# BJT - Influence of increasing $V_CE$ on $I_c$

I struggle to understand to relationship of Ic and Vce. I have the following circuit: A voltage supply at 12.4V, a LED with 0.2A current 9V, a BJT with the Ic-Vce relation shown in the picture below and a resistor to ground.

As far is I understood and measured, the LED can be considered a constant current source.

The strange behavior I do not understand is the following: If i increase the supply voltage by 1V to 13.4V, the LED's voltage remains the same, so does the voltage over the resistor. Thus the current mus still be the same, 0.2A. I measured an increase by 1V over the transistor, jumping from 1.4 to 2.4. If i now examine the Ic - Vce graph, I cannot understand why the current doesn't change.

I would be really glad, if someone could enlighten me.

• Is your I-V curve representing the same device model as the BJT in your simulation? Commented Aug 22, 2017 at 4:57

Suppose the OpAmp has gain of 1 Million. Any fluctuation of voltage across Rsense ---- the 2 ohms ---- is used in negative feedback to adjust the bipolar's base voltage in the direction needed and by amount needed, to keep the voltage across the 2 ohms at almost exactly the voltage on OpAmp Vin+.

If a 0.1 volt change in Vbase_emitter is needed, the input differential voltage (Vin+ - Vin-) need only change 100 nanoVolts.

• Thank you very much. I didn't understand the feedback before :) Commented Aug 22, 2017 at 4:17

The high negative feedback in this circuit is creating a constant current sink for the led. So the current through the led, Ic, Ie and current through the 2 ohm resistor is roughly constant regardless of your power supply. Unless you decrease the power supply such that the transistor goes in hard saturation and eventually explodes. Since the current through the resistor and led are constant then the voltage drop across them is also constant. That means that Vce is going to vary the same as your power supply voltage since Vce = Vcc - Vled - Vr2

So, the graph is really not that useful here, but if you want to see the operating point on the graph it is roughly Ib=100mA. So beta is about 20. As you increase the power supply, Ib decreases slightly and beta increases to compensate for the constant Ic=2A

The small change of roughly 10mA in Ib when you increase the power supply by 1V will result mostly in a change at the opamp output of about 2.2V which with an open loop gain of say 100000 is only about 22uV at the negative input. And since Vr2 is 4V, that will only change the 2A current by 0.0055%

The slope in the $I_{c}$ vs $V_{ce}$ curve is caused mainly by the Early effect.

It's possible that the Falstad simulator ignores this effect, or uses a very high value for the Early voltage, as it tends to use simplified models for many circuit components. (Based on a quick test on the falstad website, it looks like the only configurable BJT parameter is beta)

If you want to get better predictions of the behavior of real circuits, you need to use a simulator that includes more accurate models of circuit components.