Not another How does a transistor amplify thread

[MinnesotaState]

Not another "How does a transistor amplify" thread

I used to think solid state components such as transistors amplified signals. This may be correct, but it most certainly needs elaborating. Transistors themselves do NOT amplify signals. It is my understanding that transistors take an input voltage & ADD it to the supplied voltage.

Assume the following,

You want to amplify a guitar signal to drive a pair of loudspeakers. You're using a standard 9V. The output voltage from the guitar is 1V.

So, how do you amplify (increase) the voltage? Simple. Take advantage of the 9V & add it there.

So, 9V + 1V = 10V

You've just amplified the signal.

Is this correct?

What if your input voltage exceeds that of your supplied voltage?

So, assume your output voltage from the guitar signal is 10V & you're still using your 9V battery. Do the voltages add up? Or has the transistor reached saturation. Voltage adds up in series. What's stopping the voltage from doing so in this application? It's impossible to run a guitar signal in series to an amplifier?

Reason I ask this is because I just made the LM386. I'm near positive the the 3.5mm jack from my lap top is what they call line level, therefore the voltage is probably high. I assume that voltage exceeds the voltage from my 9V that I'm using for my LM386 amplifier ...& yet it amplifies.

 

[Jony130]

Yes, transistors themselves do NOT amplify signals. But what you need is to understands what amplification really means.
Amplifier is a device that allow as control the flow of "high power" by helps of a low power. To the Amplifier effect occurred, two things are necessary: source of energy (power supply) and a device for controlling the flow of this energy - > the amplifier.

So transistor only control the flow of a energy.

See the example of a common emitter amplifier

At first we bias the transistor in linear active region.
Next we connect the input signal 2V peak to peak AC signal.
The DC voltage at base is equal Vb=2.6V, and emitter dc-voltage is 2V.
So if we apply the ac signal to the base. The dc-voltage will be change from 2.6V + 1V = 3.6V and 2.6V - 1V = 1.6V. So the base voltage will change from 3.6V to 1.6V in "rhythm" of a ac input signal.
These changes will result that the emitter voltage will also change. From 3V to 1V. This will result the change in emitter and in collector current by (3V) 0.9mA to 0.3mA (1V). And this change in collector current will cause change in VRc voltage, between 9V to 3V. We have a three times larger change in VRc voltage because Rc is three times larger then Re resistor.

So we can say that amplifier "modulated" power supply energy in the rhythm of a input signal.

Any more questions?

 

[rbj]

i love that hydraulic or aqueduct analogy.

perhaps to answer (in a hand wave) the OP, is that the collector-base junction is reverse biased. so even though a big voltage would like to push some charge through that junction, it cannot because of the depletion region right at the CB junction. that region is less depleted when there are charge carriers available from current flowing in from the base, and the base is very thin.

 

[cabraham]

It's kind of like the following analogy. We use the term "magnifying glass" to describe a device making it easier in reading fine print. The glass is said to "magnify the print". This is not a problem if we understand the definition of the word "magnify". The glass does not literally increase the size of the print at all. It merely bends and focuses the light rays in a manner which creates a virtual image which is a scaled up facsimile of the print.

Likewise a transistor, bjt or FET, is not literally increasing the magnitude of the input current and/or voltage. Rather it controls or modulates the current delivered from an independent power source, be it a battery, ac outlet, car alternator, etc. The word "amplify" is understood to mean that the device along with an independent power supply form a signal which is the scaled up facsimile of the input signal. Two scope probes will affirm that if we view the input and output signals simultaneously, they co-exist. The output is scaled up and identical to the input, but the input still remains unchanged.

To amplify is to create a replica scaled up in magnitude. A transistor does just that. It does not literally "increase" the current/voltage of the signal.

It is very analogous to a magnifying glass that merely creates a virtual image scaled up but otherwise like the original image, which remains unchanged.

Claude

 

[MinnesotaState]

Excellent.

Thank you all. That was very straight forward.

Any recommendations as far as where to begin? Perhaps in the order of PNP, NPN, FETs, & finally BJTs before I dive into IC's.

 

[dlgoff]

http://ecee.colorado.edu/~bart/book/book/contents.htm

 

[analogdesign]

Excellent.

