zen barcode ssrs AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS in Software

Printer QR Code 2d barcode in Software AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
Recognize QR Code 2d Barcode In None
Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.
Making QR Code ISO/IEC18004 In None
Using Barcode printer for Software Control to generate, create QR Code JIS X 0510 image in Software applications.
Fig. 5-31
Denso QR Bar Code Recognizer In None
Using Barcode decoder for Software Control to read, scan read, scan image in Software applications.
Create QR Code In C#
Using Barcode generator for Visual Studio .NET Control to generate, create QR Code image in .NET applications.
Solved Problems
Paint QR Code JIS X 0510 In .NET
Using Barcode generation for ASP.NET Control to generate, create Denso QR Bar Code image in ASP.NET applications.
Creating QR Code 2d Barcode In .NET Framework
Using Barcode creation for Visual Studio .NET Control to generate, create QR image in VS .NET applications.
5.1 In Fig. 5-3, let vs 20 V, Rs 10
Creating QR Code ISO/IEC18004 In Visual Basic .NET
Using Barcode maker for Visual Studio .NET Control to generate, create QR image in VS .NET applications.
Making UPC - 13 In None
Using Barcode creation for Software Control to generate, create GTIN - 13 image in Software applications.
, Ri 990
GTIN - 128 Printer In None
Using Barcode generation for Software Control to generate, create EAN128 image in Software applications.
Print UPCA In None
Using Barcode creator for Software Control to generate, create UPC A image in Software applications.
, k 5, and Ro 3
Making Code 128C In None
Using Barcode generator for Software Control to generate, create Code 128B image in Software applications.
Data Matrix Creator In None
Using Barcode drawer for Software Control to generate, create Data Matrix 2d barcode image in Software applications.
. Find (a) the Thevenin equivalent of the circuit seen by Rl and (b) v2 and the power dissipated in Rl for Rl 0:5, 1, 3, 5, 10, 100, and 1000
Make Code 2 Of 5 In None
Using Barcode creator for Software Control to generate, create 2/5 Industrial image in Software applications.
ECC200 Recognizer In VB.NET
Using Barcode reader for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications.
(a) The open-circuit voltage and short-circuit current at A B terminal are vo:c: 5v1 and is:c: 5v1 =3, respectively. We nd v1 by dividing vs between Rs and Ri . Thus, v1 Ri 990 20 19:8 V v Rs Ri s 10 990
GTIN - 12 Creator In C#
Using Barcode maker for .NET framework Control to generate, create GS1 - 12 image in VS .NET applications.
Scanning EAN / UCC - 13 In Visual Basic .NET
Using Barcode decoder for Visual Studio .NET Control to read, scan read, scan image in .NET applications.
Fig. 5-32
Making Barcode In None
Using Barcode generation for Office Excel Control to generate, create bar code image in Office Excel applications.
Paint Data Matrix In C#
Using Barcode encoder for VS .NET Control to generate, create Data Matrix image in .NET applications.
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
Draw GS1-128 In None
Using Barcode creator for Font Control to generate, create GS1-128 image in Font applications.
GTIN - 13 Creation In Visual Studio .NET
Using Barcode encoder for Reporting Service Control to generate, create EAN-13 image in Reporting Service applications.
[CHAP. 5
Therefore, vo:c: 5 19:8 99 V is:c: 99=3 33 A The Thevenin equivalent is shown in Fig. 5-32. (b) With the load Rl connected, we have v2 Rl 99Rl v Rl RTh Th Rl 3 and p v2 2 Rl vTh vo:c: 99 V RTh vo:c: =is:c: 3
Table 5-3 shows the voltage across the load and the power dissipated in it for the given seven values of Rl . The load voltage is at its maximum when Rl 1. However, power delivered to Rl 1 is zero. Power delivered to Rl is maximum at Rl 3
, which is equal to the output resistance of the ampli er. Table 5-3 Rl ;
0.5 1 3 5 10 100 1000 v2 ; V 14.14 24.75 49.50 61.88 76.15 96.12 98.70 p; W 400.04 612.56 816.75 765.70 579.94 92.38 9.74
In the circuits of Figs. 5-4 and 5-5 let R1 1 k
and R2 5 k
. Find the gains G v2 =vs in Fig. 5-4 and G v2 =vs in Fig. 5-5 for k 1, 2, 4, 6, 8, 10, 100, 1000, and 1. Compare the results.
From (5) in Example 5.3, at R1 1 k
and R2 5 k
we have G In Example 5.4 we found G v2 5k 6 k vs 32 v2 5k vs 6 k 31
The gains G and G are calculated for nine values of k in Table 5-4. As k becomes very large, G and G approach the limit gain of 5, which is the negative of the ratio R2 =R1 and is independent of k. The circuit of Fig. 5-5 (with negative feedback) is always stable and its gain monotonically approaches the limit gain. However, the circuit of Fig. 5-4 (with positive feedback) is unstable. The gain G becomes very large as k approaches six. At k 6, G 1. Table 5-4 k 1 2 4 6 8 10 100 1000 1 G 1:0 2:5 10:0 1 20:0 12:5 5:32 5:03 5:00 G 0:71 1:25 2:00 2:50 2:86 3:12 4:72 4:97 5:00
CHAP. 5]
AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
Let R1 1 k
, R2 5 k
, and Ri 50 k
in the circuit of Fig. 5-33. Find v2 =vs for k 1, 10, 100, 1000, 1 and compare the results with the values of G in Table 5-4.
Fig. 5-33 This problem is solved by application of KCL at node A (another approach which uses the Thevenin equivalent is suggested in Problem 5.30). Thus, v1 vs v1 v2 v1 0 1 5 50 From the ampli er we obtain v2 kv1 or v1 v2 =k 34 33
Replacing v1 in (34) into (33) and rearranging terms, we obtain v2 50k 5k vs 61 10k 6:1 k 35
Values of v2 =vs in (35) are shown in Table 5-5 as functions of k. The 50-k
input resistance of the ampli er reduces the overall gain very slightly, as seen by comparing Tables 5-4 and 5-5. The feedback has made the input resistance of the ampli er less e ective in changing the overall gain. Table 5-5 k 1 10 100 1000 1 v2 =vs 0:704 3:106 4:713 4:97 5:00
Let again R1 1 k
and R2 5 k
in the circuit of Fig. 5-33. a b c Find v2 =vs as a function of k and Ri : Let Ri 1 k
. Find v2 =v1 for k 1; 10; 100; 1000; 1. Repeat for Ri 1: Discuss the effects of Ri and k on the overall gain. Show that, for k 1 and Ri 6 0; the gain of the amplifier is independent of Ri and is equal to R2 =R1 :
(a) Apply KCL to currents leaving node A to obtain v1 vs v1 v2 v1 0 1 5 Ri From the ampli er we get v2 kv1 or v1 v2 =k. rearranging terms we get Substituting for v1 in the KCL equation and
Copyright © OnBarcode.com . All rights reserved.