vb.net barcode reader source code IC = IB = 186 mA VCE = VCC IC RC + = 15 186 10 3 in Software

Creation QR Code ISO/IEC18004 in Software IC = IB = 186 mA VCE = VCC IC RC + = 15 186 10 3

IC = IB = 186 mA VCE = VCC IC RC + = 15 186 10 3
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Figure 1019 DC circuit resulting from application of rule of thumb
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Transistor Ampli ers and Switches
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Thus, the Q point for the ampli er of design C is identical to that of design B:
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IBO = 186 A
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ICO = 186 mA
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VCEO = 75 V
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3 Q point variability Now we are ready to determine the Q point variability for design C ampli er For = min = 75: IB = VBB V 474 07 = 242 A = RB + ( min + 1)RE 1,500 + 76 200 min + 1 RE min 76 200 = 77 V 75
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IC = min IB = 181 mA VCE = VCC IC RC +
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= 15 181 10 3 200 + Thus, the Q point for = min = 75 is:
IBO = 242 A
ICO = 181 mA
VCEO = 77 V
For = max = 150: IB = VBB V 633 07 = 127 A = RB + ( max + 1)RE 10, 000 + 151 200 ( max + 1) RE max 151 200 = 73 V 150
IC = max IB = 191 mA VCE = VCC IC RC +
= 15 191 10 3 200 +
Thus, the Q point for = max = 150 is:
IBO = 127 A
ICO = 191 mA
VCEO = 73 V
The change in quiescent base current, relative to the nominal value of design C (for = 100) has actually increased to 62 percent; however, the changes in quiescent collector current and collector-emitter voltage (relative to the same quantities for the = 100 design) have decreased, to approximately 5 percent This is a substantial improvement (nearly by a factor of 5!) You may wish to approximately locate the new Q points on the load line of Figure 1018
Part II
Electronics
Comments: This example may have been repetitive, but it presents some very important points about the importance of biasing and the effect of component variability in transistor ampli er design
Check Your Understanding
102 Find the operating point of the circuit of design B in Example 103 if (a) = 90 and (b) = 120 103 Verify that the operating point in design 2 of Example 104 is more stable than that of design 1 104 Repeat all parts of Example 104 if = min = 60, = max = 200, and IC =
10 mA
AC Analysis of the Common-Emitter Ampli er To analyze the AC circuit of the common-emitter ampli er, we substitute the hybrid-parameter small-signal model of Figure 103 in the AC circuit of Figure 1010, obtaining the linear AC equivalent circuit of Figure 1020 Note how the emitter resistance is bypassed by the emitter capacitor at AC frequencies; this, in turn, implies that the base-to-emitter (BE) junction appears in parallel with the equivalent resistance RB = R1 R2 Similarly, the collector-to-emitter (CE) junction is replaced by the parallel combination of the controlled current source, hfe IB , with the resistance 1/ hoe Once again, since the DC supply provides a direct path to ground for the AC currents, the collector resistance appears in parallel with the load It is important to understand why we have de ned input and output voltages, vin and vout , rather than use the complete circuit containing the source, vS , its internal resistance, RS , and the load resistance, RL The AC circuit model is most useful if an equivalent input resistance and the equivalent output resistance and ampli er gain are de ned as quantities independent of the signal source and load properties This approach permits the computation of the parameters ri , ro , and shown in Figure 108, and therefore provides a circuit model that may be called upon for any source-and-load con guration The equivalent circuit shown in Figure 1020 allows viewing the transistor ampli er as a single equivalent circuit either from the source or from the load end, as will presently be illustrated
iin + vin _ RB
IB hie hfe IB 1 hoe
iout + RC vout _
Figure 1020 AC equivalent-circuit model for the common-emitter ampli er
10
Transistor Ampli ers and Switches
First, compute the AC input current, iin = vin hie RB vin hie (1019)
and the AC base current, IB = (1020)
noting, further, that the input resistance of the circuit consists of the parallel combination of hie and RB : ri = hie RB (1021)
Next, observe that the AC base current is ampli ed by a factor of hfe and that it ows through the parallel combination of the collector resistance, RC , and the CE junction resistance, 1/ hoe The latter term is often, but not always, negligible with respect to load and collector resistances, since the slope of the collector i-v curves is very shallow Thus, the AC short-circuit output current iout is given by the expression iout = hfe IB = hfe vin = gm vin hie (1022)
and the AC open-circuit output voltage is given by the expression vout = hfe I B RC 1 hoe vin 1 1 = hfe RC hoe hie hoe
(1023)
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