Thank you all. That was very straight forward.

Any recommendations as far as where to begin? Perhaps in the order of PNP, NPN, FETs, & finally BJTs before I dive into IC's.

Maybe 95% or more of modern ICs are done with CMOS (FETs). So I would focus on those. BJTs are fun historically but they aren't as relevant as they used to be. The only time I have used them in my career is to generate a bandgap voltage on a CMOS chip.

Also, I disagree with the idea that transistors don't "amplify". From a small-signal perspective that is exactly what they do. That's why a common-emitter or drain circuit is called a "one-transistor amplifier". Of course the energy to do this comes from the supply; otherwise we'd have a perpetual motion machine.

 

[cabraham]

Maybe 95% or more of modern ICs are done with CMOS (FETs). So I would focus on those. BJTs are fun historically but they aren't as relevant as they used to be. The only time I have used them in my career is to generate a bandgap voltage on a CMOS chip.

Also, I disagree with the idea that transistors don't "amplify". From a small-signal perspective that is exactly what they do. That's why a common-emitter or drain circuit is called a "one-transistor amplifier". Of course the energy to do this comes from the supply; otherwise we'd have a perpetual motion machine.

We never said that they don't "amplify". I pointed out the meaning of the term "amplify", using a magnifying glass analogy. As long as we properly define the term "amplify" there is no problem. A transistor does indeed "amplify" provided we construe the term as follows.

When a signal generator inputs an amplifier stage consisting of one or more transistor(s) and passive parts and an independent power source, the output is a facsimile of the input signal scaled up in magnitude re current/voltage.

The transistor "amplifies" the input signal. In other words the transistor controls/modulates the power supply current delivery to a load in accordance with an input signal waveform. Thus the current delivered to the load, as well as voltage, follows the input waveform only scaled up. This is what "amplify" means.

Some construe the term to mean literally "to make bigger". The posters who said that transistors don't do that are simply pointing out that the input signal is not changed by transistor action. A good amp stage puts minimal loading on the input signal generator. The discussion is all about how to construe the word "amplify"

If the term is construed to mean that the input signal is made bigger, then no, a transistor does not do that. But if the term is construed to mean the creation of a facsimile identical to the input signal but scaled up in current/voltage magnitude, then yes, a transistor does just that.

A transistor does indeed "amplify" as long as we agree on how the term is defined.

Claude

 

[nsaspook]

Also, I disagree with the idea that transistors don't "amplify". From a small-signal perspective that is exactly what they do. That's why a common-emitter or drain circuit is called a "one-transistor amplifier". Of course the energy to do this comes from the supply; otherwise we'd have a perpetual motion machine.

Agreed.
What does a lone transistor do? It provides a gain transfer function that if used in a proper circuit can amplify (output power greater than input power) a signal. It is a controlled resistor or attenuating element but it's one with intrinsic gain.

For BJT small signal parameters we have:
Input resistance, output resistance, transconductance (the ratio of the current change at the output port to the voltage change at the input port).

That provides us with what's needed to calculate the amplifier circuit gain.

 

[cabraham]

Agreed.
What does a lone transistor do? It provides a gain transfer function that if used in a proper circuit can amplify (output power greater than input power) a signal. It is a controlled resistor or attenuating element but it's one with intrinsic gain.

For BJT small signal parameters we have:
Input resistance, output resistance, transconductance (the ratio of the current change at the output port to the voltage change at the input port).

That provides us with what's needed to calculate the amplifier circuit gain.

Also needed is current gain β. When driving a low Z load, high current gain may be needed. You need a device with sufficient β. "Amplifier circuit gain" can involve voltage gain as well as current gain, depending on the application. I just thought I'd mention it.

Claude

 

[MinnesotaState]

Holy cow I just took notes.

 

[analogdesign]

Also needed is current gain β. When driving a low Z load, high current gain may be needed. You need a device with sufficient β. "Amplifier circuit gain" can involve voltage gain as well as current gain, depending on the application. I just thought I'd mention it.

Claude

Yep. What you're really after is "power gain". That's why we call an emitter follower an "amplifier" even though its voltage gain is ideally one.

