So what is happening here is that the current through R2 splits between the transistor and the LED. The more current that is allowed to go through the transistor, the less current that will go through the LED (and vice-versa). As R3 increases in resistance, that results in less base current, which results in less collector current, which means that more current goes through the LED and it gets brighter.

After thinking about this further, I'm not sure if I do understand it completely. Suppose that the battery is capable of supplying 120 mA and the LED uses 20 mA when fully lit. Whether the transistor uses 40 or 80 mA of current, the battery would still be able to provide enough current to fully light the LED. So how exactly does the transistor limit the current supplied to the LED in this circuit? Does it have to do with the transistor and LED being wired in parallel? I guess my knowledge of circuit theory is lacking here.

 

After thinking about this further, I'm not sure if I do understand it completely. Suppose that the battery is capable of supplying 120 mA and the LED uses 20 mA when fully lit. Whether the transistor uses 40 or 80 mA of current, the battery would still be able to provide enough current to fully light the LED. So how exactly does the transistor limit the current supplied to the LED in this circuit? Does it have to do with the transistor and LED being wired in parallel? I guess my knowledge of circuit theory is lacking here.

Of course, it has everything to do with the transistor and LED being connected in parallel. If the transistor is fully on; all the current will flow through R2.

It is also possible to drive the transistor just enough so that some current passes through the transistor and the rest through the LED.

Do not concern yourself with how much current the battery is capable of supplying, it is what is being drawn that is important.

 

Recently I learned about NPN transistors in my physics class, and am having difficulty understanding them. I read about them in the AAC textbook, but some things still don't make sense to me very well.

If a transistor is used as a switch, is it true that there is an inverse relationship between the power going into the base leg and the power coming out of the emitter leg? An LDR has resistance that is inversely proportional to light, so it has the opposite of the behavior desired for a light that should only go on at night (like a streetlight). Would wiring it to the base leg of a transistor cause it to produce the desired behavior (so that the transistor would emit the most power when it is dark and the LDR is producing the most resistance)?

There were some fairly thorough basic tutorials in some 60s issues of Electronics Illustrated, AFAICR: they're archived on americanradiohistory.com

Even if the tutorials aren't what you're looking for, the magazines are a good read anyway.

 

After thinking about this further, I'm not sure if I do understand it completely. Suppose that the battery is capable of supplying 120 mA and the LED uses 20 mA when fully lit. Whether the transistor uses 40 or 80 mA of current, the battery would still be able to provide enough current to fully light the LED. So how exactly does the transistor limit the current supplied to the LED in this circuit? Does it have to do with the transistor and LED being wired in parallel? I guess my knowledge of circuit theory is lacking here.

When the transistor is turned on, what is the voltage at the collector? Notice that the same voltage feeds the LED through R1. Will that voltage turn the LED on?

Bob

 

Of course, it has everything to do with the transistor and LED being connected in parallel. If the transistor is fully on; all the current will flow through R2.

It is also possible to drive the transistor just enough so that some current passes through the transistor and the rest through the LED.

Do not concern yourself with how much current the battery is capable of supplying, it is what is being drawn that is important.

@blocco a spirale: Thanks, this is starting to make more sense to me now.

Does this mean that when the transistor is fully on, it does not draw all of the current that the battery is capable of supplying, but rather uses all of the current in the parallel part of the circuit (forming a 1:0 ratio between the transistor and LED current)?

When the transistor is turned on, what is the voltage at the collector? Notice that the same voltage feeds the LED through R1. Will that voltage turn the LED on?

Bob

@BobTPH: I'll have to build the circuit on a breadboard to measure the voltage. The way this circuit is supposed to work is that toggling the switch turns the LED on or off.

 

@mossman: I did link to the AAC textbook in post #15 (as well as in #31).

I see. My apologies. I didn't catch on to the AAC reference, and don't think of a website as being a textbook.

 

After thinking about this further, I'm not sure if I do understand it completely. Suppose that the battery is capable of supplying 120 mA and the LED uses 20 mA when fully lit. Whether the transistor uses 40 or 80 mA of current, the battery would still be able to provide enough current to fully light the LED. So how exactly does the transistor limit the current supplied to the LED in this circuit? Does it have to do with the transistor and LED being wired in parallel? I guess my knowledge of circuit theory is lacking here.

What the battery can supply isn't too important unless the currents are getting to a level that the internal resistance of the battery is a factor.

We can't talk specifics because the schematic you provided didn't include values for the all of the resistors. We could make some up, but let's work with symbolic values as far as possible.

View attachment 82735

Let's assume that the LED has a forward voltage drop of V_F at a current of I_F. So if that is the current that we want in the LED when the transistor is off (meaning that the base resistor, R3, is at a very high value), we need the LED to see a total resistance under those conditions of

\(
R_{tot} \; = \; R_{1} \; + \; R_{2} \; = \; \frac{V_{CC} \; - \; V_{F}}{I_{F}}
\)

If Vcc = 9V, Vf = 2V, and If = 20mA, then Rtot = 350Ω and we need to reduce R1 considerably otherwise we will end up with a LOT of current in the transistor when it is fully ON. We could split it up however we want, so we'll need to examine what the impact of how we split it up is. Keeping things symbolic will help us do that.

