Why does this circuit work?

Question

I'm suffering from a bit of confusion here and I was hoping that some kind soul here would be able to help me out. I want to build a small DC motor circuit just for fun, but I can't understand some aspects of the circuit (shown in the picture below)

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So basically I understand that Ic = Beta * Ib. Thus (assuming that D3 is 3.3 V and Beta is around 100), I can say that Ic = 100*(3.3/1000) = 0.33 Amps.

Then, in order to find the voltage available to the motor, I must find the voltage drop across the 33 Ohm resistor and subtract that from my power rail voltage (5 V).

So 5V - (0.33)(33) = 5V - 10.89V

Which is quite clearly ludicrous. Can anybody quickly explain to me the fault in my reasoning? I'd really appreciate it.

Answer 1

... I can say that Ic = 100*(3.3/1000) = 0.33 Amps.

Incorrect. You can say that Ic has a maximum of 0.33A (or whatever it would actually be with the correct values and formulae). If the supply isn't actually capable of supplying that much for any reason then the transistor is operating in saturation mode, where it acts as a switch. Which is exactly what we want it doing for operating a motor.

Answer 2

You can pretty much be guaranteed that your \$I_C\$ current is going to be

$$I_C\leq\frac{5V}{33\Omega+R_{motor}}\leq\frac{5V}{33\Omega}=152mA$$

Just because there's no way for it to be higher than that, no matter what you do with the transistor!

From there you can consider your base current, \$I_B\$, and its effect on the collector current.

Since the \$V_{BE(sat)}\$ is 0.7V typically, we can try

$$I_C=\beta\times \frac{3.3V-V_{BE(sat)}}{R_B}=100\times\frac{3.3V-0.7}{1k\Omega}=260mA$$

It's safe to assume (as @Ignacio pointed out) the transistor is in saturation (because \$I_C<260mA\$), and the actual \$I_C\$ is less than 152mA due to the resistance hanging off the collector.

Answer 3

Horowitz and Hill explain this memorably by means of a cartoon 'Transistor Man' who sits inside the transistor constantly watching \$I_B\$ on an ammeter and trying to keep \$I_C\$ equal to \$\beta \times I_B\$. But the only control he has at his disposal is a variable resistance connected between C and E. If he turns the variable resistance all the way down to zero and the collector current is still less than that, well, that's all he can do.

In reality his variable resistance doesn't quite go to zero - and it isn't actually a resistance - which is why checking the transistor's datasheet for the collector-emitter saturation voltage, \$V_{CEsat}\$, is always a good idea, and why you sometimes need to give a transistor more base current than the simple \$I_C = \beta \times I_B\$ equation suggests.