# Using a Transistor to "Fully" Illuminate a Lamp

Consider the following:

In the diagram, $R_C$ is resistance offered by the lamp, which is $240\Omega$. The emitter is at $0V$, though this is not marked in the diagram.

Question. I do not understand the line which reads

Thus, $100mA$ of collector current must flow through the transistor to fully illuminate the lamp.

What is meant by "fully illuminate"? Does it mean that the maximum current that can possibly flow through the lamp in the circuit is $100mA$?

If my interpretation is correct, then I am not able to undersand the reson behind it. If we choose $R_B$ very small, then the base current is high. Thus so is the collector current.

Can somebody plese clarify.

PS: I am a math major who is trying to learn some electronics during the summer vacations. I hope I haven't asked a stupid question.

• Not a stupid question. An obvious reason to put this in a text book would be to ask the student to dimension Rb correctly. Commented Jun 2, 2016 at 10:48
• Note that you probably won't find a circuit like this in real life, because it relies on the transistor's current amplification to be constant and predictable, while in real life current amplification varies greatly with temperature and other factors. Thus, BJTs shouldn't be driven close to their current amplification limit. Commented Jun 2, 2016 at 16:26
• In addition to mic_e's comment,the collector-emitter junction will have its own voltage drop too,around 0,7mV in a real life example. Commented Jun 2, 2016 at 17:50

What is meant by "fully illuminate"? Does it mean that the maximum current that can possibly flow through the lamp in the circuit is 100mA?

It's slightly sloppy wording. The reasoning employed by the question is: "If the transistor was replaced by a short circuit between the collector and emitter, how much current would flow?" In that situation, the lamp would have 100mA flowing through it. Therefore we declare that this is the maximum current that would be seen if there was a transistor there.

This figure can be used to (a) determine the base current required and (b) pick a transistor of appropriate size.

If my interpretation is correct, then I am not able to undersand the reson behind it. If we choose RB very small, then the base current is high. Thus so is the collector current.

No matter how much current you put through the base, the collector current is still limited by Ohm's law applying to the lamp.

• Suppose the base current is I_b. Then automatically the collector current is \beta*I_b. If I_b is very high, then by Ohm's law, the collector voltage will be negative. Is this not possible? I do not see a contradiction. (Sorry about the math typesetting. I am unable to write math symbols on this site. Can you also tell me how to do that?) Commented Jun 2, 2016 at 12:03
• I think it's pairs of backslash-dollar \ \$ signs: $\beta*I_b$. The collector voltage cannot go negative because the transistor would no longer be operating in the linear region, which is the only region of its operation where that holds true. The linear region is part of the "forward-active" region which requires "the base–emitter junction is forward biased and the base–collector junction is reverse biased." (per Wikipedia) Commented Jun 2, 2016 at 12:12
• So in an NPN transistor, the collector voltage should always be greater than the base voltage? Is that correct? Commented Jun 2, 2016 at 12:17
• Not quite. See en.wikipedia.org/wiki/… which describes what happens for various relative voltage levels. This ought to have been in the textbook somewhere - if it doesn't mention saturation it's a very poor resource! Commented Jun 2, 2016 at 12:19
• The book I am reading is "All New Electronics: Self Teaching Guide" by Harry Kybett and Earl Boyser. I have read this book without leaving anything out but not resembling what you just mentioned can be found anywhere till now. Can you suggest me some reading material. My main aim is to be able to make things and not just know the theory. Commented Jun 2, 2016 at 12:23

The lamp has a resistance of 240 ohms and is presumably rated for a 24V supply thus the rated current is 100 mA and therefore, to fully turn on the lamp 100 mA must flow through the transistor. This is a power delivered to the lamp of 2.4 watts.

However this will never quite be the case. The transistor, no matter how hard it tries will saturate at about ~0.1V between collector and emitter thus, only 23.9V is applied to the lamp thus, technically, it is not quite fully illuminated.

The power applied is therefore $\dfrac{23.9^2}{240}$ which is 2.38 watts.

• Technically, then, (neglecting the second and third order nasties) the lamp can never be fully illuminated unless it's driven by a 100 milliampere current source, yes? Commented Jun 3, 2016 at 15:37
• @EMFields the lamp needs 100 mA for full load according to the question. Commented Jun 3, 2016 at 15:45
• Won't a 100 mA current source fill that bill? Commented Jun 3, 2016 at 16:11
• @EMFields yes it would but i don't see that as an option here unless the supply is greater than 24V. Commented Jun 3, 2016 at 16:46
• Touché. Nicely put. :) Commented Jun 4, 2016 at 0:20

In the following drawing an analogy is made between a relay being used to turn a lamp ON and OFF, and a transistor being used to do the same thing.