Now plug the transistor back in and let's assume that it has a beta of β. The base current through the transistor will be

\(
I_B \; = \; \frac{V_{CC} \; - \; V_{BE}}{R_{3}}
\)

and, when transistor is active, the collector current will be

\(
I_C \; = \; \beta I_B \; = \; \beta \( \frac{V_{CC} \; - \; V_{BE}}{R_{3}} \)
\)

For design purposes, we can consider that the LED turns off when the voltage at the junction of R1 and R2 is equal to Vf. This is an approximation, but it is probably "good enough". At this point, there is no current in R1, so all of the current is flowing through R2 and into the transistor. Thus we have:

\(
I_{C} \; = \;I_{R2}
\,
\beta \( \frac{V_{CC} \; - \; V_{BE}}{R_{3}} \) \; = \; \frac{V_{CC} \; - \; V_{F}}{R_2}
\)

If we solve this for R3, we get the value of the base resistor that will result in the LED just turning off.

We can also develop the relationship between the base resistance and the LED current between these two points. I'll leave that for you to chew on.

 

@blocco a spirale: Thanks, this is starting to make more sense to me now.

Does this mean that when the transistor is fully on, it does not draw all of the current that the battery is capable of supplying, but rather uses all of the current in the parallel part of the circuit (forming a 1:0 ratio between the transistor and LED current)?

@BobTPH: I'll have to build the circuit on a breadboard to measure the voltage. The way this circuit is supposed to work is that toggling the switch turns the LED on or off.

Forget what the battery can supply (I think I already said that). When the transistor is fully on all the current goes through R2 and the transistor and most of the supply voltage (there will still be a fraction of a volt across the transistor) is across R2, that current ≈ supply voltage/R2

I think you are finding this confusing because you don't understand basic electronics. This will make a lot more sense if you forget about transistors for a while and get to grips with basic DC circuit theory i.e Ohm's law, series and parallel circuits etc...

 

Does this mean that when the transistor is fully on, it does not draw all of the current that the battery is capable of supplying

The presence of R2 will limit the amount of current that can flow through the transistor. For rough purposes, we can assume that when saturated the transistor looks like a closed switch and so the battery voltage will appear across R2. If this is small compared to the current the battery is capable of, then the capability of the battery is immaterial. If R2 is not at least about 5x to 10x the internal resistance of the battery, then the internal resistance needs to be taken into account.

, but rather uses all of the current in the parallel part of the circuit (forming a 1:0 ratio between the transistor and LED current)?

It's not a direct ratio because the total current (the current in R2) will not be a constant. The transistor will have to carry not only what it is stealing from the LED, but also some extra current in order to support the additional current in R2 as the collector voltage drops.

I'll have to build the circuit on a breadboard to measure the voltage. The way this circuit is supposed to work is that toggling the switch turns the LED on or off.

Earlier you were talking about the effects of increasing R3 on the LED current. Now it sounds like you have a fixed R3 and are switching it one and off. That is a much simpler problem.

Now you just want to operate the transistor as a switch. When the physical switch is open there will be no base current and the transistor will be OFF. Remove R1 and make the value of R2 such that you get the desired LED current. We've already figured that to be about 330Ω. When the switch is closed you will get some base current and you want that to be enough to fully saturate the transistor, which will result in the collector voltage being Vcesat, which is typically about 0.25V. For design purposes here we can call this zero which means that we will have the full battery voltage across R2 resulting in a current of about 30mA. To ensure saturation, we generally try to make the base current one-tenth of the collector current, so that is 3mA. The voltage across the base resistor is about 8.3V due to Vbe, but again for design purposes we can ignore that here and call it the full 9V. This means that we want the base resistor to be no more than ten times the collector resistor, or about 3.3kΩ.

If we can't get R3 to be that small, then we need to look more closely at the circuit since that value of R3 is designed to ensure the transistor fully saturates. But the LED will go off as soon as we are stealing all of it's current and bring the collector voltage down just a couple tenths of a volt. If we are willing to rely on the beta of the transistor being above a certain value, then we can put an upper limit on the value of R3 at roughly the beta times the collector resistor. If we assume a beta of 200, then that would let us get away with about 68kΩ. A more detailed analysis would make that as much as about 82kΩ.

 

When the transistor is turned on, what is the voltage at the collector? Notice that the same voltage feeds the LED through R1. Will that voltage turn the LED on?

