Q4) Assuming that all components in Fig. 3 are perfectly...

Q4) Assuming that all components in Fig. 3 are perfectly matched, find (a) The base-to-ground voltages VB1 and VB2 of each transistor (3 pts) (b) Calculate the input power, output power, and power handled by each output transistor and the circuit efficiency for an input of 12 V rms (9 pts) (c) The power delivered to the load under maximum signal conditions, and the efficiency under maximum signal conditions, (8 pts) (d) The value of capacitor Cc if the amplifier is to be used at signal frequencies down to 20 Hz (5 pts) +20 V Figure 3 1 k? Q1 1 ? Cc 1 ? Q2 1 k? 10 ?

Transcript

00:01 So here in this quotient we are given the figure where we are given this figure is somewhat like this.

00:07 Here this is vcc voltage which is plus 20 volt which is connected to a 1 kilo ohm resistor further this resistor is connected to a capacitor and let's say here is the v input here further this capacitor is connected and from for that two diodes let's say this diode is d1 and this from here is d2 are connected in series with each other then they are in connection with series with the 1 kilo ohm resistor.

00:42 Here we are having a p -n junction diode which is altered from here in this point.

00:50 Now this another end of this diode is connected to a resistor which is 1 ohm and then further of 1 ohm resistor is connected and in between them there is a capacitor that is cc is connected which is further connected to a resistor rl that is equals to 10 ohm which is from there and now in that point again a p -n junction diode is connected this is q1 from here so this is the entire circuit.

01:22 So here in the first part of this quotient we need to find out the the base ground voltage vb1 and vb2 of the each transistor.

01:32 So here what we can do is we have to apply kvl to vcc to ground loop through 1 kilo ohm resistor and diodes.

01:44 So this from here is equals to minus i multiplied by the 1 kilo ohm minus of 1 .4 minus of i multiplied by the 1 kilo ohm plus 20 that is equals to 0.

01:55 Simplifying this term from here we get the value of i that is equals to 9 .3 milliampere so this is the value of i.

02:02 So base to ground voltage that is vbr and vb1 and vb2 from here is given as vb1 is equals to 20 minus i multiplied by the 1 kilo ohm that is equals to 20 minus 9 .3.

02:17 So the value of vb1 from here is equals to 10 .7 volt.

02:22 In the same way the value of vb2 become equals to i multiplied by the 1 kilo ohm that from here is equals to 9 .3 multiplied by the 10 raised to the power minus 3 multiplied by the 10 raised to the power plus 3.

02:36 So vb2 from here is equals to 9 .3 volt.

02:40 Hence the answer to the part a of this question.

02:43 Now in the part b of the question we need to find out the input power, output power and power which is handled by each output transistor from here.

02:53 So hence we are considering about a maximum single conditioning from here.

03:00 Basically we need to find out the maximum power.

03:03 So vm from here is equals to vcc which is divided by 2 that from here is equals to 20 divided by 2 which from here is equals to 10 volt.

03:11 So what we can do here having the value of vdc, pdc that is equals to vcc multiplied by the i of cq where i cq is equals to 2 of im which is divided by pi.

03:27 So the value of rl from here is equals to vm which is divided by im.

03:31 So im from here is equals to vm divided by the rl plug into the value that is equals to vcc divided by the rl...

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