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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3682/CH3/EX3.4 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3682/CH3/EX3.4')
| -rw-r--r-- | 3682/CH3/EX3.4/Ex3_4.sce | 22 |
1 files changed, 22 insertions, 0 deletions
diff --git a/3682/CH3/EX3.4/Ex3_4.sce b/3682/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..2a1e9947b --- /dev/null +++ b/3682/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,22 @@ +// Exa 3.4
+
+clc;
+clear;
+
+// Given data
+
+// Fig. 3.13 shows output of an op-amp voltage follower
+F =2; // frequency in MHz
+Vipp= 8; // Input voltage (Peak to peak) in volts
+
+// Solution
+
+printf(' Since the frequency is given we can get time period as T = %.1f μsec. \n\n',1/F);
+
+printf(' Since, the slew rate is defined as the maximum rate of change of the output, \n so from Fig. 3.13, it can be seen that, maximum change in output is 6V in 0.25 μsec.');
+
+dVo=6;
+dT=0.25 ; // μsec
+SR=dVo/dT;
+
+printf('\n\n Therefore, the slew rate of the op-amp = %d V/μsec. \n',SR);
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