Calculating minimum resistance such that Ic is less than or equal to the maximum Ic in BJT

This is what I have been given for the question. I'm confused on how to answer this question.

I thought this was calculated just by using ohm's law where R = 5V(supply V) / .3A(max Ic) = 18. However this seems rather low. If anyone can steer me in the right direction i would appreciate it. Thanks.

• If you had, e.g., a 5 V / 1.5 W motor, it would indeed have a resistance of about 16.7 Ω. But this question is not about any realistic circuit, it just wants to you to find out what resistors you can use without overloading the transistor. In a real circuit, there would be more constraints (such as the maximum power of the resistor, or the desire not to waste power). – CL. Nov 3 '16 at 9:31

On my calculator, 5/0.3 is 16.7, not 18; but other than that you are right.

But you are right that this is a very low value. The resistor will be generating $(0.3\ \rm{A})^2(18\ \Omega)$, or 1.5 W of heat, for no particular purpose, unless the resistor represents a heater element.

Also, this circuit will not actually put the BJT into saturation unless the input signal goes above the 5 V power rail, which is a somewhat uncommon circumstance. On the other hand, if you drive the input too far above 5 V, you will forward bias the b-c junction, resulting in high current draw from the input. This makes the circuit more difficult to control correctly, even if you do have a way to drive the input above the supply voltage.

You should probably give your answer to this homework, then forget you ever saw this circuit. If you want to switch power to a resistive load, don't do it this way. Google "low-side switch" for the most common way to do it.

• Thanks for the reply, I just had no idea if what i was doing was correct. I didnt actually mean 18 that was a typo, sorry. I was just confused cause the resistance was low, and the amount of heat was high. I am in a micro-controllers class and all the resistors we've been dealing with have a 1/4W rating so 1.5W seemed high and i thought i was calculating my answer wrong. – Frank Nov 3 '16 at 3:31