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Get Full Access to Microelectronic Circuits - 6 Edition - Chapter 4 - Problem 4.48
Get Full Access to Microelectronic Circuits - 6 Edition - Chapter 4 - Problem 4.48

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# In the capacitor-coupled attenuator circuit shown in Fig. ISBN: 9780195323030 147

## Solution for problem 4.48 Chapter 4

Microelectronic Circuits | 6th Edition

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Problem 4.48

In the capacitor-coupled attenuator circuit shown in Fig. P4.48, I is a dc current that varies from 0 mA to 1 mA, and C1 and C2 are large coupling capacitors. For very small input signals, so that the diodes can be representedby their small-signal resistances rd1 and rd2, show that and hencethat , where I is in mA. Find for I = 0 A, 1 A, 10 A, 100 A, 500 A, 600 A, 900 A, 990 A, and 1 mA.

Step-by-Step Solution:
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Lecture 7: Arrays An array variable is a set of variables  share a common name and type  all linked together int x; /* 30 integers */ elements: x, x, ..., x name of array is x array size: total 30 elements in array ‘x’ it has to be an integer with known value (to determine the memory needed to store the array) each element is int type starts from 0 Array Elements #include int main( void ) { int i, n; /* n is an array of 10 integers */ for( i = 0; i < 10; i ++ ) { /*loop 10 times*/ n[i]= 2; /* set all elements to 2 */ printf( "n[%d] = %d\n" , i, n[i] ); /*display value of n[i] */ } n = 5; n = n + n; printf( "Enter an integer value: " ); scanf( "%d", &n ); printf( "Element 3 is %d\n", n); /*display element 3 */ printf( "The sixth elem

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