On the left hand side, the relay's coil is likened to the transistor's base, its common contact to the transistor's collector, and the normally open contact to the emitter.

K1 is one of those jellybean 12 volt 400 milliwatt relays and when S1 is made, current through the coil from the 24 volt source will build up a magnetic field around the coil which will attract the armature, forcing the common and normally open contacts together, which will force current from the 24 volt source through DS1 to ground, which will illuminate DS1.

Since K1 is a 12 volt 400 milliwatt relay, its coil will have a resistance of

$$R = \frac{E^2}{P} = 360\text{ ohms.}$$

So, in order to drop the 24 volts from the source to the 12 volts the relay coil wants to see, we can put a resistor with a resistance equal to that of the coil between the source and the coil.

On the transistor side, with S2 open there'll be no current into the base and out of the emitter to ground, so the resistance between the collector and the emitter will be very high, just like in the relay case with S1 open.

When S2 closes, however, current will be forced through R2 into Q1's base and out of its emitter, which will cause the collector-to-emitter resistance to fall to a very low value (like the closed contacts of a relay) and then current will be forced from the 24 volt source through DS2, and then through Q1's collector-to-emitter junction to ground.

For switching applications, like this one, the base current is usually chosen to be about 10% of the collector current in order to saturate the collector-to-emitter junction and assure a low collector-to-emitter resistance.

In this case we have (assuming a voltage drop of 300 millivolts or so across the collector-to-emitter junction) 23.7 volts dropped across the lamp's 240 ohm resistance, so that'll be a collector current of:

$$Ic =\frac{ V1-V_{CE(SAT)}}{R_{DS2}} \approx 100\text{ milliamperes}$$

10% of that is 10 milliamperes, and since the base-to-emitter junction looks like a diode, R2 is then:

$$R2 =\frac {V1-V_{BE(SAT)}}{Ib} = 2330\text{ ohms}$$

2400 ohms is, rounding up, the next E24 resistance value, so we use that.

• What, no 555 timer? Commented Jun 3, 2016 at 14:50
• @OlinLathrop: Maybe I'm missing something, but why would you advocate using a 555 in this application? Commented Jun 3, 2016 at 23:42

It is the simplest circuit possible to demonstrate how a bipolar transistor really works. I explain immediately what I mean after my own ranting about a subject that nobody noticed yet : the resistance of the light bulb is impossible to measure.

The light bulb is 2.4 watts at 24 volt, so the author calculated it's resistance as 240 ohm. If you would measure the resistance, you would find ~16 ohm because a tungsten filament at room temperature is usually 15 times lower than at operating condition.

The bipolar transistor is a current amplifier. The gain can be assumed to be 100 for demonstration. In practice, it vary from 10 (high power transistors) up to 200 for some smaller. I would not mention darlington configuration, since it is king of cheating, using two transistors to increase the gain.

The question is: which value should we set Rb to make sure the lamp is as bright as possible. The simplest answer is 0 ohms, period, I deserve 100% on the exam for that question.

The question should specify: select Rb to be as high as possible to minimize wasting energy while low enough to make sure the transistor is conducting the best it can : in other word, operate at saturation.

Assuming a gain of 100, the current between base to emitter should be 2.4 mA. If we ignore the internal loss of 0.7 volt, because the base/emitter junction is a diode, then the resistance Rb would be: Rb = 24 volt / 2.4 mA = 10 kilo ohm

In real life, we assume the worst case for the gain for a given family of transistor, we take into account the 0.7 volt bias on the base. We also want to cover the actual voltage change, ranging from 28 volts, down to 18 volt or less, depending on the shut off protection expected in a typical 24 volt vehicle.

This circuit is a nice demonstration of the DC gain of a transistor and can be used to illustrate the maximum energy that can be dissipated in any configuration or test the efficiency of a heat sink.

The short answer: if we adjust Rb (replacing Rb with a potentiometer) until the voltage across the light bulb and the voltage across the transistor is the same (12 volt), then both the light bulb and the transistor will dissipate the same amount of heat.

Measuring the current flowing thru that Rb versus the current flowing thry the collector would give the actual DC gain of the transistor.

That mid point, where the load and the transistor are set to be equal resistance is the absolute worst case. It is the sweet spot (call it the ), the point where the transistor is stressed to the maximum for heat dissipation. As soon as the transistor is asked to conduct more or less than this middle point, the energy dissipated by the transistor decrease.

