifiers we have seen so far share one characteristic: they have only one input. They are single-ended amplifiers. A signal is applied from that one input to ground. Differential amplifiers have two inputs, commonly referred to as the “plus input” and the “minus input”. A signal is applied across the two inputs. Signals applied simultaneously to both inputs with respect to ground are called “common mode” signals.

Differential Amps & CMR Suppose there is +10 mV (with respect to ground) on one input of a differential amplifier and –10 mV (with respect to ground) on the other input. Then the differential signal is 20 mV. If the diff amp has a gain of 10, the output will be 10  20 mV = 200 mV. Now suppose that +100 mV (with respect to ground) is applied to both inputs at the same time. That’s a common mode signal. Since the differential voltage is 0, and the output will be zero. Since common mode signals produce no output, differential amplifiers have “common mode rejection” (CMR). CMR is very important, as we will see.

Constant-Current Source Referring back to figure 6-17, the purpose of Q3 is to act as a “constant-current source” for the differential pair formed by Q1 and Q2. By definition, a constant-current source will conduct the same amount of current irrespective of the voltage across it. Since dynamic resistance (RD) is RD = V / I, the RD of a constant-current source is infinite. The constant-current source in the emitter circuit of Q1 and Q2 prevents common-mode signals from causing base current. There’s the CMR.

Troubleshooting The troubleshooting techniques for single-ended amplifiers: Measure DC bias levels Trace signals Isolate stages also work for differential amplifiers. In addition, the technician should look for imbalances in what should be equal DC voltage and current levels.

Diff Amps & Noise One of the main problems in using amplifiers is the presence of electrical noise and interference. A common form of “man-made” noise is “hum” from the 60 Hz power line. Noise gets mixed in with low-level signals, and when the signals are amplified, so is the noise. Like death and taxes, it’s hard to avoid noise. Most man-made noise appears on all inputs with respect to ground. It’s common mode, and will be rejected by a diff-amp. So small differential signals can be “extracted” from large common mode noise.