# How accurately does Vbe determine collector current?

#### hp1729

Joined Nov 23, 2015
2,304
hp1729 are you trying to say that the Vbe has no influance on collector current ?

How can you explain the working principle of a BJT current mirror without involving Vbe then? We can ask the same question about differential amplifier.
I am not suggest base voltage has no influence, just not an "accurate" influence. The relationship between current and voltage through a resistor in a simple circuit would be an "accurate" influence. We do not have that relationship between the base and collector of a transistor since there are two circuits involved, base circuit and collector circuit. If we have no collector load base voltage and current can change but collector current does not.
I think we would have to put this in context of the original text to realize what the author had in mind.

#### hp1729

Joined Nov 23, 2015
2,304
hp1729 are you trying to say that the Vbe has no influance on collector current ?

How can you explain the working principle of a BJT current mirror without involving Vbe then? We can ask the same question about differential amplifier.
Re: differential amplifier
Good point!

#### hp1729

Joined Nov 23, 2015
2,304
Seriously? The claim is only applicable in the active region. Not cutoff and not saturation.
More than that. Under different collector loads the change in collector current is different with the same base current and voltage. Yes, there is a relationship between base voltage and collector voltage, but I would not describe it as absolute or "accurate".

#### hp1729

Joined Nov 23, 2015
2,304
Also when I apply voltage to base, the current changes as a result. Your point being?
The point being the word "accurately".

#### joeyd999

Joined Jun 6, 2011
4,425

#### hp1729

Joined Nov 23, 2015
2,304
Current mirrors work if the two devices are identical.

Whether they are current controlled or voltage controlled is irrelevant. Ie. You can build a current mirror with current controlled devices as well, as long as they are identical.
Does the current mirror mirror current or voltage? Is the voltage on one side always going to be equal to the voltage on the other?

#### hp1729

Joined Nov 23, 2015
2,304
The reverse can be argued that if you apply a voltage on the be junction current flows through it.

That is not the primary consideration as to whether a device is current or voltage controlled.
True, of course. The problem comes up with the word "accurately". This kind of implies that there is an absolute relationship between base voltage and collector current. At some specific base voltage the collector will and must be some predictable value.

#### joeyd999

Joined Jun 6, 2011
4,425
Is the voltage on one side always going to be equal to the voltage on the other?
Wouldn't it be a 'voltage mirror', then?

#### WBahn

Joined Mar 31, 2012
26,062
I am not suggest base voltage has no influence, just not an "accurate" influence. The relationship between current and voltage through a resistor in a simple circuit would be an "accurate" influence. We do not have that relationship between the base and collector of a transistor since there are two circuits involved, base circuit and collector circuit. If we have no collector load base voltage and current can change but collector current does not.
I think we would have to put this in context of the original text to realize what the author had in mind.
And if you have no collector load then the transistor is not operating in its active region and so it doesn't apply. There is NO reasonable interpretation of what the TS said that implies that it applies to either the cutoff or the saturation region. None.

#### ian field

Joined Oct 27, 2012
6,539
Page 91 of Art of Electronics Third Edition has a passage that reads in part: "For transistors it is important to realize that the collector current is accurately determined by the base-emitter voltage, rather than by the base current (the base current is then roughly determined by beta) and that this exponential law is accurate over an enormous range of currents, typically from nanoamps to milliamps." I've breaded-boarded up the following circuit and tested it with a number of different 2N3904 and it seems like the collector current is varies from around 10mA to as high as 16mA. Is this spread normal? I would have thought the term "accurately" would have the different transistors with a few percent of each other. How do you know what the Is(T) value is?

Thanks,
Dave
View attachment 99376
Vbe is fairly close, but does suffer a little from manufacturing variations.

Maybe the current mirror is a useful example - you can get away with using discrete transistors to make a current mirror, but you get more consistent results using a monolithic transistor array where all the transistors are fabricated close together on the same slice of silicon.

In many applications; you'd design with enough nfb to make gain spread less significant - and if necessary, bypass the nfb generating resistor with a capacitor to eliminate AC nfb.

#### WBahn

Joined Mar 31, 2012
26,062
More than that. Under different collector loads the change in collector current is different with the same base current and voltage. Yes, there is a relationship between base voltage and collector voltage, but I would not describe it as absolute or "accurate".
The claim is a comparative one regarding the accuracy of using Vbe versus using Ib to predict the collector current.

