# Why does the colllector current decrese with decreasing collector-emitter voltage in this circuit?

I'm trying to understand how the circuit below works so I ran a simulation with TINA and I got the curves shown on the right (C3 is already charged). Here is a bigger one

Considering the characteristic curve of that BJT (shown below), I can't understand how it is possible for Vce to decrease with decreasing Ic. My reasoning is as follows (the lines are not the ones corresponding to the circuit above, but they are close enough to expect a similar behavior):

The red line is the DC load (Vtank = 0) and the green line is the one I would expect after Vtank (EDIT: I made a mistake here, this Vtank is 180° out of phase with respect to the one shown above, but the discussion still stands), increases from 0. The yellow dot is the initial Q point. Since the Q point must be on the green line after a certain time, and since the simulation says that both Vce and Ic decrease, I'm having trouble understanding what's going on.

Of course Ic can't simply increase either because otherwise Vbe would go down (through Re) and Ic would be immediately interrupted.

EDIT:

After trying with a lower frequency and with a higher Re I got this: Here is a bigger one.

So what seems to be happening is that locally (higher frequency) the transistor is behaving like I expected but non locally (lower frequency) the tank oscillations define the behavior. I would like someone to confirm this.

• Ant-why do you expect a shift of the Q point due to reactive elements (L and C)? The Q point remains constant. – LvW Jan 23 '15 at 8:34
• And where is the input stimulus? I think that what you are looking at is the switch on decay of the LC circuit oscillating. Also, how did you draw the red line - what fixed base current did you assume? – Andy aka Jan 23 '15 at 9:21
• @LvW he might be referring to the dynamic Q point, and Andy that's what I was thinking, at first I thought it could be an oscillator but it's missing an inductor in series with C3 to oscillate. – Vladimir Cravero Jan 23 '15 at 9:22
• The dynamic Q point would still go thru the yellow dot but the green line would have a different slope. – Andy aka Jan 23 '15 at 9:24
• @Ant, another question: Why do you expect a decreasing Vce in case of a lower Ic? – LvW Jan 23 '15 at 9:42

In the top simulation, your $I_c$ is only changing by 10uA for a 20mV. This is just the effect of the Early voltage and the finite output resistance of the transistor. From your data I calculate an Early voltage of about 80 V (very reasonable). I don't know what's going on with the LC tank, but it isn't the cause of the $I_c$ variation.