And Claude, I'm with you on the definition of amplification thing. I couldn't get it through my head that someone would think an amplifier could modify the input signal itself. Thanks for the clarification.

 

[nsaspook]

Also needed is current gain β. When driving a low Z load, high current gain may be needed. You need a device with sufficient β. "Amplifier circuit gain" can involve voltage gain as well as current gain, depending on the application. I just thought I'd mention it.

Claude

β is the product of input resistance and transconductance and yes "Amplifier circuit gain" meant power gain.

 

[meBigGuy]

To join in on the sematic debate, a transistor amplifier has gain. It can have gain greater than 1 or less than 1 depending on the design. A bipolar transistor has Beta, which can also be less than 1 at high frequencies (Ft)

A FET converts a gate voltage to a drain-source current, so its gain is considered a transconductance.

A great analog designer once told me that a bipolar transistor is actually the same, that is, a transconductance device. Variations in input voltage cause changes in collector current. We tend think of it in terms of Beta, but analog IC designers have a different view. The base voltage and current are related, of course, but the voltage was what he was concerned with.

 

[cabraham]

To join in on the sematic debate, a transistor amplifier has gain. It can have gain greater than 1 or less than 1 depending on the design. A bipolar transistor has Beta, which can also be less than 1 at high frequencies (Ft)

A FET converts a gate voltage to a drain-source current, so its gain is considered a transconductance.

A great analog designer once told me that a bipolar transistor is actually the same, that is, a transconductance device. Variations in input voltage cause changes in collector current. We tend think of it in terms of Beta, but analog IC designers have a different view. The base voltage and current are related, of course, but the voltage was what he was concerned with.

Ref bold - crackpot nonsense. Vbe does not "cause" Ic. Ie is what determines Ic. When Ie change due to an external source, Ic changes accordingly, and Vbe changes in response. Ie changes before Vbe since the junction is capacitive. Cause can never be after the effect. Vbe is not the cause of Ic. THis is a heresy that a vocal minority continually propagates. "All analog IC designers" don't think this way. The ones that do are wrong.

Claude

 

[meBigGuy]

I take it you are not a bipolar analog IC designer.

I think you meant Ib, not Ie, BTW.

Look at most common emitter models. The output current is expressed as gm*Vbe. The input is generally a voltage source acting across misc. elements, not a current source. Look at the guts of a spice model.

Wikipedia takes a different view also:

The gm of bipolar small-signal transistors varies widely, being proportional to the collector current. It has a typical range of 1 to 400 millisiemens. The input voltage change is applied between the base/emitter and the output is the change in collector current flowing between the collector/emitter with a constant collector/emitter voltage.

The transconductance for the bipolar transistor can be expressed as

g_m = \begin{matrix} \frac {I_C}{V_{T}} \frac{\mathrm{mA}}{\mathrm{mV} } \end{matrix}

where IC = DC collector current at the Q-point, and VT = thermal voltage, typically about 26 mV at room temperature. For a typical current of 10 mA, gm ≈ 385 mS.

----------------------------
Since bipolar analog IC design makes heavy use of current mirrors and is concerned with gm matching, possibly that has to do with the perspective.

 

[cabraham]

I take it you are not a bipolar analog IC designer.

You take it wrong. I've designed circuits for 35 years, analog, some of which went into IC or hybrid form, but nearly all are discretely implemented. I've designed current mirrors, sources/sinks, discrete amps, gate drivers, logic gates, transimpedance/transadmittance amps. optical detectors and drivers, rf amps, auto-gain networks, logarithmic and anti-log amps, etc. I've done a lot with analog as well as digital. Not every application requires high beta value. Sometimes, the main objective is switching, mirroring, or straight voltage gain where load current is small. If the source can only output 100 uA, but the load only draws 200 uA due to its high input impedance, the needed current gain is just 2. Any bjt can provide such gain up to half its ft value in frequency. Some applications emphasize the other parameters more than beta.

I think you meant Ib, not Ie, BTW.

No, Ie is what I meant.

Look at most common emitter models. The output current is expressed as gm*Vbe. The input is generally a voltage source acting across misc. elements, not a current source. Look at the guts of a spice model.

