# Simple curve to illustrate how much current it takes to turn on a PN2222 transistor?

It's been about 30 years since I studied transistors so please forgive my basic question.

I bought my kid an Arduino kit to give him a basic taste for how electronic circuits work. Included in the kit was a couple of PN2222 transistor. Not having a large potentiometer, I set up the following circuit, using all the large value resistors that came with the kit:

simulate this circuit – Schematic created using CircuitLab

With a jumper, I can crudely vary the current flowing into the emitter. I was very surprised to see that 10 100K resistors were not enough to turn the transistor off. I was even more surprised that and additional 10 1M resistors were not enough to shut it down, though the LED was very dim. I was also surprised to learn that the transistor consistently dropped around 3V no matter how much current was coming into the emitter. I have forgotten a lot, it seems.

I looked at the technical specs for this device that shows some plots but I honestly couldn't make heads or tails of them. How do I figure out how much resistance it will take to turn the transistor off?

And does anyone know of a simple curve that can help illustrate how the transistor is behaving in relation to the changing resistance on the emitter?

Thanks.

Photo:

• Beginning to feel like an XY problem. – Andy aka Jun 24 '20 at 16:46
• A photo of your physical setup might help us see what's going on. – The Photon Jun 24 '20 at 16:52
• You're dealing with very small currents. When taking voltage measurements, your voltmeter input resistance can be a big factor. Am curious with what instrument you measure voltage? – glen_geek Jun 24 '20 at 17:54
• The diagram you posted has no resistor to limit the current through the LED and transistor. Follow the path from + to the LED to the transistor collector to the emitter to the battery -. No resistor. If you have one in your real circuit, that's ok. It's just not in your diagram. – JRE Jun 24 '20 at 20:45
• I was very surprised to see that 10 100K resistors were not enough to turn the transistor off I'm not, because the base of the transistor is connect only to the chain of 100K resistors. Add a bias resistor connecting the base to the ground. In this setup, I would try connecting a 1M resistor from the base to the ground. This would raise the minimum voltage to light the led, but it would make the circuit more stable. – mguima Jun 26 '20 at 18:34

The transistor drops 3V because the other 2V are dropped on the LED

You ask for a simple curve to illustrate how the circuit is behaving. I can give you two. (I drew them by hand after running a simulation with a 2n2222A and a 2.1 LED to get coherent values).

The first curve shows how the BE diode responds to the current set by the Vcc - Rb circuit. Plot the diode characteristic alongside with the load line (basically the characteristic of the battery plus resistor, duly translated and flipped to consider that the current coming out of one port is the current getting inside the other port) and you will find the quiescent point at the intersection of the two curves.
Since the load line intercepts the current axis at 5V / 10 meg = 500 nA, you should expect less than that current through the BE diode. And in fact the the BE diode will work at less than 0.6V.

the current in the BE diode, with RB = 10 meg, is about 440 nA. You'll notice that by increasing RB the load line moves toward the horizontal axis pivoting aroud the VCC point, so the intersection with the BE diode characteristic will move toward lower voltages and lower currents.

Now we use the same trick on the output part of the circuit: there you have the CE terminals of the transistor 'interfacing' with the Vcc-LED circuit. Since IB is known to be 440nA, we know which one of the many curves of the transistor use (IC = beta IB - the simulation picked that for Ic = 60 uA).
You use the same load line trick, translating and flipping one of the two one-port characteristics (in this case I flipped the Vcc-LED part) and you find this

As you can see, you are near the knee part of the LED, but you still have enough current - about 60 uA - to make it look lit. And even if you raise RB, you will still be in a region where the current in the diode is not zero. Not even for IB = 0 you can take the current to exactly zero, because there is a minimum leakage current.

Using load lines helps in visualizing what's going on in the circuit: you raise RB, the input load line goes down, IB goes down and so IC but the intersection is still in a region with ILED > 0. From the simulation

it is clear that your LED is underpowered - only 1.6V instead of the nominal 2.1V. But with modern LEDs that's more than enough to get light out of them.

Caveat emptor - do not rely too much on load lines for designing a circuit because the characteristics can vary enormously in a single production batch and are temperature dependent.

If you look at the datasheet for your transistor you will certainly see graphs of collector current vs. base current, and base voltage vs. base current. You should note that the transistor is not like a toggle switch...there is no sharp threshold where it turns "on" and "off".

In your circuit, the voltage across the LED will be about 2V when it is forward biased so you should expect about 3V from emitter to collector on the transistor. What this means is that about 60% of the power consumed by your circuit is consumed by the transistor, which might cause it to become hotter than you desire.

2N2222 may have a gain of over 100. Take whatever current you have going into the base and multiply this by 100 so see approximately how much current will be pulled in by the collector.

$$\frac{5\ V}{10\ MΩ}=0.5\ µA$$ $$0.5\ µA\ \times\ 100=50\ µA$$

50 µA is enough to see in a modern LED. You could be getting even more than this.

