Common emitter amplifier
pp 100-109
4.4 amplifier circuits with transistors
4.4.1 transistor amplifier circuit with common emitter
The circuit of a common emitter amplifier is shown in Figure 4.4.1. The amplifier circuit is a common connection to the emissions because, in terms of the signal flow and its alternating loops, the emitter electrode is public, through direct or indirect (with capacitor) connection to the earth. To be easily understood function of the amplifier we consider that the capacitor impedance is very small, ie to behave as a short circuit on the AC and open circuit for DC. We also consider that the input AC voltage source (signal generator) has zero internal resistance.
The signal V is applied between the base of the transistor and ground. The connection is via the capacitor C1. The insertion of the capacitor is needed without altering the constant component of the voltage (VB) of the base.
O emitter "grounded" for the AC signal through the CE and thus achieve connectivity for alternating CE. The rotational component of the collector voltage is transferred intact to the resistance of pressure without any alteration of the operating point Q of the collector by the presence of RL. The capacitors C1 and C2 are called coupling capacitors, because through them is the coupling between
100
signal generator and amplifier and amplifier and burden. O capacitor CE, because the grounding for alternating transmission, called the decoupling capacitor leakage.
Figure 4.4.1 Amplifier transistor in connection with CE
The analysis in the continuum has been presented in the previous paragraph. As for the AC analysis, the equivalent circuit shown in Figure 4.4.2
Figure 4.4.2 The equivalent circuit of the AC amplifier sch.4.4.1
We observe that the equivalent circuit is derived from that
sch.4.2.5 where the diode has been replaced with equivalent anti-
tion of b · re, re, where the resistance is given by,
re = ^ 4.4.1
The current IC is in mA and the result (re) in ohms.
101
The re resulting from low signal approach, which appears in the characteristic input sch.4.4.3. For the AC is not used now, DC, but the second, called AC current gain of the transistor and is defined as the quotient of the change in collector current to base current change.
Figure 4.4.3 Typical input of the amplifier: (a) AC voltage applied between the base and the emissions, (b) The current generated at the emitter and (c) calculation of the re
In the equivalent circuit, Figure 4.4.2, the resistance which the presence
basis risk, or (rB) is beta times the re (r, = b · re). The anti-
tion is connected in parallel by R1 and R2. On exit, RC and RL
connected in parallel and the current flows through the source b · iB.
Watching the progress of the signal we find that the AC
trend, which applies to the input of the amplifier causes a stream-
iB but which leaks the emitter diode. This current is equal to
yi
'B = Fn 4A2
This current is enhanced due to the gain of the power-tranzi
blinds, and a collector displays AC current iC, which-given a
rom the equation
ic = b · iR 4.4.3
102
AC Power Output 2
AC Input
AC Output Voltage
Figure 4.4.4 Typical input-output. Shows the AC line and load
the operating point Q
If the equivalent output impedance, which results from the parallel connection of RC and RL, is rL, then the voltage across the ends and therefore the load impedance will be,
4.4.4
We observe that when a voltage is applied to the HI input is shown a tendency Y ^ in amplifier output. These trends have a phase difference between the 180 °, ie the waveform signal to the output is inverted in relation to this entry.
In an amplifier directly interested in the relationship between output voltage and input voltage. For this reason is defined as a gain (profit) of the voltage amplifier AV, as a quotient of the AC voltage output (Y) to the AC input voltage (HI), namely:
4.4.5
In the case where the capacitor is not grounding the CE program, the relationship that gives the gain becomes:
103
Example 4.4.1
The amplifier is given in Figure 4.4.1 VCC = 20 V, u ^ 10 mV,
R., = 82 KO, R2 = 18 KO, RE = 1 KO, RC = 2,2 KO and KO RL = 4,7. Be calculated
excellence or base resistance, the gain and the output voltage when the density
the CE is connected to the circuit when disconnected.
Solution
This example uses the same DC circuit
in example 4.3.1. So we use components of the examples
Tosh 4.3.1, where we have already calculated the collector current IC = 2,9 mA.