The "gm*vbe" is equivalent to "hfe*ib". How does one compute "vbe". The hybrid pi circuit has a small signal equivalent resistance representing the b-e junction called "r_∏". This resistor is in series with rbb' (base spreading resistance), re (emitter resistance, and Re (emitter degenerating resistance). To compute vbe we start with vin, the signal source, and use voltage divider principle. But what is r_∏? It is hfe/gm. So r_∏ is related to hfe and gm meaning that one cannot compute gm*vbe w/o a value for hfe.

Expressing the output current as hfe*ib gives the same result. Vin signal source across the sum total of rbb', r_∏, re, and Re results in ib the signal base current. Multiplying by hfe gives ic, the same result as gm*vbe.

Wikipedia takes a different view also:

The gm of bipolar small-signal transistors varies widely, being proportional to the collector current. It has a typical range of 1 to 400 millisiemens. The input voltage change is applied between the base/emitter and the output is the change in collector current flowing between the collector/emitter with a constant collector/emitter voltage.

The transconductance for the bipolar transistor can be expressed as

g_m = \begin{matrix} \frac {I_C}{V_{T}} \frac{\mathrm{mA}}{\mathrm{mV} } \end{matrix}

where IC = DC collector current at the Q-point, and VT = thermal voltage, typically about 26 mV at room temperature. For a typical current of 10 mA, gm ≈ 385 mS.

----------------------------
Since bipolar analog IC design makes heavy use of current mirrors and is concerned with gm matching, possibly that has to do with the perspective.

All transistors, bjt and FET, have transconductance as well as current gain. It is the power gain that makes active devices so useful. Because a bjt has transconductance does not make it voltage controlled. Because a FET has current gain does not make it current controlled. As far as which parameter is more useful, I covered that above, it depends on the application. Current mirrors are not employed for high current gain, but rather to create a facsimile of a current in another location. High beta minimizes error, but not the most important parameter.

However, an analog designer may wish to make a 1-stage broadband amp that must present a high input impedance to the signal source while driving a load that demands substantial current. Beta is all important here. WHen I design a FET gate driver w/ bjt emitter follower, the beta value at high current levels of 1 or 2 amp is very important.

Your diatribe about "gm" is valid only in the small signal domain, not when a bjt is used for switching. If I toggle a bjt from cutoff to saturation, gm is not useful, but beta is, as well as storage time, reverse recovery time, etc. People who challenge the current control model only use small signal models as their "proof". The current control model is equivalent since gm*r_∏ = hfe. But when bjt incurs large signal operation and/or switching mode, the parameter "gm" has no use. In fact in switch mode, current control is not helpful, charge control is employed.

I know that some IC designers who work a lot with mirrors and small signal amps do not pay much attention to hfe. In some apps, any bjt has sufficient hfe value, and other requirements get the attention. That in itself does not make vbe the control signal. The change in ie and ib take place before vbe changes. THe ic value changes in response to ie changing. Then vbe eventually catches up.

Ie/Vbe/Ie are correlated, they cannot exist separately, a change in 1 cannot happen w/o the other 2 being impacted. But I must stress that vbe is not controlling ic, ie is what controls ic. The transistor action equation is Ic = alpha*Ie. Comments/questions welcome, I will elaborate if you wish.

Claude

 

[analogdesign]

Debating how a BJT "really" works is interesting, but these discussions seem to me to be about distinctions that make no difference. All of my professional work has been with CMOS ICs so the only real exposure I've had to BJTs are awful substrate BJTs used in bandgap references (even though they have beta around 1 (!) they work fine for bandgap generators).

My classwork and labwork back in school, though, convinced me that while, physically, the BJT is a current controlled device (transistor action is precipitated by minority injection into the base region) in practice it can for 90% of applications be treated as a voltage controlled device with a finite input resistance (r_pi). Either viewpoint (current vs. voltage) works almost all the time.

What I find really interesting is subthreshold operation of MOSFETs. In that case the output current is exponentially related to the Vgs of the device, like a BJT! That region is used in micropower applications because gm/I (a bang-for-the-buck figure of merit) is higher in subthreshold than in saturation.

One last question for Claude. I don't have any experience in discrete board-level power switching designs... but I'm curious why you would use a BJT to switch a heavy load when power MOSFETs are available. It seems to me the charge storage in the base of the BJT would limit the switching frequency. Since there is no analogous effect in a MOSFET, wouldn't a MOSFET be faster? What benefit does a BJT have?