Bob

I assembled the circuit on a breadboard and took voltage and current measurements with a multimeter. I used 2.2kΩ and 44kΩ resistors for R2 and R3, respectively (approximately the values that were used in my physics book). These were the values that I got:
With the switch open:

• LED (lit): 1.824V, 3.62mA
• Emitter-base junction of transistor: None
• Collector-base junction of transistor: 1.87V, 0.008mA

With the switch closed:

• LED (unlit): 0.019V, 0.045mA
• Emitter-base junction of transistor: -0.670V, -0.204mA
• Collector-base junction of transistor: -0.651V, -0.18mA

 

Trans-istor. Trans-Resistor. Geddit? Current controlled voltage source. A typical BJT (Bipolar Junction Transistor) -- has about 4MOHM of DC resistance between the emitter and the collector. By applying (n) current at the base junction of the BJT, this in turn changes the resistance of the emitter to collector junctions. So it could be perceived as an current controlled variable resistor. So we take (n) small amount of current at the base to control a much larger amount of current at the collector OR the emitter. If an voltage is applied to the collector, the emitter voltage tends to follow the base voltage less 0.6V drop. This is called an emitter follower...

@MCU88: Your post helped me understand things better, I think. Somehow I had missed it before in the multitude of replies.

Thinking of a transistor as a "current controlled variable resistor," I think I can understand how one works when used as a common emitter amplifier. When no current at all is applied to the base junction, its resistance is small enough that the LED emits light. But when a current is applied, it resists enough that the LED cannot light up.

 

The presence of R2 will limit the amount of current that can flow through the transistor. For rough purposes, we can assume that when saturated the transistor looks like a closed switch and so the battery voltage will appear across R2. If this is small compared to the current the battery is capable of, then the capability of the battery is immaterial. If R2 is not at least about 5x to 10x the internal resistance of the battery, then the internal resistance needs to be taken into account.
...

@WBahn: I appreciate your trying to help, but I think I'm probably wasting everyone's time by prolonging this discussion, since:

I think you are finding this confusing because you don't understand basic electronics. This will make a lot more sense if you forget about transistors for a while and get to grips with basic DC circuit theory i.e Ohm's law, series and parallel circuits etc...

so I think it would be best for me to study the basics of circuit theory.

 

@MCU88: Your post helped me understand things better, I think. Somehow I had missed it before in the multitude of replies.

Thinking of a transistor as a "current controlled variable resistor," I think I can understand how one works when used as a common emitter amplifier. When no current at all is applied to the base junction, its resistance is small enough that the LED emits light. But when a current is applied, it resists enough that the LED cannot light up.

No, no, no,...... when no base current is applied resistance is high and when base current is applied its resistance is low.

 

No, no, no,...... when no base current is applied resistance is high and when base current is applied its resistance is low.

Ah, yes, because of the inverse relationship between the input signal and output voltage. I don't know what I was thinking exactly.

 

@WBahn: I appreciate your trying to help, but I think I'm probably wasting everyone's time by prolonging this discussion, since:

so I think it would be best for me to study the basics of circuit theory.

Yes, you're tying this whole thing in knots simply because you don't understand the basics.

 

Thinking of a transistor as a "current controlled variable resistor," I think I can understand how one works when used as a common emitter amplifier. When no current at all is applied to the base junction, its resistance is small enough that the LED emits light. But when a current is applied, it resists enough that the LED cannot light up.

Thinking of it as a current-controlled variable resistor is useful in a coarse fashion, but it does NOT behave like that once you look at it closely. It acts much more like a current-controlled current source, which is quite different in many situations.

When there is no base current, the collector-emitter "resistance" is very high which means that very little current flows through the transistor. But in the circuit you are looking at, that means that all of the current must go through the LED. Similarly, when current is applied the resistance drops and all the current goes through the transistor which means that very little current can flow through the LED.

 

Ah, yes, because of the inverse relationship between the input signal and output voltage. I don't know what I was thinking exactly.

Once again, no. There is no inverse relationship; base current causes a proportional amount of collector current to flow. There is nothing inverse going on.

 

Once again, no. There is no inverse relationship; base current causes a proportional amount of collector current to flow. There is nothing inverse going on.

That was disputed earlier in this thread, but for common emitter amplifiers, the AAC textbook says (as I mentioned in post #31):

For light exposure levels somewhere between zero and maximum solar cell output, the transistor will be in its active mode, and the output voltage will be somewhere between zero and full battery voltage. An important quality to note here about the common-emitter configuration is that the output voltage is inverted with respect to the input signal. That is, the output voltage decreases as the input signal increases. For this reason, the common-emitter amplifier configuration is referred to as an inverting amplifier.

 

Ok, yes a common emitter amplifier is an inverting amplifier if you look at what's happening with the voltages. But your LED circuit makes more sense if you think in terms of current because, when the transistor switches on it effectively steals the LEDs current and the LED extinguishes because it is a current operated device.

 

Ah, yes, because of the inverse relationship between the input signal and output voltage. I don't know what I was thinking exactly.

That depends on what you define as the "input signal" and the "output voltage". What if the output voltage is the voltage across the collector resistor? Then as the input current (if that is your input signal) increases the output voltage would increase.

When statements like, "the common emitter amplifier is an inverting amplifier," are made they are in the context of a specific circuit topology and specific definitions of input and output signals. You can define different input/output signals and get very different relationships even with the exact same circuit.