Problem regarding a transistor state in a NAND circuit

[diredragon]

Homework Statement

Calculate the output function and draw a graph of output voltage in the function of the input voltage of the following digital circuit:

Homework Equations

3. The Attempt at a Solution [/B]
By analysis of what happens when inputs A and B are on the logical "1" i have concluded that this is a NAND circuit and its function is $Y=\overline {AB}$.
To draw a graph i need to do a complete analysis and find the points which are important to draw on the graph. I have done my calculations up to some point and i will post here the best so you can understand where i got and what i don't get. I will point out stuff i don't quite understand by making the text bold.
Since the circuit is NAND i will place $V_b$ on an inactive level in this case $V_{CC}$ so i can only consider $V_a$
I start when $V_a=0V$
$Q_1$ is saturated due to non existent collector current and $Q_2$ is cut-off because it doesn't have enough voltage on the base and so is $Q_3$. Transistor $Q_4$ and diode $D_1$ have enough voltage to be working but have negligible current so out first $V_Y = V_{CC} - V_{BE4} - V_{D1} - R_3i_{b4} = 3.8V$ since the current is negligible due to $Q_3$ being cut-off.
Next logical step in the circuit happens when the $Q_2$ just begins to work. That happens on $V_a = V_{b2} - V_{ces} = 0.4V$ Nothing changes in the output and it stays $3.8$.
Next $Q_3$ turns on and it does so when it's base voltage is exactly $0.6V$. To check what happens with $Q_1$ at this point i look at the currents of $Q_2$ and compare with currents of $Q_1$.
$i_{e2} = 0.6mA => i_{b2} = 11.7uA << i_{b1} = 950uA$ whch means it's still saturated.
Voltage $V_a = 1.1V$ when $Q_3$ turns on and $V_Y = V_{CC} - R_3(i_{c2} - i_{b4}) - V_{be4} - V_{d1} = 2.84V$ assuming the base current of transistor $Q_4$ is $0$ and approximating the collector current as equal to the emitter current of transistor $Q_3$.
A question: Before the transistor Q3 turns on we had 3.8V for the output as there was no current through R2. When Q3 on in this resulting 2.84V calculation we could say that the base current was still 0 as the transistor had just turned on and does not immediately have current though it's collector?
Here is the graph of the circuit when everything is drawn so you get the idea of what it's about.
$V_{OH} = 3.8, V_{OL} = 0.2$

 

[Tom.G]

i have concluded that this is a NAND circuit and its function is Y=\overline {AB}.

I agree.

Q1 is saturated due to non existent collector current and Q2 is cut-off because it doesn't have enough voltage on the base and so is Q3. Transistor Q4 and diode D1 have enough voltage to be working but have negligible current

I agree.

so out first VY=VCC−VBE4−VD1−R3ib4=3.8V

What is the V_CE of a saturated transistor? How would that affect the above formula?

Va=Vb2−Vces=0.4V

I agree.

Voltage Va=1.1VV_a = 1.1V when Q3 turns on

I agree. For later use, what happens to Vb4 as Q3 turns on?

and VY=VCC−R3(ic2−ib4)−Vbe4−Vd1=2.84V

Please check the signs in the above formula. I think there is a problem associated with R3.

assuming the base current of transistor Q4 is 0 and approximating the collector current as equal to the emitter current of transistor Q3.

If Q4-B current is 0, how is there a voltage at Y when Q3 starts to conduct?

we could say that the base current was still 0 as the transistor had just turned on and does not immediately have current though it's collector?

I'm having trouble with that! How can a transistor be "On" with zero current flowing?

I suggest you calculate the output voltage with Q3 cutoff (which was your above approach), then with Q3 saturated. Be aware of what Q4-B is doing during this transition.

Hint: Q2,3,4 look an awful lot like an audio output stage.

 

[diredragon]

What is the V_CE of a saturated transistor? How would that affect the above formula?

Depends on which transistor your asking for. If you mean $Q_3$ its $V_{ces}$ and it just sets the output to $0.2$.

I agree. For later use, what happens to Vb4 as Q3 turns on?

Decreases as the loop in which both Q3 and Q2 are saturated and work, both diode and the transistor can't work.

Please check the signs in the above formula. I think there is a problem associated with R3.

It's a typo, its +ib4 but it doesn't matter as it's zero in this case.

If Q4-B current is 0, how is there a voltage at Y when Q3 starts to conduct?

I'm having trouble with that! How can a transistor be "On" with zero current flowing?

That would be the voltage $V_ϒ = 0.6V$ i assumed. I was looking at the case where the voltage is just lower than that to be negligible in the calculations but large enough for the current to not go through it yet. Its because a voltage of 0.59 is almost negligible in the calculation but it's lower than necessary for current to flow. Any further increase createst current. Since I'm dealing with discrete values it's necessary.

For your last suggestion i did calculate and then chose my points $V_{OH} = 3.8V$ with Q3 cutoff and $V_{OL} = 0.2V$ with Q3 saturated.

 

[Tom.G]

With the exception of the comment about Vces below, these all tie together to get an improved result. For the moment we'll ignore the "How can a transistor be "On" with zero current flowing?" I think it falls out when the rest is straightened out.

so out first VY=VCC−VBE4−VD1−R3ib4=3.8V

What is the VCE of a saturated transistor? How would that affect the above formula?

Depends on which transistor your asking for. If you mean Q3 its Vces and it just sets the output to 0.2.

I assume you mean V_CEsat there. V_CES is breakdown Voltage of the C-E junction with the Base 'S'horted to Emitter, or V_BE=0. (Extra comment: V_CEX would be with a specified circuit between the Base and Emitter.)
But I was really referring to Q4 since that helps determine V_Y. If you have a circuit with only a 5V supply, R3, R4, and Q4, what is the voltage at Q4-E?

Voltage Va=1.1V when Q3 turns on

I agree. For later use, what happens to Vb4 as Q3 turns on?

Decreases as the loop in which both Q3 and Q2 are saturated and work, both diode and the transistor can't work.

I was trying to ask what happens during the transition between Q4 saturated and Q3 saturated. i.e when Q2 is operating in its linear region.