#### WBahn

Joined Mar 31, 2012
26,062
True, of course. The problem comes up with the word "accurately". This kind of implies that there is an absolute relationship between base voltage and collector current. At some specific base voltage the collector will and must be some predictable value.
No! It is merely claiming that the value predicted by using Vbe will be more accurate than the value predicted using Ib.

#### WBahn

Joined Mar 31, 2012
26,062
Does the current mirror mirror current or voltage? Is the voltage on one side always going to be equal to the voltage on the other?
That's irrelevant to the question of how it works.

The source side of the circuit produces a voltage that is then applied to the mirror side of the circuit. It does NOT produce a specified current into the base of the mirror transistor.

Again, draw up a circuit that uses a current-controlled current source model of a BJT and then use that to explain the behavior of a current mirror.

#### crutschow

Joined Mar 14, 2008
25,260
I don't want to go down this rabbit hole again (and it's been transversed several times before on these forums), but the solid-state physics and equations of the BJT clearly indicate that it is a voltage controlled device with a transconductance gain (gm), and that the base current is actually a by-product of the voltage across the base-emitter junction.
That view is useful when calculating the small-signal AC gain of BJT amplifiers (or the operation of current-mirrors).

That being said, the black-box operation of the BJT makes it also appear that it is a current operated device with a current gain (β or Hfe).
That view is generally more usefully when calculating the large-signal characteristics of the device such as the bias point in AC applications or its operation as a switch.

So I think the best view is that the BJT can be seen as either voltage-controlled or current-controlled, and which black-box characteristic works best depends upon the application.

#### hp1729

Joined Nov 23, 2015
2,304
No! It is merely claiming that the value predicted by using Vbe will be more accurate than the value predicted using Ib.
Now there's an interesting point. Time to sit down and build a circuit to test the stories.

#### WBahn

Joined Mar 31, 2012
26,062
I don't want to go down this rabbit hole again (and it's been transversed several times before on these forums), but the solid-state physics and equations of the BJT clearly indicate that it is a voltage controlled device with a transconductance gain (gm), and that the base current is actually a by-product of the voltage across the base-emitter junction.
That view is useful when calculating the small-signal AC gain of BJT amplifiers (or the operation of current-mirrors).

That being said, the black-box operation of the BJT makes it also appear that it is a current operated device with a current gain (β or Hfe).
That view is generally more usefully when calculating the large-signal characteristics of the device such as the bias point in AC applications or its operation as a switch.

So I think the best view is that the BJT can be seen as either voltage-controlled or current-controlled, and which black-box characteristic works best depends upon the application.
And it is worth noting that treating β as a constant (even an unknown constant) is so unreliable that circuits that do so are generally categorized as "bad". Instead, we design circuits, as much as possible, in which all the matters is that β have some minimum value (and sometimes some maximum value) over the range of operation. Why? Because even if we measure the β of a specific transistor at a specific operating point and hold everything else except base current the same, we don't have a very good predictor of collector current at other base currents far removed from the one it was measured at. Why? Because β itself is a function of current. So we design bias circuits so that the target current is not achieved through the β, but rather through the interaction of Vbe with the other components in the circuit with our primary sop to β being merely to ensure that the base current can be at least large enough so that it can support the desired collector current.

#### dannyf

Joined Sep 13, 2015
2,197
"collector current is accurately determined by the base-emitter voltage, rather than by the base current (the base current is then roughly determined by beta) and that this exponential law is accurate over an enormous range of currents, typically from nanoamps to milliamps."

I think it meant something less sophisticated. Because of the exponential relationship between vbe and ib this ic, a small change in vbe produces a large changed in ib/ic.

So for a given range of device parameter variations, a small range of vbe can counter it (= hitting the same ic) vs. A large range of ib.

That is, if you want to fine tune ic, it is easier to do it via ib.

The word choice of "accurately" is poor.

#### cabraham

Joined Oct 29, 2011
82
Page 91 of Art of Electronics Third Edition has a passage that reads in part: "For transistors it is important to realize that the collector current is accurately determined by the base-emitter voltage, rather than by the base current (the base current is then roughly determined by beta) and that this exponential law is accurate over an enormous range of currents, typically from nanoamps to milliamps." I've breaded-boarded up the following circuit and tested it with a number of different 2N3904 and it seems like the collector current is varies from around 10mA to as high as 16mA. Is this spread normal? I would have thought the term "accurately" would have the different transistors with a few percent of each other. How do you know what the Is(T) value is?

