I've been trying to figure this out forever: How does a BJT amplify?

[asfastasdark]

Say we have this simple little thingy here... This is pretty much your basic BJT boost circuit.


How does that work? I never understood this, for example: How can the 6VDC go towards the collector with the signal going away from the collector to the output? How can two currents go alongside each other in opposite directions? And what does the emitter resistor do? I've asked it a million times, but I still don't know... 

[AndrewCE]

You should first know that this amplifier is inverting. When the input current is minimal, the Collecter-Emitter current is minimal, so the output current is at maximum. And vice versa.

[geertjacobs]

Going by your name, I assume you understand dutch.
http://users.khbo.be/peuteman/fotoelektronica.htm
-> chapter 8: transistors
-> chapter 9: transistor circuits

Lots of reading though 

[R.G.]

How does that work?

The transistor is a kind of variable trap door/valve for current. The collector-base inside the transistor is a reverse biased diode and normally lets no current through at all. The base-emitter interferes with the ability of the collector-base to prevent current from flowing. This is proportional to how much current is injected into the base. So the more current that goes into the base, the more current the collector lets through. The ratio of how much collector current gets through for a unit of current into the base is the current gain of the transistor.

Since there is a total of 6V being supplied, the 6V must be divided among the collector resistor, the transistor, and the emitter resistor. All three of those must all up to 6V.

Whenever a current goes through a resistor, it causes a voltage to be generated. This is Ohm's law; the voltage generated by current flowing is V = I*R, the product of the current times the resistance. If there is no current flowing through the transistor (that is, no base current) then there is no voltage across either the collector resistor or the emitter resistor. That means the voltage across them must be zero, so the collector voltage must be 6Vdc. The emitter voltage must be zero, since there is no voltage generated by current to pull it up from zero volts.

Now when you put base current into the transistor, it lets some current leak from collector to emitter. For a high gain transistor, we can ignore how much base current is going in, and simply say that the collector current is equal to the emitter current; this happens to be a very good approximation for most purposes. So some current goes through, and that causes a voltage to be generated across each of the emitter and collector resistors. The battery voltage didn't change, which means that the total of the voltages across all three parts (two resistors and a transistor) must still add up to 6V. So the voltage across the transistor goes down by the exact amount that the voltage on the resistors increase. When the base current increases until the collector current equals the battery voltage divided by the sum of the two resistors, the battery voltage is then dropped totally across the resistors, and the transistor can't make any more current flow - that's all the battery can force through those two resistor values - and the transistor is said to be saturated. It's like a closed switch.

And now we can have some idea of voltage gain. For this one, you're simply going to have to trust me. The base is internally connected to the emitter by a silicon diode. This lets enough base current flow to pull up the emitter voltage to keep the emitter one diode drop, about 0.5 to 0.V, lower than the base. So the emitter follows the base. Always, unless the power supply or some external factor keep it from following the base. This is why the "emitter follower" circuit has a gain of 1 - it HAS to. Anything else either doesn't work at all or damages the transistor.

So the "gain" from base to emitter is one to a very close approximation. Base voltage goes up, enough current flows in the base to bring the emitter up by the same amount. This is true even if there is no collector voltage. However, in a powered circuit like this, there IS voltage and current available at the collector, and the base current lets collector current through and that raises the emitter as well. This just makes the emitter follow the base even better than it would if there were no collector power available. What is happening is that the current from the base causes a many-times-larger collector current to flow; that current also flows in the emitter resistor, and that raises the emitter. This is in a direction to raise the base and decrease the base current, decreasing the collector current flow. It's negative feedback. Raising the base voltage causes more base current to flow, causing more collector current to flow, causing more emitter current to flow (the emitter and collector currents being the same current, after all), raising the emitter, and decreasing the base-emitter voltage, which is what is letting base current in. It's negative feedback. It self balances with the emitter following the base very closely and one diode-drop lower.

So far we have a gain of one from base to emitter. What's happening at the collector?