Nice discussion guys!

 

[cabraham]

Debating how a BJT "really" works is interesting, but these discussions seem to me to be about distinctions that make no difference. All of my professional work has been with CMOS ICs so the only real exposure I've had to BJTs are awful substrate BJTs used in bandgap references (even though they have beta around 1 (!) they work fine for bandgap generators).

My classwork and labwork back in school, though, convinced me that while, physically, the BJT is a current controlled device (transistor action is precipitated by minority injection into the base region) in practice it can for 90% of applications be treated as a voltage controlled device with a finite input resistance (r_pi). Either viewpoint (current vs. voltage) works almost all the time.

What I find really interesting is subthreshold operation of MOSFETs. In that case the output current is exponentially related to the Vgs of the device, like a BJT! That region is used in micropower applications because gm/I (a bang-for-the-buck figure of merit) is higher in subthreshold than in saturation.

One last question for Claude. I don't have any experience in discrete board-level power switching designs... but I'm curious why you would use a BJT to switch a heavy load when power MOSFETs are available. It seems to me the charge storage in the base of the BJT would limit the switching frequency. Since there is no analogous effect in a MOSFET, wouldn't a MOSFET be faster? What benefit does a BJT have?

Nice discussion guys!

But my point was that this model works only in small signal mode. The r_∏ value is determined by hfe and gm, where gm is determined by current Ic. In the switching mode, gm is never used. The gm value is obtained by computing dIc/dVbe at a specific value of Ic. Ic is exponential wrt Vbe so a linear slope is valid only if Ic deviates a very small amount from quiescent value. If Ic swings from saturated (fully on hard) to cutoff, gm varies wildly and is never used for analysis.

As far as bjt parts used for switching, storage time is not bad if the bjt is used as an emitter follower. An EF never goes into saturation, never incurs a forward biased collector-base junction. When driving a MOSFET gate, EF bjt stages are often used in gate driver IC chips. In discrete gate driving, a small signal FET or bjt is used. The FET is usually common source, CS, which inverts polarity. The bjt stage is usually EF, non-inverting.

A bjt totem pole in common emitter is a poor choice for gate drive due to the issues you mentioned. An EF does not suffer from these limitations.

Claude

 

[analogdesign]

But my point was that this model works only in small signal mode. The r_∏ value is determined by hfe and gm, where gm is determined by current Ic. In the switching mode, gm is never used. The gm value is obtained by computing dIc/dVbe at a specific value of Ic. Ic is exponential wrt Vbe so a linear slope is valid only if Ic deviates a very small amount from quiescent value. If Ic swings from saturated (fully on hard) to cutoff, gm varies wildly and is never used for analysis.

Point taken. You're quite right about this. I guess I don't think about things in those terms much since I do ICs for signal processing and am (almost) always worried about small-signal performance.

 

[cabraham]

Point taken. You're quite right about this. I guess I don't think about things in those terms much since I do ICs for signal processing and am (almost) always worried about small-signal performance.

I understand and agree. Like I said, an active device, FET or bjt, can be used for voltage gain, current gain, or both (power gain). Applications vary and usually we don't need both in unison. If I want to actuate a relay needing 2.0 amp of current and 3.3 volts, with a micro-controller whose I/O outputs 3.3 volts but only 2.0 mA of current, I need unity voltage gain, but large current gain.

A FET or bjt can be employed. A FET has very large current gain at low frequency, and is usually the part of choice. Here, gm for the bjt or FET is not even considered, just the required base/gate drive current.

But we seem to agree that different applications will make differing demands on the active device. Anyway, I think we have given the OP some good food for thought. I will clarify if needed. THanks to you and all for your valuable input.

Claude

 

[meBigGuy]

It's ironic you would label my brief statements as crackpot and a diatribe. Chill out, dude.

The unapproximated Ebers–Moll equations are used to describe the three currents in any operating region of a transistor. Those equations are based on the transport model for a bipolar junction transistor. If you look at them they are all exponential relationships of Vbe/Vt. (look in wikipedia bipolar transistor page)

The current through a diode is exponentially related to the voltage relative to Vt (V/Vt). This is the way the junctions work. Transistor currents work the same way. If you looks at the above models you see that the currents are a result, not a cause.

