Electronics The application of bipolar transistors

Transcription

1 Electronics The application of bipolar transistors Prof. Márta Rencz, Gergely Nagy BME DED October 1, 2012

2 Ideal voltage amplifier On the previous lesson the theoretical methods of amplification using bipolar transistor were overviewed. Before we can start to analyse amplifiers, we need to understand a few basic concepts. An ideal voltage amplifier: has no input current, has an output voltage that is proportional to its input voltage: v o = A v i, can be modelled with a voltage controlled voltage source.

3 Non-ideal voltage amplifiers I. A non-ideal voltage amplifier: has an input current that is proportional to its input voltage, has an output voltage that is proportional to its load. This non-idealities can be modelled with resistors: Input resistor: R i, Output resistor: R o. The gain (A) is called the nominal gain.

4 Non-ideal voltage amplifiers II. Amplifiers connected in series can be calculated using the extensions seen on the previous slide. Amplifier are usually built up of several stages. In the circuit below: the amplifier is driven by a voltage source v s that has an internal resistance (R S ), that can be seen as the output resistance of the previous stage. The amplifier has an input (R i ) and an output resistance (R o ) and is loaded with R L.

5 Calculating the gain The voltage of source is divided by R S and R i. The input voltage of the amplfier: v i = R i R S +R i v s The amplified voltage is also divided by two resistors (R o and R L ): v o = R L R R o+r L A v i = A L Ri R o+r L R S +R i v s Thus the real gain is always smaller than the nominal value. If R i R S and R o R L then the real gain is very close to nominal gain.

6 The Bode plot I. The Bode plot is widely used in control theory. It can be used to depict the transfer function of systems. The transfer function: A(ω) = V out /V in. The Bode plot consists of two plots: the absolute value and the phase shift of the transfer function. If the input signal is: V in = V 1 cos(ωt) and the output is V out = V 2 cos(ωt + ϕ) then the transfer function s absolute value is: V 2 /V 1, phase shift is: ϕ. The amplitude and the phase shift are depicted as a function of the frequency: the amplitude is usually plotted on a log-log scale, the phase shift is usually plotted on a log-lin scale, the frequency is on a logarithmic scale in both cases.

7 Bode plot II. The amplitude s unit is db (decibel). If the quantity in decibels is voltage or current, then if it is power, then A db = 20 lg(a), A db = 10 lg(a). The reason for 20 as a coefficient is that in linear systems: P V 2 or P I 2.

8 Bode plot III. The bode plot of an operational amplifier: The nominal gain (A N ): 100 db= The cutoff frequency (f C ): 30 Hz GBW: 3 MHz Important points of the Bode plot: Cutoff or break frequency (f C ): where the gain goes below the nominal gain minus 3 db (A 3 db = 2 A N 0.7 A N ). Transit frequency (f T or GBW ): the point where the absolute value of the gain becomes 1.

9 Common-emitter amplifier I. V CC = 12 V R B1 = 47 k, R B2 = 5.1 k R E = 470, R C = 4.7 k C B = 22µ, C E = 470 µ, C C = 22 µ The calculation: 1 first the DC operating point has to be found, 2 in order to determine the gain and the input and output resistances, the small-signal model of the circuit is used.

10 Common-emitter amplifier II. The operating point The operating point is found using an approximation. The input characteristic equation is approximated with a DC voltage source (the same as with diodes). The voltage is: V BE 0.7 V

11 Common-emitter amplifier III. The operating point The output characteristic equation is approximated as follows: in the normal active region (V CE > V CES ): I C = B I B in saturation: V CE = V CES where V CES is the saturation voltage V CES = V.

12 Common-emitter amplifier IV. The OP calculation I. The device is assumed to be operating in the normal active region. 1 The base-emitter junction is substituted with a supply voltage 0.7 V. 2 One of the transistor s currents is determined (the others can be expressed using B and the KCL). 3 V CE is calculated. 4 If V CE > V CES the assumption about the operating mode was correct, which means that the OP calculation is done. If the device is in saturation: 1 The value of I B is correct but V CE = V CES. 2 In saturation I C B I B, I C has to be calculated using V CE = V CES. In amplifiers the transistor operates in the normal active region, in switching application it operates in saturation when switched on.