The emitter voltage wiggles around  following the base, it's voltage always one diode drop lower than the base. If it's voltage is base voltage minus a diode drop, it's current has to be that same voltage divided by the emitter resistor. This same current flows through the collector resistor, so the voltage at the collector drops  by an amount equal to the emitter current times the collector resistor. So a voltage increase at the base causes the same voltage increase at the emitter, and it also causes a voltage DROP at the collector equal to the ratio of the collector and emitter resistors, since the same current goes through each of them. And because the collector drops when the base (and therefore emitter also) rises, the gain from base to collector is negative. It inverts the phase of the signal.

In the circuit you show, the collector resistor is 100 ohms, the emitter resistor is 1500 ohms, so the "gain" from base to collector is - 100/1500 or -1/15. The circuit you happened to pick out is a bad example. It has a gain less than unity from base to "output" with the output from the collector. Worse, it's called an "emitter follower" but does not take the output from the emitter. I don't know of any good reason to use that circuit. Perhaps there is a mistake and the output should really be shown as coming off the emitter.

I never understood this, for example: How can the 6VDC go towards the collector with the signal going away from the collector to the output? How can two currents go alongside each other in opposite directions?

The 6Vdc does not go towards the collector. Current from the 6Vdc source goes toward the collector. The signal going away from the collector is a voltage in most cases, not a current. If you add to this a capacitor and an external load resistor, then the collector can pull current from BOTH the 6Vdc source and from the external load when its current increases. When its current decreases, current from the 6Vdc source "recharges" the capacitor leading to the external load. In no case does current flow two ways at once. The collector is simply pulling down or releasing up a certain amount of current. That current comes from the 6Vdc if that's the only thing in the circuit, or from the 6Vdc plus the external load if one is connected.

And what does the emitter resistor do?

It provides the necessary loading and feedback for the base voltage and current.

I've asked it a million times, but I still don't know... 

Keep asking until it makes sense! 

[SonicVI]

Thanks RG!

[asfastasdark]

Yeah, I realized later that that circuit made no sense at all, with nothing flowing into the base from the battery... but I'm still kind of confused... what do you mean if the emitter "follows the base"? Or what does "raising the emitter" mean?

[doitle]

This sort of makes me want to speed up my plan to do a fully writeup start to finish designing a single transistor boost. I just need to get back down to Peoria and get my notebooks... I was planning on going step by step through all the design equations for building a boost and explaining how you can design for all the aspects, input impedance, output impedance, gain, high and low frequency rolloff... Making a nice PDF out of it and all.

[R.G.]

Yeah, I realized later that that circuit made no sense at all, with nothing flowing into the base from the battery

That wasn't a problem - I simply assumed that there was a suitable power source running the base; a dc voltage level with some signal on it.

Or what does "raising the emitter" mean?

In the description, I'm using the terms following and height (upwards, downwards, etc.) as a metaphor for voltage above 0V, which I took to be the bottom/negative side of the battery. So when something "goes up" or "rises" that means that its voltage increases in the positive direction. If the voltage at the emitter was 0V, then changed to 2V, I would describe that as "the emitter rose by 2v". The description of altitude is in terms of volts.

what do you mean if the emitter "follows the base"?

Literally, as the base voltage rises, the emitter rises at the same time, but at a little lower voltage.

A resistor has a linear relationship of voltage and current. That is, if you double the voltage across it, the current doubles. This holds true from nano-volts on up to voltages and currents that incinerate the resistor. A silicon diode has a NON-linear relationship of voltage to current. As a diode's voltage increases from zero volts, the current rises too, but much more slowly than linearly. Then at some voltage, the current begins to increase every-more-rapidly and the diode can conduct almost any current without going more than a little above that voltage. The turn-on voltage for diodes depends on the material they're made from, but once you know the material, the turn-on voltage is pretty constant. Silicon diodes hardly conduct at all until they get to about 0.45 to 0.5V, then there's a slight rise in current. This rise in current gets rapidly bigger from 0.5 to about 0.7V, at which point, the diode will conduct amperes of current without increasing its forward voltage much. (NB - for you fact-checkers: yes, I am oversimplifying this to help him learn)

