Urgent Exam help!!What will be o/p of the circuit????

[raghu.khanna]

hi neone plizz tell me what will be out put of this circuit..BZ1 is an ultrasonic receiver(40KHZ)..Its urgent

 

[abated_mars]

yaa It is a part of Ultrasonic receiver circiut ..I am also searching for its answer..If u get to know plizz do let me knw

 

[laktronics]

Hi,

The AC gain of the circuit is 1+R5/R4 = 101.
The DC gain is positive unity since the -ve input is AC coupled to ground and R5 prvides full feed back to input.
R1, R2 is a divider network dividing 5V supply and applies 2.5V D.C to non inverting input of op-amp and the same voltage appears at the output since the DC gain is unity. So, when no signal is applied, the capacitor C3 charges through R8, but is clamped to about 3V (2.5V plus D1 drop). When signal is applied in series with the 2.5V DC bias, it is amplified by the AC gain of 101 and a signal appears at op-amp output swinging around 2.5V. For +ve swings, the capacitor fails to charge through the large time constant caused by R8, but for -ve swings, it discharges through diode and op-amp output. Thus, when signal is applied, the voltage across C3 falls to zero, unable to get charged through R8 during short positive cycles of the applied signal.
In effect, you get a logic high when no signal and a logic zero when signal is applied.
Regards,
Laktronics

 

[10kangstroms]

Excellent and thorough answer, Laktronics!

EDIT: i just realized that the gain-bandwidth product of LM324 is 1MHz. At 40kHz, the gain will only be about 25, not 101 as calculated. The circuit should use an op amp with a GBW of at least 10MHz to realize something close to the calculated gain.

 

[laktronics]

Hi,
Thank you 10kangstroms for pointing out the open loop gain limitations of LM324, you are very right, the AC gain will be reduced to below 25 in this case.
Regards,
Laktronics.

 

[raghu.khanna]

Wont the Dc level change and act like an envelop for the signal coming from BZ*?

 

[laktronics]

Hi,
The DC level does not change since C1 bypasses the AC signal appearing across the DC bias. Remember, the the DC bias has a source resistance of 50K, that is , R1|| R2. You can determine the net voltage appearing at the + input of the op-amp using superposition therom. Thanks to C1, the AC signal appears directly to the op-amp input and it will have a source impedance of Rz1|| R3.
Amplitude Modulation as I understand, is a process of multipying two signals where as in the present case what happens is a DC level shifting of the AC signal.
Regards,
Laktronics.