# NPN BJT transistor: base voltage / emitter current

One side in me says: i only want the most practical simplification to work with those parts. And the other one is saying: no no! You really need to understand the real thing....

I think, both views are correct and necessary!
Only if you understand "the real thing" you are able to decide if and under which circumstances you are allowed to make simplifications.
Example: In many cases, opamps may be treated as ideal voltage-controlled voltage sources - but not always!. And it is up to YOU to decide if simplifications for a specific application are possible or not.

That's where I have my difficulties. I had often simplifed the diode (pn-junction) to the extreme states of being OPEN or CLOSED. I thought, when the voltage across the two terminals of a forward biased diode is high enough (above the forward threshold voltage), it gets "transformed" into a short circuit with constant voltage drop (or a "constant voltage source"?). - This picture is flawed, I know, because it does not describe the relationship between current and applied voltage, which is non-linear... As you've stated already, the current as a function of voltage is an exponential one (Shockley's equation). - But normally, the simplification often works fine, because the diode is used above its threshold voltage, and the current is only limited (controlled) by a adequate resistance in series.

As you said: "...the simplification often works fine". Often - but not always. And it is absolutely necessary to know about the real physical behaviour.
In oscillators, we use diodes for dynamic gain regulation - and we need the knowledge about the exponential characteristic for estimating the oscillator amplitudes.

But ...maybe this should be stated the other way around: the voltage across the diode is controlled by the current that is "allowed" to pass through it, by the resistor. The voltage as a function of the current? The larger the series resistor, the less current can flow, thus the voltage drop across the diode falls below the threshold voltage. It does merely conduct anymore.

Yes - this another example for a "simplified view": Current causes voltage drops. But this is not true - in reality it is ALWAYS the voltage with allows a current to exist. But sometimes such a view helps and can simplify calculations. But we must never forget that it is a simplification!

Physically this does not make much sense though! The current is the result and the not the cause, of the potential difference required across the diode terminals, to "squeeze" the depletion layer inside the diode together, so that the charge carriers can pass. (?)

Yes - correct.

That's where my BJT model breaks and is probably totally inaccurate. I think of the base-emitter voltage VBE as constant, and so I can know where the emitter sits (voltage).

Why do you assume the voltage VBE as constant? This is another good example for my comments above: This simplification of a constant VBE is allowed only if you know under which conditions this is allowed. What are the conditions? Answer: Emitter degeneration ! In this case, the value of VBE (0.6 or 0.7V) plays a minor role only because of all other uncertainties (in particular, parts tolerances). I have tried to explain this using a simple graph (see my post#22 ). In the graph, I have used the name "working line" - it is better to say: "stabilization line".
But realize: The exact value (between 0.6 and 0.7V) does not matter too much - but VBE must exist and must have a value in this region! And the base current IB is only a result of this "opening voltage" - not the cause!.

Physically, it makes absolutely sense! The voltage VBE controls IB.

VBE controls IE and - at the same time (because of IE=IB+IC) - also IB and IC. This is the contents of Shockleys formulas.

But do I need to take that variable base-emitter VBE into account, when looking at BJT circuits?

Yes - at first, because it is the reason for using an emitter resistor and, secondly, each signal voltage at the base causes a change of VBE which is transferred to a change in IC.
And the amount of IC change can be derived from the transconductance gm which is the SLOPE d(IC)/d(VBE) of the IC=f(VBE) curve. The transconductance gm is the key parameter for calculating the voltage gain of an amplifier stage.

I don't really know what I'm talking about, to be honest. It would be great to have a handy tool (model) that makes building/understanding of active circuits easier. - LvW, What would you suggest instead of that "linear BJT model" posted earlier?

I would propose to use no "model" at all. For designing or analyzing an amplifier you need only four basic rules:
* Ohms law,
* IE=IC+IB
* Transconductance gm=IC/Vt (as a result of Shockleys equation)
* VBE=(0.6...0.7)V, to be estimated.
Please, have a look again to my post#33. Here I have listed the 5 basic steps for designing a BJT-based amplifier stage. It was not necessary to make use of any model.