The base-emitter of a transistor looks like a diode. If you raise the base from 0V, not much current flows until you get the base about 0.45 to 0.5V higher than the emitter. Then a little current flows, and as you raise it more, rapidly more current flows. If the emitter is tied to ground/0v, this looks much like a simple diode, not much transistor action as far as just the base and emitter are concerned. If you put a resistor between the emitter and ground though, you can start to see the transistor letting current in through the collector in proportion to the current going in through the base.

With an emitter resistor to 0V, the emitter would "follow" the base upwards in voltage at about 0.6V less than the base even if the collector of the transistor was not connected to anything. In this case, all of the current would be coming through the base to the emitter. But if the collector is connected to a power source so it can let current through, it lets current flow to the emitter at a rate that's current-gain times the base current at any instant. That extra current from the collector causes the emitter voltage to rise above 0V more than it would have if only the base was providing emitter current. So the emitter voltage rises, and that reduces the voltage between the base and emitter, reducing the base voltage a little. This has the effect of making a tiny base current control a much larger current flowing from collector to emitter. And the limited nature of voltage across the base-emitter diode means that the emitter will follow the base to higher or lower voltages, always about 0.45 to 0.7V lower than the base.

[asfastasdark]

OK, this is starting to make sense, but I have the feeling I'm still only getting parts of what you're trying to explain, so...

Explain to me, what exactly is the purpose of the power source connected to the collector? How does that work? And what exactly is flowing, voltage or current, and how would I know?

Also, you talk about the base-emitter, the collector-base, but also the collector-emitter junction, which makes me wonder how it all is connected, are the three pins all connected to each other internally or something? I guess a diagram would help...

Either way, this is already starting to make a lot more sense to me, so thanks  !

Edit:

I never understood this, for example: How can the 6VDC go towards the collector with the signal going away from the collector to the output? How can two currents go alongside each other in opposite directions?

The 6Vdc does not go towards the collector. Current from the 6Vdc source goes toward the collector. The signal going away from the collector is a voltage in most cases, not a current. If you add to this a capacitor and an external load resistor, then the collector can pull current from BOTH the 6Vdc source and from the external load when its current increases. When its current decreases, current from the 6Vdc source "recharges" the capacitor leading to the external load. In no case does current flow two ways at once. The collector is simply pulling down or releasing up a certain amount of current. That current comes from the 6Vdc if that's the only thing in the circuit, or from the 6Vdc plus the external load if one is connected.

Wait... if a current is flowing through, say, a wire, wouldn't that current also have a voltage? Doesn't the little "thingy" of electricity going through the wire have both a measurable current and measurable voltage?

[R.G.]

Explain to me, what exactly is the purpose of the power source connected to the collector?

It is the source of all power for this circuit. Without it, nothing happens. It serves the same purpose as an electrical power generating station - without that, your lights don't work the computer doesn't work, etc.

How does that work?

The circuit does not say what kind of power supply it is, so we are free to suppose that it's a battery, that being the simplest power source in terms of operation. A battery (or an electronics power supply which mimics a battery) is a source of voltage (which is electrical pressure) and current (which is electrical flow). These are analogous to water pressure and water flow. Pressure is the force moving water. Pressure can exist with or without flow. But water cannot flow without pressure of some kind moving the water. Voltage (electrical force) can exist without current flowing. Current (which is, literally the number of electrons moving per unit time) cannot exist without some source of voltage/force to move the electrons.

And what exactly is flowing, voltage or current, and how would I know?

Charge carriers, almost always electrons are flowing, and you would know by reading, asking questions, and learning. As a start, read http://geofex.com/Article_Folders/How_It_Works/hiw.htm

Also, you talk about the base-emitter, the collector-base, but also the collector-emitter junction, which makes me wonder how it all is connected, are the three pins all connected to each other internally or something? I guess a diagram would help...