If the transistor base current is adjusted to drive more current than that mid point, then the light bulb generate more heat. For example, If the transistor base/emitter current is 0.9 mA and the gain is 100, thus allowing 90 mA thru the collector/emitter junction (and lets say the light bulb is still 240 ohm despite not fully on, a resisting less like tungsten do), then the total resistance is: R total = 24 volt / 90 mA = 266 ohm. The bulb is assumed to be 240 ohm, so the transistor act as a resistance in series of: 266 ohm - 240 ohm = 26 ohm

The transistor represent roughly 10% of the total resistance. The light bulb dissipate about 10 times more heat when the brightness is close to maximum.

Going to saturation, the point where the transistor conduct the best, this point where the voltage between Collector and Emitter is minimum, the heat generated by the transistor is minimal.

On the opposite range, when the transistor conduct no current at all, beside leakage so low that it is a waste of time to even mention it, there is no power dissipated in the transistor. This is why engineer always try to operate transistors as binary switch. They use the linear region only when they don't know better (when the component count for low cost win over the cost for saving energy).

Back to the worst case, the point where the voltage on the collector is half the power supply value, if the transistor is set to drive less current starting from that mid point, then the heat generated by the transistor also decrease. In that case, the transistor still generate more heat then the light bulb, but the total energy keep decreasing as less current is allowed between the collector and emitter.

Brief, for those who feel like these kids who spin the tires of their friend cars, now you know how to stress their transistors. They hate conducting electrons in that linear half ass mode, not fully on and not fully off is killing them... literally.

Ignoring the transistor for a moment .... A handy insight, which I apply when I'm reasoning about circuits like this, is that current is mostly what makes things happen, and your job is to arrange the voltages so you get the currents you want. A bulb glows brighter, or an electromagnet is stronger because you're sticking more current through it.

The truth of the bulb is that the manufacturer has made this bulb that glows at "full" (ie useful and sustainable) brightness with 100mA current, and they've designed it to present a resistance of 240 Ohms when powered on which conveniently means you can plug it into your cheap 24V voltage supply. For convenience, they print 24V on it, which means "put 24V across it, and you'll get 100mA for full sustainable brightness". The brightness itself is related to the power (rate of energy dissipated), which is the product of Volts and Amps.

You could put a higher voltage across it, and you'd get more current and more brightness. But then you're also dissipating more heat, and thus probably pushing the bulb's materials beyond their limits: it's a bit like overclocking a processor, you might get more performance, but push too hard and the thing will degrade or fail. Similarly, you could put 12V across it, and you'll get 50mA - much less bright, cooler, probably last a very long time. Note that halving volts give half the amps, which means a quarter of the power.

Moving on to the transistor. As you note, the transistor C-E current is approximately a linear function of $I_{be}$. So as you decrease Rb, $I_{be}$ gets bigger, and Ice gets much bigger. If the transistor's beta is 100, then allowing $I_{be}$ =0.1mA would set $I_{ce}$ to be 10mA and that's how much current would flow through the bulb and C-E. In that respect, C-E is like a current-powered tap/faucet; if you do the sums with those values, it's also like a 2.16kOhm resistor. As the $I_{be}$ increases, the transistor's apparent C-E resistance decreases, more current flows.

Loosely speaking, when the tap is fully open, the C-E is now allowing all the current there is to pass through, the C-E's resistance has become negligible, and the dominating resistance in the circuit is the bulb. The transistor is now "saturated" - opening the tap any more doesn't let more current through.

Some notes:

• beta is a lousy parameter, very hard to control. If you assume beta=100, make sure the circuit will also work for beta=50 and beta=500. Two devices from the same packet can differ by a factor of two, and it changes with temperature and everything. It's a quick-and-dirty param for quick estimation, and much transistor design is all about neutralizing the ill-effects of such real-world characteristics. A good design doesn't depend much on beta.
• when it really matters, you can buy integrated transistors which have matched characteristics.
• there are more accurate models than beta (see Ebers-Moll model), which treat the Ice as an exponential function of the Vbe. It approximates a linear function for constant temperature over the range we've been talking about.
• thinking about a transistor as a programmable resistor is a good way to get over comprehension bumps (like "why won't the voltage go negative"). But no electronics engineer thinks that way - with practice, you start getting used to thinking about controlling currents and voltages. A resistor is a way to turn a voltage into a current, etc.
• not all devices' outputs depend on power, although "things that use heat" (bulbs, heaters) do. Magnets (and therefore magnet-powered motors) depend only on current. Some devices (like FETs) are essentially controlled by voltage, because they depend on an electric field strength. So maybe a more accurate rule is to say "I arrange voltages so I get the voltages and currents I want".
• get hold of Horowitz and Hill, The Art of Electronics and read the first two chapters. Make sure you understand every single example circuit. And then read the rest of the book...