13 Common-emitter amplifier V. The OP calculation II. In DC capacitors are susbstitued with an open circuit. The transistor is assumed to be operating in the normal active region. If the base current is neglected, the base potential is simply the supply voltage divided by the R B1 and R B2 : V B = V CC R B2 R B1 + R B2 = 1.17 V

14 Common-emitter amplifier VI. The OP calculation III. The emitter s potential is equal to that of the base minus the forward voltage of the B-E junction (0.7 V), thus the emitter current is: I E = V B V BE R E = 1 ma If the base current is neglected then I C = I E = 1 ma. V CE can be determined by writing the KVL for the V CC -R C -BJT-R E loop: V CE = V CC I C R C I E R E = 6.8 V V CE = 6.8 V > V CES = 0.1 V

15 Common-emitter amplifier VII. AC calculation I. At the input signal s frequency the capacitors can be treated as short circuits (their impedance is very small the effective 1 resistance of a capacitor is ωc, thus it is inversely proportional to the frequency of the signal). The small-signal model is created according to the following rules: 1 The DC supply voltages are substituted with short circuits. 2 Capacitors are also substitued with short circuits. 3 Non-linear devices are substitued with their small-signal models.

16 Common-emitter amplifier VIII. AC calculation II. The parameters of the small-signal model: β = B = 150, r e = V T I E = 26 1 = 26 Ω (V T = 26 mv) The input and output voltages should be expressed with i b : v in = (β + 1) r e i b v out = β i b R C where the negative sign is due to the opposite direction of the current and the output voltage. The gain can be expressed using the two equations above: A = vout v in = β β+1 RC re = 180

17 Common-emitter amplifier IX. AC calculation III. The input resistance: three resistors in parallel can be seen from the input r in = R B1 R B2 (β + 1) r e = 2.1 kω The output resistance: only R C can be seen form the output as the current source is an open circuit r out = R C = 4.7 kω

18 The emitter follower I. The output resistance of the common-emitter amplifier is a large value, while in an ideal amplifier it is a very small value (so that most of the output voltage is dropped on the load resistor). The circuit below, which is also called analog buffer, copies its input to its output (it gets shifted by a forward voltage) but its output resistance is much smaller. V B is the DC offset of the input signal. V E = V B V BE = V B 0.7 I E = (V B V BE ) /R E V CE = V CC I E R E

19 The emitter follower II. The current of the R E resistor is i RE = i b + β i b = (β + 1) i b The gain can be calculated by expressing v in and v out with i b : A = vout v in = (β+1) R E (β+1) (r e+r E ) = R E r e+r E 1 The input resistance similarly: r in = v in i in = (β+1) (re+r E) i b i b = (β + 1) (r e + R E ) The output resistance (calculation not given): r out = r e + R G β+1 where R G is the resistance of the driving stage. Thus this circuit divides the input resistance by β.

20 The emitter follower III. An example V CC = 12 V R B1 = 47 k, R B2 = 5.1 k R E1 = 470, R C = 4.7 k C B = 22µ, C C = 22 µ R E2 = 3.3 k, C E2 = 22 µ A common-emitter amplifier is amended by an emitter follower stage. The OP of Q2 s base potential: V B2 = V CC I C1 R C1 = = 7.3 V The emitter current can now be calculated: I E2 = V B2 V BE = 2 ma R E2 The output resistance: r out = ug i g = r e2 + R C1 β+1 44 Ω

21 The switching operating mode The transistor switches the current of the load represented by R C and is controlled by its base potential. If the input is 0 V, there is no base current, hence the current of the load is zero as well. If the input voltage is above V BE0, then: I B = (V CC V BE) /R B. If the resistances are appropriate, the transistor gets into saturation: V CE = V CES. The current of the load: I C = (V CC V CES ) /R C

22 The switching operating mode Example I. V CC = 5 V V CES = 0.2 V V BE = 0.7 V If V in = 5 then I B = Vin V BE R B = If V in = 0 then I B = = 4.3 ma V CE has to be calculated to determine the operating mode of the transistor: V CE = V CC R C I C = 640 V < V CES thus the transistor saturates. V out = V CE = V CES = 0.2 V I C = V CC V CES R C = = 4.8 ma

23 The switching operating mode Example II. In this example an load of 1 kω is switched on and off. It can also be seen as the input being inverted: this circuit is also a logic inverter.