\(
a. v_l=0V \\
v_l=0V => \text{Q1. Q2 is OFF} \\
V_E,V_B=0V \text{since no IC1 current flowing and drop across 1kOhm is 0V}\\
b.v_l = 3V; \\
\text{for Q1} \\
V_B - 0.7V = V_E \\
\text{for Q2} \\
V_{EB} = 0.7V => \text{Q2 OFF} - eq1 \\
\text{Applying KVL to Base circuit of Q1} \\
3V - 10k\Omega*I_B - 0.7V - 1k\Omega*101I_B =0 - eq2 \\
I_B = 0.0207mA ; V_B = 3V - 10k\Omega*0.0207mA = 2.98V \\
c.v_l = -5V; \\
\text{Q1 is OFF} \\
\text{Q2 Active} \\
\text{Applying KVL to Base circuit of Q2} \\
-101I_B*1k\Omega - 0.7V - 10k\Omega * I_B +5V = 0; \\
I_B = 4.3V/111k\Omega = 0.0387mA;
V_E = 0V - 101*0.0387mA *1k \Omega (I_E*\beta) => -3.91 V \\
V_B = -3.91 - 0.7V = -4.61V \\
d. v_l = -10V
\text{Q1 is OFF} \\
I_B = 9.7V/111k\Omega = 0.087mA; \\
V_E = 0V - 101*0.087mA*1k\Omega = -8.82V \\
V_B = -9.52V \\
\text{Q2 is not active but saturation} \\
V_E = -5V + 0.2V = -4.8V \\
V_B = -4.8V - 0.7V = -5.5V \\
\)
After quite an effort i calculated the values, are my justifications correct for the states of Q1 and Q2 and the values?

 

hi V123,
This plot will help check your results.
Vin sweeps down from +3V down to =10V.
E

 

Sure I will check mistakes are there I think.

 

As the circuit is symmetrical, you can ignore Q2 and just analyze Q1 emitter follower circuit with Vin at 0 V, 3 V, 5 V, and 10 V.

 

hi V123,
This plot will help check your results.
Vin sweeps down from +3V down to =10V.
E

View attachment 364151

Hi Eric,

It would be interesting to see how this circuit responds if we start the sweep at 0v and keep it there momentarily, or maybe even for a long time.
What is interesting is in the perfectly ideal circuit, this circuit would be completely symmetrical. In the real world even of Spice, there could be very slight differences in both the speed of each transistor and the base emitter drops. Since current can flow into the base, it looks like one or the other transistor can turn on first, before the other, or one could conduct more current than the other for a brief time. That brief time could make Ve rise (or fall) slightly, which makes it look like one transistor will turn on first and keep the other one off.
Didn't do an analysis of this yet, but that's just a hunch for now.

 

View attachment 364141
View attachment 364142

\(
a. v_l=0V \\
v_l=0V => \text{Q1. Q2 is OFF} \\
V_E,V_B=0V \text{since no IC1 current flowing and drop across 1kOhm is 0V}\\
b.v_l = 3V; \\
\text{for Q1} \\
V_B - 0.7V = V_E \\
\text{for Q2} \\
V_{EB} = 0.7V => \text{Q2 OFF} - eq1 \\
\text{Applying KVL to Base circuit of Q1} \\
3V - 10k\Omega*I_B - 0.7V - 1k\Omega*101I_B =0 - eq2 \\
I_B = 0.0207mA ; V_B = 3V - 10k\Omega*0.0207mA = 2.98V \\
c.v_l = -5V; \\
\text{Q1 is OFF} \\
\text{Q2 Active} \\
\text{Applying KVL to Base circuit of Q2} \\
-101I_B*1k\Omega - 0.7V - 10k\Omega * I_B +5V = 0; \\
I_B = 4.3V/111k\Omega = 0.0387mA;
V_E = 0V - 101*0.0387mA *1k \Omega (I_E*\beta) => -3.91 V \\
V_B = -3.91 - 0.7V = -4.61V \\
d. v_l = -10V
\text{Q1 is OFF} \\
I_B = 9.7V/111k\Omega = 0.087mA; \\
V_E = 0V - 101*0.087mA*1k\Omega = -8.82V \\
V_B = -9.52V \\
\text{Q2 is not active but saturation} \\
V_E = -5V + 0.2V = -4.8V \\
V_B = -4.8V - 0.7V = -5.5V \\
\)
After quite an effort i calculated the values, are my justifications correct for the states of Q1 and Q2 and the values?

Hi there,

It looks like you set the diode voltage drops to 0.7 volts but I do not see that in the problem specifications.
Did you just assume that or was that given to you through some other channel?

It would be interesting to see what happens if the drop is 0v too as it is presumed to be in some other circuit problems.

 

Hi there,

It looks like you set the diode voltage drops to 0.7 volts but I do not see that in the problem specifications.
Did you just assume that or was that given to you through some other channel?

It would be interesting to see what happens if the drop is 0v too as it is presumed to be in some other circuit problems.

