# BJT LT spice DC transfer characteristic

Why is it that the DC transfer characteristic isn't flattening out at -5 v, so it looks like a Z-shape? I did a DC sweep from -10 V to +10 V.

The input wave is a 2V sine wave for the graph below:

• Which source are you sweeping? – Linkyyy Dec 7 '18 at 21:39
• Im sweeping VAC – David Dec 7 '18 at 22:28

It seems that you're sweeping VAC from -10V to +10V while holding VCC and VEE steady at 5V each. Note that since you're driving the base with a voltage source, R1 and R2 have no effect on the circuit whatsoever.

The first part of your curve (from -10V to a little over -5V) is where the Q1 is cut off completely. There's no current through RC, and no voltage drop across it, so Vc sits at +5V.

The next part (from -5V to -4V) is where Q1 is starting to conduct. As the current through RC increases, the voltage drops, until Q1 is saturated. At this point, Vc is a little bit less than Vb.

From this point on, Q1 is completely saturated, and Vc tracks Vb with a small offset that represents the forward drop of the B-C junction. This can drive Vc even beyond VCC, as you can see.

• can u help me fix it. What should I do? – David Dec 7 '18 at 22:19
• Fix what? It's working exactly as designed. If you think it should do something different, you're going to have to specify what that is. – Dave Tweed Dec 7 '18 at 22:21
• the gain is the slope of the slanted line right? – David Dec 7 '18 at 22:28
• Well, yes -- in a way. In the first part (cutoff), the gain (slope) is zero. In the second part (active), it is approximately -16.5 (using -Rc/Re as an approximation). In the last part (saturation), it is slightly less than +1. – Dave Tweed Dec 7 '18 at 22:34
• I updated the question. Can u explain how the vin vs vout graph is related to what you said about the DC transfer...For this vin vs vout graph, the input wave is 2 v sine wave. – David Dec 7 '18 at 22:47