I'm not saying that the currents are not part of the process. I'm saying that the currents are a result of Vbe/Vt, and that's what causes things to happen. Obviously Ic = aplha*Ie, but look at what causes Ie. It's exponentially related to Vbe/Vt and the reverse saturation current.

Take a deep breath and try to reply succintly. I have a short attention span.

 

[analogdesign]

It's ironic you would label my brief statements as crackpot and a diatribe. Chill out, dude.

The unapproximated Ebers–Moll equations are used to describe the three currents in any operating region of a transistor. Those equations are based on the transport model for a bipolar junction transistor. If you look at them they are all exponential relationships of Vbe/Vt. (look in wikipedia bipolar transistor page)

The current through a diode is exponentially related to the voltage relative to Vt (V/Vt). This is the way the junctions work. Transistor currents work the same way. If you looks at the above models you see that the currents are a result, not a cause.

I'm not saying that the currents are not part of the process. I'm saying that the currents are a result of Vbe/Vt, and that's what causes things to happen. Obviously Ic = aplha*Ie, but look at what causes Ie. It's exponentially related to Vbe/Vt and the reverse saturation current.

Take a deep breath and try to reply succintly. I have a short attention span.

I think the issue is between what the device is *really* doing (whatever that means) vs. what it is doing effectively.

The Ebers-Moll model does a, but it's just that, a model (in fact it's a pretty black-box model). The fact that the currents are drive by Vbe/vt doesn't mean the transistor works that way, it just means it is an effective way to model the device.

A BJT is structually two back to back diodes, sure, but if it were not fundamentally a current-based device, how come the base has to be so narrow?

This is all semantics and doesn't really matter, but I guess it's fun to think about the differences between physical reality and our models of it.

 

[meBigGuy]

I'm going a little over my own head here, but let me try to relate what I have heard. The internal workings of the junctions are based on action potentials, fermi levels, etc and these potentials control charge injection and recombination, etc. Sure the transistor exhibits current gain at low enough frequencies, but the currents are due to Vbe/Vt.

To a certain extent this is like arguing whther forcing current into a resistor causes a voltage or vice versa. But, when you place those junctions together you produce action potentials that must be overcome in order for carriers to flow. Vbe/Vt and Ise are very fundamental concepts in that physical process.

 

[meBigGuy]

If you think about how bipolar transistors are often used, the Beta model is an excellent approximation. There is generally a "large" base resistor and Vbe is dealt with as a constant. By ignoring the change in Vbe one can easily calculate the input current. Makes life simple. But it is Vbe/Vt that causes the currents. And temperature changes Vt.

 

[cabraham]

Current is not a "result" of voltage. The E-M equations do not indicate that voltage is in control. The paper in 1954 published by Ebers and Moll shows a schematic model of the bjt as 2 current controlled current sources, where Ie controls Ic. Any p-n junction has an exponential I-V relation per Shockley's equation (SE). Let's take a look at SE:

Id = Is*exp((Vd/Vt) - 1)

This is SE for any p-n junction, bjt, LED, diode etc. The diode current Id is the isolated variable whereas the diode voltage Vd is the one being acted upon by the exponential function. This is where much confusion originates. The function argument Vd is not necessarily the controller or causal variable. The isolated variable Id is not necessarily a result or effect.

Id and Vd are simply correlated and the equation does not imply causation either way. Most physics/electronics texts present and deal with both inversions of this equation, the 2nd form being:

Vd = Vt*ln((Id/Is) + 1)

Again Vd could be the forcing quantity sometimes and the result other times, likewise for Id. Examine Ohm's Law (OL):

V = I*R.

Here V is the isolated variable. Are we to assume V is always a result of I? I say not for the same equation is equally true as follows:

I = V/R, R = V/I

Picture an inductor in a shorted loop (superconductor). Current I circulates forever. A superconducting switch is opened and the loop now connects to a finite resistor R. The current I is already established, charges are in motion. But they enter R and electrons collide with lattice ions. In doing so some electrons in conducting band lose energy and drop down into valence band. Photons are emitted when this happens and the resistor emits heat, as expected.