So,
AV =-L = ^ 4.4.6
AC resistance and the base is,
4.4.7
The study of Eq. 4.4.6 and Eq. 4.4.7, where the term RE is much higher in re, found that when removed the capacitor CE-based resistance increases and the voltage gain decreases. The change in these two sizes is important.
Figure 4.4.4 shows the typical output of the amplifier. For AC or AC load line resulting from the rL and has a different slope from the DC load line, but again through the point Q. Moreover distinct phase reversal between input and output signal, ie the minimum current-
Tosh eisooou corresponds to a maximum voltage esooou and vice versa.
IC 2,9 mA
104
With the CE connected to the circuit resistance of the base is
r, = b · re = 100 x 8.6 = O 860O
The voltage gain is
A = ^ =: = j = i500O 1744
If u = | = re = 8,6 O = 174.4
and the output voltage
u; ^ = AV · u | = 174,4 x 10mV = 1,744 V
Without the CE resistance of the base is
r, »b · (re + RE) = 100 x (8.6 + O 1000) = 100.86 KOS
The voltage gain is
A = yL = rL = 1500 15
u n | re + RE 8,6 Z + 1000 "
and the output voltage
u; ^ = AV · u | = 1,5 x 10mV = 15mV
4.4.2 Equivalent circuit of hybrid parameters h
The description of an equivalent circuit for the transistor was continuous in paragraph 4.2.2. In the case of AC, the analysis of complex circuits often presents difficulties. For this reason, the equivalent circuits are used, which is more streamlined.
Towards the equivalent circuit of a circuit (eg a transistor amplifier), you must first draw the equivalent circuit of the transistor itself, if it works in linear region. The determination of equivalent circuit is in the following way.
Equivalent
h1 h2 quadrupole circuit
Equivalent circuit quadrupole
Figure 4.4.5 a circuit diagram of a tranzstor as quadrupole
105
The hybrid parameters are briefly the following concepts and notations:
ht1 hj impedance input shorted output transfer ratio h12 hr reverse-open circuit voltage at the entrance
h21 hf ortho-ratio transfer power to the shorted output
h22 ho Conductivity output open circuit at the entrance
Using the new notation the equations (4.4.8) writes:
h1 =
We believe in principle that there is a circuit that is alternating for the same behavior as that is the transistor, and that this circuit is placed in a box. Furthermore we believe that this box has four terminals (see sch.4.4.5), that is a quadrupole. Because this circuit at that
the application of voltage or injection currents, has two entrances 1 and 2, called the two-door. O determination of the equivalent circuit can be done by means of alternating voltages h1 and h2 and the currents i1 and i2 of the inputs. With proper selection of size and those measured as specified Z equivalent circuit (channel currents and measuring voltages), the Y equivalent circuit (application trends and measurement of currents), etc.
If, for making the equivalent circuit of the transistor h2 applied voltage and the current i1 and measure the voltage and current h1 i2 then calculated the four hybrid parameters and the equivalent is called hybrid equivalent.
In the case of the hybrid equivalent of the measured data, ie the voltage and current h1 i2, will depend on the voltage applied to the h2 input 2 and the current i1 flowing in input 1 through the following equations:
Depending on the type of wiring transistors added a second indicator: e for the JV, the KB for b and c for PA connection.
According to equations (4.4.9) and the concepts of hybrid parameters the equivalent circuit includes an input voltage source, the iGy2 connected in series with a resistance, fy, which flows through the current of i1. The equivalent output circuit includes a current source, the hfi1, which is also connected to output conductance, ho. The full equivalent circuit input-output is shown in Fig 4.4.6. Considering the above, we find that: The hybrid circuit is input by the equivalent Thevenin (constant voltage), while the hybrid circuit is output by the equivalent Norton (constant intensity). Thus, the full hybrid equivalent circuit is mixed (hybrid) and includes one in a Thevenin and Norton equivalent circuits in.