A bipolar transistor is a piece of silicon with (at least) three different regions inside it. It starts as an ultrapure crystal of silicon, and is then doped with different carefully selected kinds of impurities; these impurities are concentrated in three different regions of the chip of crystalline silicon. The doping impurities can cause the silicon to have an excess of negative charge carriers, or an excess of positive charge carriers. These are know, prosaically enough, as N-type and P-type regions. You can conceive of an NPN transistor as a sandwich of three regions: N-type, P-type, and N-type again; this is an NPN transistor, and called that for that reason. Obviously, you can also make a PNP transistor by reversing the "stacking order". I will describe only NPN devices; PNP is the same, but all voltages and current directions are reversed.

What's important is that each of these regions gets one lead. The middle one is always the base. The other two are the collector and emitter, and the junctions between them are the collector-base junction on one side and the base-emitter junction on the other side.

In the very first transistors, the collector and emitter were identical, but we quickly started making them asymmetrical because they do different things. When you set up an NPN transistor, one of the two non-base leads is more positive than the base. This is the "collector". The junction between the collector and base regions is reverse biased because the collector N region is more positive than the base P region. This forms a reverse-biased diode junction, and no current flows. The lead that's more negative than the collector and base is the "emitter".

If you don't do something to the base, no current flows into the base, and so the reverse biased collector junction conducts nothing, and nothing happens. The transistor is off. If you pull the base voltage more negative than the emitter, then BOTH the collector and emitter junctions are reversed and the still nothing happens. But when you increase the base voltage with respect to the emitter voltage, at some point (0.45 to 0.5V, remember) current starts flowing into the base, forced in by the tiny voltage turning the base-emitter junction on.

The collector is still more positive than the base (in normal operation; let's stay with that for a while) so current should not flow through the reverse biased collector base junction. But charge carriers are being injected into the base region. These get injected into the base part of the collector-base junction and they poison the ability of the collector-base junction to hold off the pressure from the voltage on the collector trying to force current through. And some current slips through the collector-base region and flows through the base to the emitter, which is the most negative lead. If you construct (dope) the collector-base junction and base-emitter junction correction, more than one charge carrier slips through the collector base junction for every charge carrier that's injected into the base. The multiple of how many more slip through to the ones injected into the base is the current gain of the transistor.

To get the advantages of high voltage withstanding, you want the collector-base junction to be lightly doped on the collector side. To get easy base current flow, you want the base-emitter junction to be heavily doped. To get high gain, you want a very thin base region. Getting all this at once in the same little chip of silicon took the best and most expensive engineers and semiconductor physicists decades.

Wait... if a current is flowing through, say, a wire, wouldn't that current also have a voltage?

And you win a cigar! Yes! You are absolutely correct, and this is something that trips up many people. Wire has a resistance, and current flowing down the wire requires a voltage to force it through the wire. What is different is that wire has a much, much lower resistance than other parts, so it's customary to ignore it. There are situations where you can't ignore it, though. Copper wire is usually rated in milliohms (that's 1/1000th of an ohm) per foot, so one ampere of current causes millivolts of signal; looked at another way, to cause 1V to appear across 1milliohm, you need one thousand amperes. So the voltages are tiny, but they are there.

Doesn't the little "thingy" of electricity going through the wire have both a measurable current and measurable voltage?

No. What it has is charge. The "thingy" of electricity is the electron. This is a subatomic particle that's almost infinitely tiny, but has an electrical charge. To all intents and purposes, the charge of the electron is the most fundamental unit of electrical charge. The proton has an equal but opposite charge, but is about 2000 times more massive. All ordinary electricity is the movement of electrons. The movement of 6.242 × 10¹⁸ electrons past some point in one second is one ampere of current. The electrical pressure/force required to get them to do that is voltage. The voltage needed to force that many electrons past one point depends on the resistance of the path. If the resistance is one ohm, then one volt will force one Coulomb (that's a name for that many electrons) through a circuit in one second. If the resistance is one thousand ohms, then you need one thousand times more voltage/force to get the same current.

