3) For the constant current source circuit Show that IR =...

3) For the constant current source circuit Show that $I_R = I_{C2} = I_Q$ $B_1 = B_2 = B$, $V_{BE1} = V_{BE2}$ when $B$ is large For Bias stable circuit $R_{Th} = 0.1(B+1)R_E$ $R_{Th} = \frac{R_1R_2}{R_1+R_2}$, $V_{Th} = V_{CC}\frac{R_2}{R_1+R_2}$ $I_C = \beta I_B$, $I_E = (\beta+1)I_B$

Transcript

00:01 So here we have to design an ammeter stabilizer network from here.

00:04 So here we are given the value of vcc that is equals to 20 volt.

00:08 I of c is given that is equals to 10 milliampere.

00:11 We are having the value of beta that is equals to 120 and rc is equals to 4 times of re.

00:16 So we are considering about icq.

00:18 So icq from here is equals to 0 .5 of ic which is multiplied by the s of 80 and rc we are having that is equals to 4 times of re.

00:27 So the value of vce from here is equals to 0 .5 of vcc and the value of vce become equals to 0 .5 which is multiplied by the 20 that become equals to 10 volt and the value of ic become equals to beta that is divided by 1 plus beta which is multiplied by the ie.

00:46 So solving this is a value of ic.

00:49 So solving for ie.

00:52 So ie from here is equals to 121 multiplied by the 0 .5 multiplied by the 10 multiplied by the 10 raised to the power minus 3 that is divided by 120.

01:03 Solving the term we get the value of ie from here that become equals to 5 .0416 multiplied by the 10 raised to the power minus 3 ampere and the value of ic from here i sorry ie from here is equals to f of i multiplied by the 0 .416 milliampere and ic become equals to frm of ampere.

01:27 So these are the values.

01:28 Now we are considering that vc.

01:30 So vc from here is equals to vcc minus ic multiplied by the rc.

01:35 Plugging into the value vc become equals to 20 minus 5 multiplied by the 10 raised to the power minus 3 times of rc.

01:42 Let's say this is the equation number 1.

01:44 So the value of vce become equals to vc minus ve that is this is the value of vce.

01:51 Plugging into the value that is 0 .5 multiplied by the 20 is equals to vc minus ve is equals to ie minus multiplied by the re that is 5 .0416 multiplied by the 10 raised to the power minus 6 of re.

02:02 So solving the term from here so vc become equals to 10 plus 5 .0416 which is multiplied by the 10 raised to the power minus 3 of re.

02:13 Again plugging into the value we can write it as that 10 from equation 1 10 plus 5 .0416 multiplied by the 10 raised to the power minus 3 of re is equals to 20 minus 5 multiplied by the 10 raised to the power minus 3 of 4 of re...

Question

3) For the constant current source circuit Show that $I_R = I_{C2} = I_Q$ $B_1 = B_2 = B$, $V_{BE1} = V_{BE2}$ when $B$ is large For Bias stable circuit $R_{Th} = 0.1(B+1)R_E$ $R_{Th} = \frac{R_1R_2}{R_1+R_2}$, $V_{Th} = V_{CC}\frac{R_2}{R_1+R_2}$ $I_C = \beta I_B$, $I_E = (\beta+1)I_B$

Please give Ace some feedback