(In case you prefer to have a model, I would recommend the small-signal pi-model with voltage-controlled current source ic=gm*vbe. Dont forget that such a model is small-signal model only which is valied for one single operational point only)

Last edited:
1rel
Those who swear by voltage control make assumptions not supported by science. This question extends beyond bjt devices. if a resistor is used as a heater, is it a voltage controlled heat source or current controlled heat source? Does the voltage across R determine current, or vice-versa, or can it be either depending on conditions? PhD profs disagree. You can surf web sites from MIT, Stanford, Caltech, Case Western Reserve, & find published positions supporting either side. Back to bjt, here are some issues with voltage control that science refutes.
1) The Shockley diode equation is often introduced to students as Id = Is*exp((Vd/Vt)-1). This expression emphasizes to beginners that large changes in current take place while the change in voltage is quite small. But when a diode is forward biased, we do not put a CVS (constant voltage source) directly across it, it would thermally run away. Saturation current Is is temperature dependent. As temp increases, Is increases drastically. A CVS would output a current equal to Is*exp( (Vd/Vt)-1), resulting in a temp rise, which raises Is, which raises Id, raising power, raising temp, raising Is, etc. Id inceases until destruction usually. Vt is thermal voltage = nkT/q, which increases with temp, but not enough to offset Is increase, so diode runs away thermally.

But if we bias the diode with CCS (constant current source), it is thermally stable since Vd = Vt*ln ((Id/Is)+1). Increases in temp will increase Is, but Vd goes down since Is is in denominator, hence power goes down & device is thermally stable. Of course power ratings from diode maker must be observed. This why we always drive p-n junctions with current, not voltage, i.e. "current driven". But a CVS plus a resistor is about as good, since current cannot run away. Any increase in current increases the resistor voltage drop, which reduces Vd, so that current cannot run away. There is seldom an argument that diodes, LED, bjt, SCRS, etc., must be "current driven", never voltage driven, because of thermal stability. To be continued.

Claude. :-)

Last edited:
Averagesupernova
Gandes
For this time FETs are much reliable like MOSFETs. Fet basics, In this type of transistors, the gate is insulated from the channel with the dielectric layer. The area marked “N+” is heavily doped “N” type semiconductor. In case of E MOS transistors with a voltage of UGS = 0, the channel is blocked (its resistance takes the value of MΩ and the ID current doesn’t flow). By increasing the UGS voltage channel increases its conductivity and after reaching a certain value called UT threshold voltage, the flow of the ID drain current was possible through the channel.

dlgoff
Gold Member
For this time FETs are much reliable like MOSFETs. Fet basics, In this type of transistors, the gate is insulated from the channel with the dielectric layer. The area marked “N+” is heavily doped “N” type semiconductor. In case of E MOS transistors with a voltage of UGS = 0, the channel is blocked (its resistance takes the value of MΩ and the ID current doesn’t flow). By increasing the UGS voltage channel increases its conductivity and after reaching a certain value called UT threshold voltage, the flow of the ID drain current was possible through the channel.
From the OP's profile:
Last Activity: Oct 19, 2016

jim hardy
1rel - I know that some books prefer a view which they call "simple view" or "application-oriented view".
But this is simply false. Viewing the BJT as current-controlled does not simplify anything. This subject has really something to do with "religion".
Some guys have learned that the BJT would be current-controlled, they have followed this view for some years - and now they are unable or unwilling to accept that this may be not correct.
I think it makes no sense to stress this subject further, however, without going deep into the physics of the transistor, it is completely sufficient to realize the following simple considerations:
* For a pn-diode, the current follows the applied voltage according to the well-known exponential relation, OK?
* There is no reason that the pn junction between B and E should not follow the same function.
* Therefore, we "open" the B-E junction with a voltage of - lets say - 0.6V. As a result, there will be an emitter current IE with a certain value (of course, assuming a suitabe collector voltage). Of course, a small part of this current goes into the base node (IB).
* Would you agree that the emitter current will be larger if we increase the voltage up to VBE=0.7V ?
* Hence, the amount of emitter current depends on the applied voltage VBE. Thats logical, is it not?
* Now - do you see any reason why this current - suddenly - should depend on the base current IB and NOT on VBE?
____________________
Regarding your last question (oscillator) I would recommend to open a new question.
This oscillator has many interesting aspects worth to be discussed.
You make assumptions refuted by observation. Let us restrict this discussion to a simple 2 terminal diode device, a simple p-n junction.