Voltage is electrical pressure.
Charge is the number of electrons
Current is the MOVEMENT of a number of electrons.
Resistance is the ratio of the voltage needed to force a current to flow. The unit of resistance is the Ohm, which is one volt per ampere.

[brett]

Hi
It seems that you might benefit from a "mental model" or even a physical model of the characteristics of electricity.
Ohm's law is described in many different ways on the web.  It is one of the fundamental processes that you might like to learn about.

My mental model of electricity is the "buckets of water".  I seem to recall Mr Hammer using this analogy.  High voltage is like high water pressure, as can be produced by a bucket up very high(1).  A high bucket (+100V) connected to a low bucket (0V) by a thick hose (low resistance, 1 ohm), lets more water (current(2)) flow than a thin hose with high resistance.  Charge is the amount of water in the bucker.  Electrical energy is the combination of bucket size and bucket height.

In these terms, a transistor can be thought of like a water pump.  When a small amount of water flows from the base to emitter, a large amount of water flows from the collector to emitter.  In a simple BJT common source aplifier, there's a large pipe from the base via the emitter to the ground (e.g. 1k) and a much smaller pipe from the base to the +ve supply via the collector (e.g. 100k).  As RG said, almost exactly the same current flows through both connections.  So where will the pressure (voltage) be least?  In the collector pipe (collector resistor).  When will it be least?  When there is a large current flow through the system.  Turning the transistor valve on more makes the collector pressure (=voltage) lower (until it "saturates").  Turning the transistor valve off increases the pressure (=voltage).  If the transistor is totally off, there is no current flow and the pressure at the collector is the same as in the power supply +ve.

I'm sure there must be graphics of these things on the web.

Apologies if this just confuses things.
cheers   



1. equating electrical potential with gravitational potential.
2.  I presume that electrical current is called current because early researchers had this "water flow" model in their heads.  Anybody know the facts?

[frank_p]

With the graphs and water analogy:
http://www.diystompboxes.com/smfforum/index.php?topic=75652.msg616803#msg616803

[asfastasdark]

When a small amount of water flows from the base to emitter, a large amount of water flows from the collector to emitter.

OK, this, I think, may be the main thing I don't understand... In a simple one-transistor boost circuit (for example, EHX LPB-1), the circuit's input (that is, the guitar sound coming from the pickups) is fed into the base, right? Well... how can it go to the output (which is connected to the collector) if both the base and the collector go towards the emitter, which is connected to ground? 

[R.G.]

OK, this, I think, may be the main thing I don't understand... In a simple one-transistor boost circuit (for example, EHX LPB-1), the circuit's input (that is, the guitar sound coming from the pickups) is fed into the base, right? Well... how can it go to the output (which is connected to the collector) if both the base and the collector go towards the emitter, which is connected to ground? 

Ah. Yep, that's a biggie.

You need to understand the difference between current flow and signal flow. The *currents* in the base and collector do, in fact, both flow toward the emitter.

However, current isn't what you're looking for. You want to amplify a *signal*.

That schematic left out a very important item - how the base is biased, or set up in a ready-and-waiting condition so signals can be amplified. If the base is not held somewhere in the middle of the power supply so that a medium, but not all-it-can-conduct current flows in the collector, the thing can't amplify. Exactly how to do that is outside what we're discussing now, so just imagine that somehow, in an unspecified way, there is a resistor connected to the base, and that resistor is connected to another DC voltage source. This other source is not the same voltage as the 6Vdc source, but something between the 6V power supply and zero volts, and it's just enough to hold the base at whatever voltage is needed to keep half of the maximum possible current flowing in the collector.

With the bias set up, the collector can either increase its current (and thereby lower its voltage because of the more current in the collector resistor) or decrease its current (and thereby raise its voltage because of less current in the collector resistor.)

