How to plot the Ib vs VBE and IC Vs VCE for a transistor with IB as the variable resistor?

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Joined Oct 21, 2014
Hey everyone,
I've been stuck on this problem for a while, but I am supposed to to plot Ib vs VBE and IC Vs VCE with IB as the variable resistor using a simulator (I'm using circuit maker) for three separate transistors (2N2222A , 2N3904, 2N44401)). I have set up a circuit to bias the transistor with VCE- 5V and Ic=1mA, however, the graphs I am getting do not seem to make sense. I've tried to put a current source of 1mA at the base and voltage source at the collector to find the IC vs VCE and then a 10V voltage source at the base and a 5V voltage source with a 1k resistor at the collector to find the Ib VS VBE. However, none of the graphs seem to make sense to me and seem very different than what it should be looking like. It would be so great if someone can help me out! All help is appreciated :)



Joined Aug 23, 2012
Since there is no Rb in series with b of bjt, the transistor could be damaged by 10V.
10V through a Rb to b of bjt so you can get a Ib.


Joined Mar 31, 2012
For Ib vs Vbe, you need to leave the collector open. The point of the curve is to measure the diode characteristic of the base-emitter junction.

For the Ic vs. Vce, you need to set a value for Ib (a current source should work fine) and then vary Vce (a voltage source should work fine). You repeat this for several different values of Ib. You need to see if your simulator will do what are generally called parametric sweeps.


Joined Mar 31, 2012
I can't tell what you are doing from the plots you give -- I'm not familiar with your simulation package or how it shows information on the plots.

What is your top plot?

Is your plot X-axis for the values of Vce, which is NOT the output of your voltage source. Remember, Vce is the voltage difference between the transistor collector and emitter. If you connect your voltage source directly between these two points (i.e., get rid of the resistor or, if you need it for current sensing, make it zero ohm) then they are the same.

Until you get your basic curve working right, just fix the current source driving the base to something like 1 mA or so. Then do a sweep on the voltage source from 0V to about 15V and do it in smaller increments, such as 0.25 V or 0.1 V.


Joined Oct 2, 2009
I used LTSpice, which understands the .DC... syntax (think of it as nested FOR loops), used by Berkley Spice since 1973. I have no idea how to do that with circuit maker, which seems to have gone out of its way to make this difficult...
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