I am trying to find the maximum current allowed through the load for a BJT transistor but I am having trouble understanding what I need from the datasheet. I know my base current is 1 mA but I do not understand what DC current gain I would use to try and obtain the current through the load (which I think is equal to collector current).

• If it's not just a little less than the load would take with its negative terminal grounded this is an awful circuit. Commented Feb 11 at 1:41
• jbiscan, You need to discuss the load itself. Details matter. Also need to discuss the microcontroller, itself. Sometimes, details there also matter. Also, need to discuss the application as how and how fast you intend operating this also matters. And so on. Everything matters. So write more. Please. Commented Feb 11 at 2:54

If you want a guaranteed very low voltage drop for the transistor, you can draw 10mA according the datasheet you linked:

It will have less than 200mV of voltage drop with a 10mA load and 1mA Ib.

If you don't mind if the transistor drops 1V (guaranteed maximum) you can draw 70mA (causing 70mW dissipation in the transistor and robbing the load of 1V of voltage).

Both those guarantees are at 25°C. If your application must operate at lower temperatures than normal room temperature, reduce the above numbers to compensate for the lower gain as shown the typical (that means not guaranteed) numbers in the below curve- and note that they are at the much higher 5V Vce so at the higher currents the transistor would quickly burn up).

This particular transistor is not very good at high currents, I would not use it much above about 50mA, when much better ones are available for a similar cost.

• Thank you for the helpful information. Would you expect the maximum load current to decrease if I decrease the MC supply voltage via the formula involving collector current, beta, and base current?
– anon
Commented Feb 11 at 3:42
• If the base gets less current then the maximum load current would be less, yes. But you can reduce the base resistor to compensate. Commented Feb 11 at 3:43
• Right, thank you. One last thing I was wondering if I want to achieve that min voltage drop of 200 mV, does that mean I just need to grab the right resistor between Vcc and the junction?
– anon
Commented Feb 11 at 3:50
• Assuming you have the 1mA (or a bit more) base current, and a 12V source, a load of 1200Ω or higher between the collector and +12 will be okay. Commented Feb 11 at 4:00

You'd be looking for hFE, the current gain. In your datasheet, fig. 1.

Note that this varies in general, and especially for your transistor, a lot with temperature.

Note that these curves are showing a fixed 5 V voltage between collector and emitter. That's not going to be desirable for just every load! But they do tell you that if you inject 1 mA of base current, you get 5V across the collector-emitter junction is your load is actually drawing ca 220 mA.

If you need better regulation, especially in the face of warming up, try to make your circuit independent of the exact gain. This usually calls for feedback and more than one transistor.

• Thank you for the response. Where do you see in the sheet that 1mA of base current gives you a load/collector current of 220 mA?
– anon
Commented Feb 11 at 2:10
• Figure 1, room temperature. The gain is roughly 220. Again, at that current you get a CE voltage of 5 V (as specified in the same figure) Commented Feb 11 at 2:11
• Gotcha, I see the 220 at 1 mA there but that figure seems to be for the collector current vs. Hfe and not base current?
– anon
Commented Feb 11 at 2:12
• Ah if I misread the figure's horizonal axis, please edit my answer. I'm on phone now. Commented Feb 11 at 2:14

It will help to picture the two scenarios representing the state of the circuit with the transistor switched completely on, and with the transistor completely off. Here's are some example configurations:

simulate this circuit – Schematic created using CircuitLab

When the transistor is switched on, it's (naively) like a closed switch between collector and emitter, a state shown in circuit C. Conversely, when Q1 is off, that "switch" is open, as in circuit B.

The first thing I'd like to correct is that base potential $$\V_B\$$ is capped to about +0.7V. This occurs when $$\V_{IN} >> 0V\$$, so if $$\V_{IN} = +5V\$$, then base current $$\I_B\$$ (through R1) is:

\begin{aligned} I_B &= \frac{V_{R1}}{R_1} \\ \\ &= \frac{V_{IN} - V_B}{R_1} \\ \\ &= \frac{(+5V) - (+0.7V)}{5k\Omega} \\ \\ &= \frac{4.3V}{5k\Omega} \\ \\ &= 860\mu A \end{aligned}

You asked what current gain is required, but you don't control that. Current gain is set by the transistor, which you lookup in the datasheet, or measure. The parameter is $$\h_{FE}\$$, but I and many others call it $$\\beta\$$.

Let's say $$\\beta = 100\$$.

Knowing base current $$\I_B\$$ and current gain $$\\beta\$$, you can work out the approximate maximum collector current you can expect from your transistor:

\begin{aligned} I_C &= \beta I_B \\ \\ &= 100 \times 860\mu A \\ \\ &= 86mA \end{aligned}

I say approximate, because that value of $$\\beta\$$ is only valid for certain conditions. Near saturation, for instance, $$\\beta\$$ will drop off dramatically, but we'll account for that in a moment.

I also said "maximum". That's because no matter what the load, and what the supply potential, you can never have more collector current than this. You can have less, but not more, meaning that you have to ensure that your load won't ever require more than 86mA. If your load needs more than 86mA, the transistor will need more base current.

Typically calculations go the other way; you start by understanding what current your load requires, then choose R1 to provide the necessary base current. That might go like this: you have a load that will pass $$\I_C=12mA\$$ with the full 12V supply across it. That's the situation shown in circuit D. Then minimum required base current will be:

\begin{aligned} I_B &= \frac{I_C}{\beta} \\ \\ &= \frac{12mA}{100} \\ \\ &= 120\mu A \end{aligned}

Remembering that $$\\beta\$$ reduces near saturation, we'd better double that value to ensure saturation. Then our choice of R1 will be:

\begin{aligned} R_1 &= \frac{V_{R1}}{I_B} \\ \\ &= \frac{V_{IN} - V_B}{I_B} \\ \\ &= \frac{4.3V}{2 \times 120\mu A} \\ \\ &= 18k\Omega \end{aligned}

With a simple resistive load like R2, calculating the required collector current is easy, but be aware that not all loads are so well behaved. Your calculations for minimum $$\I_B\$$ and R1 can only start once you know the largest $$\I_C\$$ that you will require, and depending on the load, that might not be so trivial.

• Thank you for the helpful response. So your beta value of 100 was more so for illustrative purposes since it changes so much because if you were able to have a larger one you would increase both the base and collector currents?
– anon
Commented Feb 11 at 14:40
• @jbiscan I repeat, you don't choose collector current, the load does. You aim for a minimum base current based on the expected load current and $\beta$. The value of $\beta$ has no bearing on how much current the load demands. Using a transistor with higher $\beta$ only means you may reduce the amount of base current required to satisfy the load's demands. Higher $\beta$ cannot possibly increase current through a load which already has the full supply voltage across it. Once the transistor "switch" is "on", it can't get any more "on". Commented Feb 11 at 15:06
• @jbiscan In my circuit D, the load is R2=1kΩ. The maximum current you could ever get through that load, given a supply of 12V, is 12mA, because 12V is the largest voltage you could ever find across it. It doesn't matter what the transistor $\beta$ is, that's the upper limit, 12mA. Commented Feb 11 at 15:17