**LAB 5 FREQUENCY RESPONSE OF OPAMP**

Goal:

To construct a simple op amp and find its,

1) 3-dB frequency

2) Open loop bandwidth

3)Unity gain frequency

4)Phase lag at unity gain and

5)Phase margin

Set up:

For our differential pair, we need to give two out of phase signals one each at the inverting and the non-inverting terminals.

First we need to have two out of phase signals, for which we need to construct an inverting amplifier with gain 1.

So, today’s experiment consequently has two stages to be constructed.

1)Inverting Amplifier with gain 1 and

2)A simple differential amplifier.

First we are constructing the inverting amplifier and then connecting it to the simple differential amplifier.

Construct the following inverting amplifier,

Procedure:

1)Input a sine wave of 100mVpp - 100Hz – zero V DC offset at A.

2)Adjust R1 till you get the maximum output voltage swing at the drain of the transistor.

3)Now adjust the R2 to get the 100mVpp sine wave at B.

4)The waveform at **B would be 180 degrees** out of phase with that of A.

Your waveforms would look as follows,

Now let us move to construct the second stage.

**Differential Stage**

Procedure:

1)Ground A and B.

2)Vary **Rb, so that X = 0V and Y = 0V. Make sure that Y = 0V, rather than X because Y** is where we are going to tap the output for this stage.

3)Now disconnect A and B from the ground and connect it to the **A and B of the first stage** (the inverting amplifier), respectively.

4)Now observe the waveform across X and Y. i.e. in oscilloscope probe, connect **Y to the signal line and X** to the ground line.

Note: At any stage of the experiment, if you see the output waveform appearing clipped either on positive or negative side, you can always **adjustthe bias resistorRb** to get the unclipped waveform.

5)Now sweep the input frequency from 100Hz to 10MHz.

6)Note the output amplitude for 30 points, including 100Hz, 1KHz, 10KHz, 100KHz, 1MHz and 10MHz.

7)Also note down the phase lag observed at the output for those 30 points. You have to **visuallymeasurethe phase lag** between the input and the output waveforms employing the zero crossing trigger of CRO.

For example, let us assume that we have connected the input to a 100Hz sine wave, which means that the sine wave will have 10 ms as its time period.

Now lets assume that the output is lagging by 90 degrees, in which case the waveform is going to be like the following,

Now when you employ the dual channels of the CRO, you will observe as,

When you notice the waveform, 10 ms corresponds to 360 degrees of the waveform (i.e. one full cycle). Now when you observe the difference in time between the two zero crossings, it would be 2.5 ms.

So, when 10 ms corresponds to 360 degrees, then 2.5 ms would correspond to

(2.5/10) x 360 = 90 degrees.

Thus you need to visually measure the phase lag between the input and the output.

8)Now draw the following graphs in the logarithmic graph sheets.

a)Frequency response plot

b)Phase response plot

The **gain in dB is given by 10 log (Vout/Vin)**

F3dBis the point where the output magnitude is 0.77 times the input magnitude.

Funityis the point where the output and the input have the same magnitude.

The bandwidth is same as the F3dB point.

The phase lag at unity gain frequency is the -

The **phase margin is (180 - **)

From the above graphs, report the following results.

RESULTS:

6) The 3-dB frequency is ___

7) The open loop bandwidth is ___

8)The Unity gain frequency is ___

5)The phase lag at unity gain point is___

6)The phase margin is ____

7) Prove analytically why the phase is -45° at the three dB frequency.