Now, what are we considering to be "signal"? For most audio work, signal is an alternating voltage. We connect a signal source to the base through a capacitor, which blocks DC but lets the alternating signal through. We already know that increasing the base *voltage* causes the base current to increase, and also that decreasing the base *voltage* causes the base current to decrease. We also know that base current increasing/decreasing causes the emitter and collector currents to increase/decrease, but bigger by a factor of the current gain of the transistor. And those changes in collector or base current cause the collector voltage to drop and the emitter voltage to rise (and vice versa) because of the increased/decreased current.

==> The voltage changes on the emitter or collector are the output signal <==

The currents that flow through the collector and emitter cause the collector and emitter voltages to change because they cause a voltage change on the collector and emitter resistors. We can take that amplified *signal voltage* out of either the collector or emitter (or, in fact, both at the same time) by using another capacitor to let the alternating signal out while blocking the DC conditions at the collector or emitter.

So the input signal causes a current which flows to the emitter. The input current changes cause collector current changes, but those are converted to a signal voltage in the collector resistor.

[brett]

Hi
it is amazing how hard it is to amplify an input signal, and NOT create differences (other than volume) between the input and output signals.

So some people spend $1000s on their hi-fi systems.  When they listen to the music and distortion on their CDs, they're getting very accurately reproduced music and distortion.
cheers

[alt234]

The puzzling part is when you connect an output capacitor and load resistor to the collector. When the output voltage swings positive, current flows from the right hand plate of the capacitor, through the load resistor to the battery negative terminal via ground. When the output voltage swings negative, does the current flow in the opposite direction i.e. back up through the load resistor to the right hand plate of the capacitor ?  If this is true, what is the full path taken by the current ?

[R.G.]

The puzzling part is when you connect an output capacitor and load resistor to the collector. When the output voltage swings positive, current flows from the right hand plate of the capacitor, through the load resistor to the battery negative terminal via ground. When the output voltage swings negative, does the current flow in the opposite direction i.e. back up through the load resistor to the right hand plate of the capacitor ?  If this is true, what is the full path taken by the current ?

What is happening is that the voltage across a capacitor cannot change instantaneously. Sitting at no signal conditions, the capacitor has some voltage in the middle of the power supply on its collector side. It has zero volts on the other side, pulled down to zero volts by the load resistor. The cap is charged to some voltage, and therefore has a net charge inside it. When the collector moves up, the collector side plate must move up with it. The load-side plate must follow the collector-side up, because the voltage cannot change instantly.

In fact, the speed which the voltage across the capacitor can change is determined by the capacitor equation: I = C* (dv/dt), or put another way, the speed of voltage change (dv/dt) is equal to I/C. The current available to charge the capacitor is limited by the load resistor and the voltage across it. If the voltage on the collector steps up by 1V, then the whole capacitor steps up by 1V, and the load resistor instantly have 1V across it. That lets the load resistor pull down on the load side of the cap with a current of I = 1V/R, and the capacitor voltage starts increasing as the load resistor again pulls down on its outside terminal.

If the collector then steps down by 2V, the cap voltage is now too big, and it pulls the load resistor terminal below ground. Again, the load resistor starts conducting current out of the cap into ground and pulling the cap's load side UP towards 0V. So the cap can pull the load resistor both positive and negative with respect to ground because it carries a bucket of charge. The capacitor's voltage is always "sagging" in the direction to relax the capacitor, but if the signal frequency is fast enough to pull the cap both positive and negative equally and faster than it can "relax" from the load resistor, the average voltage across the cap remains constant.

We know this is how this works, because we can calculate the frequency where the cap starts "eating up" the applied voltage by sagging. The lower the frequency, the more the cap will charge to the applied collector voltage minus zero volts. The higher the frequency, the more the voltage on the cap will stay the same because it doesn't have time enough to change, and the more of the changing voltage which appears across the load resistor.