In order to commence forward current, you need NOT "open up" the junction with 0.65 volts. A diode is connected to a 1.0 kohm resistor & 12 volt battery with a switch in series. The switch is open initially. Diode current Id, & voltage drop Vd, start near zero. The junction barrier voltage at 25 C room temp is around 25.7 mV, the thermal value, Vt.

The switch gets closed. What happens? Not what you think. Positive charges move towards the anode, negative charges towards cathode of diode. Vd is at 25.7 mV. However, charges conduct very well through the diode, which shoulld not surprise us. Positive charges, or *holes* inside the p type anode, move easily through the p type anode, as they are majority carriers. Likewise, when electrons reach the n type cathode, they conduct easily being majority carriers.

The diode voltage Vd is still at 25.7 mV, yet current conducts readily through the diode. No surprise, in the forward direction, p type anode readily conducts holes, & electrons easily move in n type cathode.

After holes travel through anode, they cross junction & enter cathode, then recombine. Holes do not get very far in the cathode, being minority carriers. Likewise electrons in cathode cross junction, enter anode, recombine, being minority carriers. A depletion region is formed by these stored minority charges. A local E field is created, forming a potential barrier. Vd changes as a result of this new current Id. Eventually equilibrium is reached. Shockley studied junction diodes & described the I-V relation as follows:
Id = Is*exp((Vd/Vt)-1). Another form of this equation is as follows:
Vd = Vt*ln((Id/Is)+1). Here is the dispute in a nutshell.

Does the value of Id control Vd? Or vice-versa? Current control deniers, or voltage supremacists if you prefer, insist that Vd determines Id. Simple observation in lab with probes & scopes negate this view.
I have posted sims from Spice in threads showing that current Id changes first, then Vd catches up. A lab test gives the same result.

Of course many will dispute me. But independentt verification is available. Please search using key words *diode reverse forward recovery*.
SE, Shockley equation, holds under steady state, not transient. A reverse biased diode has a Vd negative in value, with SE predicting tiny value of reverse current. However, before Vd goes forward, Id forward current exists. For a moment, current is positive with negative junction voltage. This condition is forward recovery, eventually the junction voltage Vd, changes to 0.65V, & equilibrium is attained.

The voltage across the diode junction is literally determined by forward current. During forward recovery, SE does not hold. Vd does NOT determine Id. It is opposite.

Reverse recovery can be studied using the search engines. My point is that the current control deniers accept as religion that junction forward voltage controls the forward current. This is mere faith. A lab test easily demolishes this.

I will post simulations & reference material Tuesday. Cheers.

Claude Abraham :-)

Hello Claude - do you really intend to start the discussion again?
I know your position and you know my position. Still - I am convinced that you are wrong!
It is already the 2nd line of your contribution which is questionable: Why are you using a 1k resistor in your fictitious "test arrangement"?
Why dont you connect the diode to a test voltage?
I know your answer ("in reality, nobody will put an ideal voltage source across a pn junction"), however, this is a bad argument because we are dicussing physical properties and not practical circuits. In another thread (another forum?) I have asked several questions to you regarding dimensioning of practical circuits (based on, of course, voltage control). But I never got an answer from you. So - let`s stop it at this point.

Regards
LvW

PS: In your contribution (Dec, 28, 2016) you completely misssed the point of discussion. While speaking about physical control mechanisms we should not mix this discussion with practical design aspects (thermal run-away effects). From the physical point of view it is the VOLTAGE across the pn junction that controls the current through it - however, it is a completely other question if it is practical to use an ideal voltage source for this purpose or a voltage source in series with a suitable resistor. Of course, even in this case, we have a control voltage across the junction (principle of volage division).

Last edited:
Law- I simmed a voltage source directly across a pźn junction, & my statements were affirmed.
You are in denial of forward & reverse recovery. Take a simple full wave rectifier network of 4 diodes in a bridge. An a.c. voltage source powers it with caps & load a free the bridge.
Each half cycle the diodes transition, 2 of them from forward to reverse, vice-versa for the other 2. A scope & probes will affirm that a diode in reverse bias has a large negative Vd, with a small negative Id. When polarity of the input source changes, the forward current Id commences pos itively while Vd is still negative.
This is published in the Power Integration company handbook, late 1990s. Forward recovery is the term. If the diode was reverse biased at -10 volts, then polarity changed, with Id transitioning to +1.0 amp, the po wet dissipated is momentarily 10V*1.0A=10 watts. This is very short in duration.
The new value of Id results in Vd transitioning to 0.7 volts, & steady state dissipation becomes 1.0A*0.70V=0.70 watts.
Just search under the key words *diode recovery forward reverse*.
This is well known. In order to change the diode current, you think that the voltage must change first, then current follows. That is wrong. Voltage changes as a result of current. Diode will conduct full forward current with any voltage, +0.7V, +0.1V, 0V, -2.5V, -10V, -50V, etc. But only momentarily.