Figure 4.4.6 Full hybrid equivalent circuit
For a transistor which is a common emitter connection (CC) the hybrid equivalent circuit, the parameters h and their corresponding indicators, shown in Figure 4.4.7. Indicators of trends and patterns have been replaced with the 1 b, corresponding to the base and 2 with c corresponding to the collector.
Figure 4.4.7 Full hybrid equivalent circuit of a transistor in connection JV
107
Example 4.4.2
The amplifier is given in Figure 4.4.1 hie = 1 KO, hre = 2x10-4,
hfe = 120 and hoe = 25 mA / V. If u = 10 mV, Rr = 2,2 KO and KO RL = 4,7, be
calculate the voltage gain and the output voltage.
Solution
In this example we
r = RC'RL = 2,2 KO KO x 4.7 "15KO
rL RC + RL 2,2 KO KO + 4.7 "1.5
Using equation (4.4.11) without the negative sign,
we
D 120 x 1500
If
n 1000 + (1000 x 25 x -120 x 10-6 2 x 10-4) x 1500
120 x 120 1500 x 1500
+ 0.001 x 1000 1500 1000
and the output voltage
u, = AV · yi = 180 x 10mV = 1,8 V
Using the h-parameters to calculate the voltage gain of an amplifier common emitter wiring (Fig. 4.4.1) is straightforward. The calculation starts from the definition of gain, ie the first part of the equation (4.4.5)
If TT ^ = 4.4.10
which express the input voltage versus output voltage and current HR input ib. Finally the voltage gain of the amplifier is given by the equation
- Hfer,
A = fL_L 4411
V hie + (hieh0e-hfehre) rL ''
The negative sign in (Ex. 4.4.11) indicates 180 ° phase difference between output and input signal.
Finally, the hybrid parameters of transistors are provided in technical brochures of the manufacturers.
108
SUMMARY 4.4
• In a CE amplifier the signal is applied at the base and taken
from the collector through capacitor couplers.
• The emitter of the transistor can be "grounded" for the alternating-
only signal via a decoupling capacitor.
• In a CE amplifier we gain power and voltage gain
greater than one
• In a CE amplifier gain power and we have defined as
the ratio of signal strength esodou to the signal strength
entry. The power gain is also equal to the product of de-
handle current and voltage.
• Not using the emitter decoupling capacitor is
effect of reducing the voltage gain and increased
resistance of the base.
• The AC load line is determined by the resistance and collector
load resistance, which has a different slope from DC
straight load.
• The AC load line intersects with the line load in DC operating point Q.
• The hybrid equivalent circuit of a transistor consists of an input circuit in a circuit and Thevenin output by Norton.
• The hybrid equivalent circuit used for simpler and easier solution to a transistor circuit in accordance with the technical data sheet manufacturers.
109
QUESTIONS
- Exercises 4.1-4.4
4.1. The emitter diode is usually
Good polarized a b c No reverse biased conduct
D. Operation in decay
4.2. For a transistor which operates in the active region, the collector diode is
Good polarized a b c No reverse biased conduct
D. Operation in decay
4.3. The power gain of a transistor, a wiring CE, is equal to the ratio
a current collector to
b current emitter to collector current
c current base current to the base
d collector current to emitter current of the collector current
4.4. Draw CE circuit which apply: VBB = 20 V, VCC = 30 V, IC = 8 mA, VCE = 8 V and
, SG = 100.
4.5. To map out the line below the load circuit.
4.6. If the collector voltage of +10 V to reduce what would happen in a straight load circuit of the KLA. 4.5.
4.7. If the collector resistance doubled what happens in the straight
1.2 KO
Figure 4.5
burdens (ask.4.5)
4.8. A CE amplifier operates in the active region of a
b in the saturation region c d Under cutoff decay
4.9. In the CE amplifier behaves as a capacitor CE
a open circuit for
b alternating short for direct
c open circuit to power the circuit
D. Short for alternating 4.10. CE amplifier in the presence of a CE capacitor increases the voltage gain b reduces the voltage gain
c increases the power gain
d not affect the operation of the amplifier
