Temperature changes also change the hFE of a transistor. It causes "thermal runaway".

Originally Posted by Karkas

is there another factor?

Silicon transistors have a max allowed emitter-base voltage (when the voltage is the reverse of what is normal). It is 5V to 7V. If the voltage is exceeded in a circuit continuously or for many times then the junction is damaged and the hFE is permanently reduced.

Thanks for the answer, it will be useful. But wouldn't it be 0,5 to 0,7 V?
I mean the base emitter voltage.

Normally the base-emitter junction of a transistor is forward biased so it conducts like a diode with a 0.7V voltage drop.
But if the base-emitter junction is reverse biased then it breaks down with what is called Avalanche Breakdown at 5V to 7V.

The datasheet for every silicon transistor lists the max allowed reverse voltage called "VEBO, Max Emitter-Base voltage" and is 6V for the 2N3904.

Originally Posted by Karkas

Hi, this is for an electronics lab report.
I've made three graphics for the output characteristics (for Ib=5µA, 15µA and 40µA) and thre for the input characteristics (for Vce= 2V, 4V and 6V) for an NPN transistor (2N3904), and calculated the hfe for two ponits Ib= 15µA; Vce= 2V and Ib= 40µA; Vce= 6V, the first was 220 and the second was 232, of course these values read from the output characteristics graphic, i know that hfe will change if I give a higher base current, and if I have a higher collector emitter voltage, but what other factors can make it change how it changed in my practice?

Thanks.

Did you notice that the graph of output characteristics is non-linear? What method are you using to calculate hfe? This quantity is defined as the tangent of the operating point on the curve. Any two points on the curve will have different tangents.

Well there's your answer.