The electrons dropping to valence ionize the atoms and a local electric field is generated. This E field repulses incoming electrons. The integral of this field is voltage drop across R. So the current entering R produced V. I already existed while V did not. I consists of electrons moving, they collide with ions, drop down from conduction to valence, ionize atoms resulting in E field and voltage drop V. I produced V in the R.

A cap has been charged to V. There is no I. But when R is put across the cap, I begins. V produced I. It's a 2-way street. In the L case, V is IR. In the C case, I is V/R. It works either way.

SE does not say that Id is controlled by Vd. Likewise for bjt. In order to increase ic, ie must increase. The collector can only collect the electrons that the emitter has already emitted. Contrary to what was earlier posted, vbe does not make ie increase, but vice-versa.

A bjt has ie, vbe, ib, and ic at steady state value. A signal generator increases its magnitude (microphone, strain gauge, antenna), so that ib/ie entering the b-e junction increase. The signal source sees the cable impedance and the signal has not yet reached the junction. Electron flow has increased due to signal increase. When electron increased flow rate reaches junction and cross, what happens? The depletion zone incurs an increase in charge carrier number. Recombination continues but more electrons build up in the interim causing the depletion region to ionize more and increase the E field and barrier potential.

So the increase in vbe is a consequence of increase in ib/ie. Vbe is the result. I will elaborate if needed, but Ebers-Moll is useful equation for computing bjt amp stage transconductance. Ic = alpha*Ie gives us transistor action across junctions. Ic = beta*Ib gives us current gain.

The bjt (FET as well) are very useful because they give both gains, I and V. Ebers-Moll provides the base for computing voltage gain, beta for current gain. Some apps need 1 or the other, some need both. BR.

Claude

 

[meBigGuy]

Lets see. I said that analog IC designers think in terms of gm, you call me a crackpot, I copy a short quote from wikipedia, you long windedly accuse me of a diatribe and invoke your 35 years of experience, I suggest ebers-moll, and you (again long windedly) say one can look at it either way. Is that a fair representation? Am I a crackpot? Who produced a diatribe? What about the below statement, meant to increase perspective, is so wrong?

"A great analog designer once told me that a bipolar transistor is actually the same, that is, a transconductance device. Variations in input voltage cause changes in collector current. We tend think of it in terms of Beta, but analog IC designers have a different view. The base voltage and current are related, of course, but the voltage was what he was concerned with."

Maybe the next step is to look at the physics and see how Vbe/Vt relate at that level.

 

[meBigGuy]

I've been talking to the analog IC guys here and the consensus view is that, for small signal, gm rules, and for large signal, consider Vbe constant and Beta rules. (which isn't so different from what is being said). But I take offense at being called a crackpot.

But, I have an additional bias because in the semiconductor physics you must overcome the base emitter potential differences to send carriers into the base, which, to me, makes it potential based. But again, that is the same as my tendency to think that it is a voltage across a resistor that "causes" a current.

 

[analogdesign]

I've been talking to the analog IC guys here and the consensus view is that, for small signal, gm rules, and for large signal, consider Vbe constant and Beta rules. (which isn't so different from what is being said). But I take offense at being called a crackpot.

I'm an analog IC guy myself and I agree with this view. Like I said before, I only work with CMOS so this large signal stuff is new to me as well.

But, I have an additional bias because in the semiconductor physics you must overcome the base emitter potential differences to send carriers into the base, which, to me, makes it potential based. But again, that is the same as my tendency to think that it is a voltage across a resistor that "causes" a current.

Very interesting. I have a somewhat different view. I agree that the voltage across a resistor is what causes a current (since the electro-motive force on the electrons causes carrier drift). However, I took some advanced solid-state physics in grad school and what I came away with was, fundamentally, the transistor action "amplifies" minority carrier injection in the base region of the BJT. Without the minority carrier current you can have no BJT.

As an aside, the FET is called "field effect" in order to differentiate it from a bipolar or current-controlled device. A FET is truly voltage controlled since there is no DC path for gate current (ideally) while a BJT won't function without base current. So does the Ic cause base current or does the base current case Ic? What came first, the chicken or the egg?

