Unformatted text preview: Carlton University Tutorial for LAB 2 Prof Ram Achar Farhad Ramezankhani Morteza Nabavi Winter 2011 1 Inverting Amplifier
For the circuit shown below find a relationship between and . Assume opamp as an ideal one and .What is the function of this circuit? Figure 1 Solution: Since Opamp is ideal, and It means that is virtually grounded. If we calculate currents through , we can write: V1 V2 Figure 2 Since Opamp is ideal, so input current to OpAmp is zero. It means that, . Hence: Since the ratio of the gain is negative, the circuit is referred to as an “Inverting Amplifier ”.
2 Summing Amplifier
For the circuit below, find the relation between output and inputs. Explain the operation of the circuit. Opamp is ideal and . Figure 3 Solution: Like the previous example in Figure1, we can write the currents in as : Figure 4 3 Or : This circuit amplifies each input with a certain weight and then adds them up together. Hence this is referred to as “Summing Inverting Amplifier” Differentiator Circuit
For the circuit below, write output voltage as a function of time. Consider the Opamp to be ideal. It is used to amplify a sinusoidal input at 1MHZ by a factor of 5 and is 1KΩ, what is C? What is the mathematical operation performed by this circuit? Figure 5 4 Solution: Figure 6 OpAmp is ideal, Hence we can write: and It means that, this circuit works as a “Differentiator” and the gain is “ example, if we want to have a gain factor of 5, then we have: ”. For the Correspondingly: There is a “π/2 (or 270o)” phase shift between input and output.
5 The graph of input and output is illustrated in figure below.
Input & Output Vs. time 5 4 3
Input and Output(V) Output 2 1 0 1 2 3 4 5 0 0.2 0.4 0.6 0.8 1 time(s) 1.2 1.4 1.6 1.8 x 10 2
6 Input Figure 7 Integrator Circuit
The circuit below is used to amplify a sinusoidal input by a factor of 10. Opamp is ideal and = 10ns. Compute the frequency of sinusoid. C has initially no charge. Figure 8 6 Solution : Like all previous examples we can write: Figure 9 So gain factor here is and the circuit works as an “Integrator”. There is a “ “(or 90o) phase shift between input and output. Input and output voltages are plotted in the same graph in Figure10 to show this phase shift. 7 Input and Output Vs. time 10 8 6
Input and Output(V) Output Input 4 2 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 time(s) 1.2 1.4 1.6 1.8 x 10 2
7 Figure 10 Low Pass Filter (modified integrator)
The below circuit is called “Low pass filter”. Calculate R2 and R1 to have 3 dB frequency at 10 KHz and gain 2 in pass band. C is 20nf and Opamp is ideal. Figure 11 8 Solution: Why do we call this circuit a lowpass filter? C1 is open circuit C1 is short circuit virtual ground Hence at higher frequencies the output becomes zero and at lower frequencies there is a gain of , this means that this circuit does not pass high frequency inputs to the output. In the other word this circuit is a “Low Pass Filter “. Writing the current equations: shorted to negative input of Opamp which is Pass band gain = 2 9 High Pass Filter Circuit (Modified Differentiator)
In circuit below = 10KΩ , find C and is the function of the circuit? to have = 1KHZ and pass band gain = 3. What Figure 12 Solution: There is no path between and so =0 gain. Circuit is like an “Inverting Amplifier” with Hence this circuit allows higher frequencies passage and blocks lower frequencies, it means that it is a “High Pass Filter” . If we follow the same procedure as we did in low pass filter, we can write: Pass band gain = 3 10 ...
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This note was uploaded on 02/15/2011 for the course SYSC 3600 taught by Professor Adsd during the Spring '10 term at Universidad Alfonso X El Sabio.
 Spring '10
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