Claude

In order to change the diode current, you think that the voltage must change first, then current follows. That is wrong. Voltage changes as a result of current. Diode will conduct full forward current with any voltage, +0.7V, +0.1V, 0V, -2.5V, -10V, -50V, etc. But only momentarily.
Claude

Voltage changes as a a result of current? Really surprising!
Just one simple question: Connect a battery voltage of 0.5 volts directly (without any resistor) across a pn diode (forward direction).
Is there a current as a result of the voltage? Yes or no?
Do you really dispute the validity of Shockleys equation Id=f(Vd)?

Last edited:
Voltage changes as a a result of current? Really surprising!
Just one simple question: Connect a battery voltage of 0.5 volts directly (without any resistor) across a pn diode (forward direction).
Is there a current as a result of the voltage? Yes or no?
Do you really dispute the validity of Shockleys equation Id=f(Vd)?
Yes & no. The 0.5 volts is generated via reduction & oxidation in the chemical process, which results in actions (positive) being forced towards the positive terminal, & anions [negative) forced towards negative terminal. The battery voltage comes into existence via internal current. The charges inside the battery move against the E field. Thus battery internal current produces the terminal voltage, since moving charges constitute current.

Of course, any load connected across battery, diode in this case can draw current as a result of the electric field which is related to voltage. But this external current tends to decrease the battery voltage. As electrons exit battery negative terminal, pass through diode, then enter battery positive terminal, voltage would decrease but redox reaction inside battery propels ions against the E field to restore the terminal voltage.

A similar example is to use a charged cap instead of battery. Say a cap is charged to 0.65V, & placed across diode, forward biased. The cap will discharge & diode current results. But the cap loses charge & energy, eventually discharging, then current ceases. Unlike the battery, the cap has no redox going on inside.

Battery generates a voltage via current internal, which is generated via redox. External lies current cam be said to happen due to battery voltage, but the internal current in battery maintains this voltage. If diode current is 1.0 amp, the battery redox reaction must generate 1.0 amp internal to maintain the voltage at the terminals. So both are active, current & voltage.

But remember that the moment the diode is connected the junction barrier voltage is 25.7 mV, thermal value. The current is 0.5V battery value minus 25.7 mV, divides by silicon bulk resistance. As current continues, minority carriers build up the depletion region until the barrier potential settles to a value under 0.5V. The steady state condition results in a steady current equal to 0.5V minus Vd, divided by Rdiode.

Of course a current can happen due to a voltage. But you always omit the fact that that voltage required a current to come into existence. Just as a charged cap can bias a diode into conduction, so can an energized inductor.

Many reading this have seen a diode placed across a solenoid or relay coil. It is reverse biased when power source drives coil. When power is switched off, inductor current continues into the catch diode, & dissipates. The current in the inductor consists of electrons in the conduction band. Upon reaching the diode So material, electrons collide with lattice ions. Electrons fall from conduction band to valence band, a lower energy state. Photons are emitted per Planck law E = if.

Also as the current passes through the diode, voltage builds up in the barrier. Switching power converters display this.

I have stayed forever that a voltage CAN give rise to a current. A current CAN give rise to a voltage. We must examine the conditions for each example.

I have tried posting simulation results before, but my files are already posted in another thread, & a duplicate file won't upload. I'll rerun & post later today.

Yes. Voltage changes due to current, & vice-versa.

Claude

Of course a current can happen due to a voltage. But you always omit the fact that that voltage required a current to come into existence. Just as a charged cap can bias a diode into conduction, so can an energized inductor.
In this thread, we are speaking about currents in electronic circuits (driven by an applied constant voltage source) - and, in this context, it is of no importance HOW this voltage is generated.
This brings to an end my participation in is thread. We should agree not to agree.

(If you can't explain it simply, you don't understand it well enough. A. Einstein)

berkeman
Mentor
Hopefully the OP's question has been answered. Time to close this thread.

dlgoff and davenn