In my experience, the whole "Is a BJT voltage or current controlled" is one of those impossible to answer questions that gets people worked up (and is why you were called a crackpot, which of course you are not). It's similar to PC vs. Mac, vi vs. emacs, Bud vs. Miller, Cubs vs. White Sox, Coke vs. Pepsi... these questions will never be answered.

So, use what works! As we both agree, from a design standpoint the models makes sense thinking about BJTs as voltage-controlled current sources, so do it!

 

[meBigGuy]

One of the guys I talked tomade the opservation that if you think about it in terms of current, you have to deal with charging depletion regions, etc until the voltage "appears", where as if you just "put" the voltage at the junction, its done.

But of course you are right that its all about the amplification caused by the minority carrier injection that creates the currents, so it is definitely a current amplifier in that sense.

6 of one, half dozen of the other. As long as we keep learning, it's all fun.

 

[cabraham]

π

Debating how a BJT "really" works is interesting, but these discussions seem to me to be about distinctions that make no difference. All of my professional work has been with CMOS ICs so the only real exposure I've had to BJTs are awful substrate BJTs used in bandgap references (even though they have beta around 1 (!) they work fine for bandgap generators).

My classwork and labwork back in school, though, convinced me that while, physically, the BJT is a current controlled device (transistor action is precipitated by minority injection into the base region) in practice it can for 90% of applications be treated as a voltage controlled device with a finite input resistance (r_pi). Either viewpoint (current vs. voltage) works almost all the time.

What I find really interesting is subthreshold operation of MOSFETs. In that case the output current is exponentially related to the Vgs of the device, like a BJT! That region is used in micropower applications because gm/I (a bang-for-the-buck figure of merit) is higher in subthreshold than in saturation.

One last question for Claude. I don't have any experience in discrete board-level power switching designs... but I'm curious why you would use a BJT to switch a heavy load when power MOSFETs are available. It seems to me the charge storage in the base of the BJT would limit the switching frequency. Since there is no analogous effect in a MOSFET, wouldn't a MOSFET be faster? What benefit does a BJT have?

Nice discussion guys!

I generally use FETs for high power switching. Sometimes a bjt can be cheaper, but the speed is very limited due to minority charge storage times. I think we agree on that point. Also, the hybrid FET/bjt structure known as the "IGBT" provides lower saturated Vce drop than a FET, Being bipolar in nature at the output. But a bjt input would need too much base drive due to beta limitations at high currents. So the FET input results in low gate drive even w/ large cllector currents.

The FET, bjt, and IGBT all provide useful functions and which is most suitable varies with application. For cost sensitive applications, a bjt is really hard to beat. I use FETs more than bjt or IGBT for high power switching, at voltages below 600V, and currents below 30, 50 or more amps depending on voltage. I agree that FETs are really good for switching. Did I explain it well?

Regarding the voltage control model working well 90% of the time, that depends on how often you use a bjt as a small signal linear amplifier. I use the bjt as a switch at least as often as an amp. I also use it as a large signal amp, i.e. emitter follower. In these cases the gm model cannot provide results that make sense, but alpha and beta can. In the switching mode even current control does not work, I use charge control because it accounts for storage time, delays, rise & fall times, stored charge, etc.

The charge control model is best when examining the internal physics of bjt, better than current control. I've always stated that current control is an oversimplified model that works at low speed and linear amplification region, small or large signal. For small signal gm*vbe is equivalent to ib*hfe. Because of r_π being defined as vbe/ib, it is equal to hfe/gm. Since hfe=ic/ib, and gm=ic/vbe, then:

hfe/gm = (ic/ib)/(ic/vbe) = vbe/ib = r_∏.

So in order to compute vbe as a fraction of vin the signal at the input, you must compute r_∏ which requires hfe to determine. So even if you use a voltage based computation, the current gain hfe is needed for computing the gain and collector current. Even a voltage model of the amp stage cannot produce an answer unless hfe is included.

At small signal values, gm can be regarded as constant, not so in large signal mode. Hence ic=gm*vbe gives the right answer as does ic=hfe*ib. But in large signal mode only the current control method works.

For switching and/or high speed, only charge control is reliable.

Claude