- #1

diredragon

- 323

- 15

## 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##