• Yes, this roughly matches up with the readings I was getting last night. Note that the circuit has 11 M ohms, not 10. So the calculation is as simple as just doing 5V / 11 M Ohms? How do I factor in the base-emitter resistance into the calculation? – StevieD Jun 24 '20 at 19:43
• Base to emitter is a voltage drop, rather than a resistance, you subtract that from your 5 V (about 0.7 V). Otherwise, yes, just divide that voltage by the total series resistance (11 MΩ in your case). $$\frac{(5\ V-0.7\ V)}{11\ MΩ}$$ – evildemonic Jun 25 '20 at 14:43

I know the following may look "a little complicated," but it's not that bad. Let me present it and explain it a bit, first. But it is a very rough model of the NPN BJT when it is operating in active mode:

simulate this circuit – Schematic created using CircuitLab

In active mode, the voltage between the emitter and base will be roughly somewhere in the range, for small signal BJTs anyway, of $$\500\:\text{mV}\$$ and $$\900\:\text{mV}\$$. The smaller voltages in small signal BJTs are usually associated with smaller collector currents. And usually about $$\700\:\text{mV}\$$ in small signal BJTs is associated with collector currents in the roughly "few mA" area. A change in collector current of a factor of 10 will imply a change in the base to emitter voltage of about $$\60\:\text{mV}\$$ (discounting that little $$\r_b\$$ showing in the diagram, anyway.)

I've included a small base resistor, as this is pretty typical in small signal BJTs. Usually, you don't even need to worry about it as the base resistor you place externally is almost always a lot bigger.

I've also included a tiny little battery near the emitter. If it helps you, just bypass it in your mind with a wire. It's not critical. I just added it because it represents the thermal voltage which varies with operating temperature. At room temperature, it is about $$\26\:\text{mV}\$$. All of the base and collector currents (which are the emitter current) pass through this little battery. And sometimes you may read about a so-called "dynamic resistance" of a BJT. If so, that's this little battery voltage divided by the emitter current and it's called $$\r_e\$$. So, in some cases, folks will place $$\r_e\$$ instead of the little battery I show. But they are one and the same thing. Just FYI. But don't let it confuse you, either. It's effect isn't important in your case.

The key take-away from the above is that the collector "looks like" a current sink in the NPN BJT, when it is in active mode. Active mode is anytime the voltage at the collector is equal to, or above, the voltage at the base. In your case, because the LED only has about $$\2\:\text{V}\$$ across it, the collector voltage will always be about $$\3\:\text{V}\$$. That pretty much ensures "active mode" operation. So the above model almost certainly applies.

Now, one more small detail to add. The LED itself also will have small variations in its voltage drop with respect to current in the LED. But like the base-to-emitter case I mentioned above, a factor of 10 change in its current will only lead to perhaps $$\100\:\text{mV}\$$ to $$\150\:\text{mV}\$$ change across it. So it doesn't change its voltage, much. Just a little -- even when huge changes in current take place. So the collector voltage will move around a little bit, but not a lot, as you play around with the base resistors in your experiment.

Most modern small signal BJTs will have $$\\beta\ge 100\$$. When I run through batches of them and from different families too, I usually find $$\120\le \beta \le 280\$$. We don't know what you have, of course. But it's probably somewhere in that range.

All that this means is that when you supply different base currents via your resistor experiment changes, whatever that value is will be multiplied by that $$\\beta\$$ value and become a collector current -- which also will happen to be the LED current, too. (For reasons that should be obvious.) In short, the LED is pretty much always "on" to some degree, with the degree of "on" being related to that calculation.

If you want to turn it completely off, you'll need to nail the base down close to zero volts (say, below $$\500\:\text{mV}\$$.) Any more positive voltage will supply sufficiently useful base current into the base to make for a likely-visible collector current because of the $$\\beta\$$ value multiplying it.

Continue your experiments, for example, but first by adding a $$\10\:\text{k}\Omega\$$ resistor between the base and the emitter in your circuit. This will create a divider and will allow you to present smaller voltages to the base -- enough smaller, in fact, that you'll be able to turn it off, completely, during your experimentation.

Not really an answer but a personal comment:
In material science, base current is due to the leakages and the re-combinations. It is an illusion/approximation that the base current controls anything. It is always about the subtle change of Vbe, which in turn is intertwined with Ic, very similar to the voltage and the current thru a diode. In a way, BJT is a BaseEmitter diode with collector being the avalanche region that will take everything the BaseEmitter passes on.

A very good insight, just 1 min long watch:
Key understanding of a PN junction: https://youtu.be/O_ntPMM0EQA?t=2165
BJT is in reality voltage controlled: https://youtu.be/KcTlDFAGkJI?t=1937
Follow the rest of the lectures to understand why Ib is NOT controlling anything.