Thank you for the reply, I assumed 0.7V, but i am stuck up with Vl = 0V itself i will post the queries i have in some time.

 

I simulated the circuit, i have made a mistake in the voltage sign (5V instead of -5V) and got wrong values for Vl=0V, i have corrected and it is now matching with the results. Sorry for confusion and thank you for support.

 

View attachment 364141
View attachment 364142

\(
a. v_l=0V \\
v_l=0V => \text{Q1. Q2 is OFF} \\
V_E,V_B=0V \text{since no IC1 current flowing and drop across 1kOhm is 0V}\\
b.v_l = 3V; \\
\text{for Q1} \\
V_B - 0.7V = V_E \\
\text{for Q2} \\
V_{EB} = 0.7V => \text{Q2 OFF} - eq1 \\
\text{Applying KVL to Base circuit of Q1} \\
3V - 10k\Omega*I_B - 0.7V - 1k\Omega*101I_B =0 - eq2 \\
I_B = 0.0207mA ; V_B = 3V - 10k\Omega*0.0207mA = 2.98V \\
c.v_l = -5V; \\
\text{Q1 is OFF} \\
\text{Q2 Active} \\
\text{Applying KVL to Base circuit of Q2} \\
-101I_B*1k\Omega - 0.7V - 10k\Omega * I_B +5V = 0; \\
I_B = 4.3V/111k\Omega = 0.0387mA;
V_E = 0V - 101*0.0387mA *1k \Omega (I_E*\beta) => -3.91 V \\
V_B = -3.91 - 0.7V = -4.61V \\
d. v_l = -10V
\text{Q1 is OFF} \\
I_B = 9.7V/111k\Omega = 0.087mA; \\
V_E = 0V - 101*0.087mA*1k\Omega = -8.82V \\
V_B = -9.52V \\
\text{Q2 is not active but saturation} \\
V_E = -5V + 0.2V = -4.8V \\
V_B = -4.8V - 0.7V = -5.5V \\
\)
After quite an effort i calculated the values, are my justifications correct for the states of Q1 and Q2 and the values?

It's hard to follow your work because it is all crammed together. At least separate each part of the problem with a blank line. Also, don't put multiple results on the same line since people look at the beginning of the line and see, for example, I_B and they expect that line to be about I_B. If they are looking for V_B, they will tend to move on.

Also, you don't provide all of the answers that you were asked for. For instance, for V_i = 3 V, what did you get for V_E? Based on the 2.98 V you got for V_B, it appears that you would have gotten 2.28 V for V_E.

So, recalling that most answers can be verified starting from the answer itself, let's check if this is correct.

If

V_E = 2.28 V

then

I_E = 2.28 V / 1 kΩ = 2.28 mA

This means that

I_B = I_E / (ß+1) = 2.28 mA / (100 + 1) = 22.57 µA

But this is not in good agreement with your result of 20.7 µA.


You REALLY need to get in the habit of checking your own work.

Knowing that your result is wrong, I'm willing to bet you can track down your mistake pretty easily.

You ARE going to make these kinds of mistakes. You are human. We all make them. It is YOUR responsibility to develop and USE techniques to catch them.

One of the useful approximations that can be made in this circuit is that the transistor gain makes the emitter resistor look, to the base, like a resistor that is ß times larger (actually, ß+1 times larger). So that means that the circuit looks like a voltage divider producing an output of

V_E = (3 V - 0.7 V) * (10 kΩ / (10 kΩ + (100 + 1)*(1 kΩ)) = 2.093 V

 

The V_B voltage when V_l is 3V
\[ V_B = 3V - 10k\Omega*0.0207mA = 2.793V \\ V_E = 2.793V - 0.7V = 2.093V \\ I_E = \frac{2.093V}{1k\Omega} \\ I_E = 2.093mA \\ I_B = \frac{2.093mA} {101} \\ I_B = 0.0207mA \]
I will verify others as well. I will ensure that i verify my results before i post, thank you for support.

 

hi V123.
This LTS sim will enable you to check your calculations.
Just change the Vin and Vsample points on the asc file and the read the Spice Error Log,
E
I have included the V123Calc2.asc file

 

It would be interesting to see how this circuit responds if we start the sweep at 0v and keep it there momentarily, or maybe even for a long time

Hi Al,
The results are almost identical for a swept Vin and a stepped Vin.

E

 

Hi Al,
The results are almost identical for a swept Vin and a stepped Vin.

E
View attachment 364179

Hi Eric,

Oh thanks for the graph and circuit and .asm file too.

I am sorry I forgot to mention that the experiment I was suggesting was with the ideal base emitter voltage of 0v. That would not be practical anyway though because if we call it 0v then we can't have a variation in Vbe anyway. I suppose we could have a delay time difference, but it's not useful to do that now I don't think.

With Vbe=0.7 then we get no conduction ever though either base emitter diode when the input is 0v because the output is also biased at 0v.