Thanks,
Dave
View attachment 99376
Page 91 of Art of Electronics Third Edition has a passage that reads in part: "For transistors it is important to realize that the collector current is accurately determined by the base-emitter voltage, rather than by the base current (the base current is then roughly determined by beta) and that this exponential law is accurate over an enormous range of currents, typically from nanoamps to milliamps." I've breaded-boarded up the following circuit and tested it with a number of different 2N3904 and it seems like the collector current is varies from around 10mA to as high as 16mA. Is this spread normal? I would have thought the term "accurately" would have the different transistors with a few percent of each other. How do you know what the Is(T) value is?

Thanks,
Dave
View attachment 99376
Let's not lose sight of the TS's question -- which parameter, Ib or Vbe, more accurately predicts Ic (in the linear region). Note specifically that the TSdid not ask whether a BJT is voltage-controlled or current-controlled.

To predict something we need a prediction equation. We really only have two available:

$$i_c \; = \; \beta i_b$$

or

$$i_c \; = \; I_s \( e^{\frac{v_{be}}{\eta V_{th}}} \; - \; 1$$
\)

We all know that beta is not a constant and that it varies with collector current (among other things, such as temperature which we are assuming is being held constant). The variation with current, however, is more than enough to render it suspect if we are relying on it being constant in order to predict the current. The other option involves two parameters neither of which is significantly affected by current.

So if you were to take a real transistor (even though we are leaving out the Early voltage) and use these two equations to predict the collector current within the linear region, which would give you the more accurate prediction. Answer -- the second.
The circuit above is not a good practice, the bjt likely dies. With bjt, one must remember it is current driven. Attempting to voltage drive a base-emitter junction is certain doom for the device. The people who insist that Vbe is what "controls" Ie/Ib/Ic are simply assumimg that current exists because of a "driving voltage". Voltage does not in general drive current nor vice versa.
There are 3 equations that describe Ic in terms of input quantities Ib, Vbe, & Ic as follows:
1) Ic = beta*Ib
2) Ic = alpha*Ies(T)*exp((Vbe/Vt) - 1)
3) Ic = alpha*Ie

The most accurate at predicting Ic is clearly equation 3). Alpha describes Ic in terms of Ie. This relation holds over a huge range of currents from nanoamps up to 100's of mA, even amps for power devices. Also, alpha shows very little temperature variation. However equations 1) & 2) deserve some attention.

Equation 1) describes Ic in terms of Ib, with parameter "beta". Devices are available where beta is fixed over 3 to 5 decades of Ic. But there are variations from part tp part, known as "specimen" variations". Also beta can vary from low to high temp by a factor of 2 to 3 or so. Even if flat beta parts are used, and devices hand selected after beta testing, temp variation remains.

So if a general purpose bjt, like a 2N2222A, or any other, has 1.0 mA of base current, what is Ic. At low temp, it is around 50 to 200 mA. At high temp, 100 to 300 mA. So it can vary by a factor of 6 or so. Good texts show students how to implement networks that set emitter current Ie to a known value, then depend on alpha for Ic. Alpha is very reliable, more so than beta.

Alpha = beta/(beta+1). In the previous example, with beta at 50 min, alpha = 0.98 min. At beta of 300 max, alpha = 0.997 max. Beta varied 500% (from 50 to 600), yet alpha changed by a mere 1.73%. Clearly equation 3) is more predictable an indicator of Ic than equation 2). I am a practicing EE, and since the 70s I exploit this property.

How about equation 2)? If I know that Vbe = 0.65 volts, can I safely estimate Ic with good accuracy? The Ies current (base-emitter junction saturation value), is a strong function of temp, and varies with specimen. Over the temp range, Ies can vary from femto amps to nanoamps or even microamps for power parts. Also, Vt changes with absolute temp, i.e. Vt = nkT/q. For a 0.65V value of Vbe, Ic can range from microamps to amps. Beta is known to be variable by a factor of up to 6 or so, maybe 10 if you vary current in the collector. But the Ic as a function of Vbe can literally vary by a million (not a typo, I said a million).

Equation 3) is the basis for bjt applications. If we know Ie, then Ic is extremely predictable. If I told you Vbe = 0.65 V, you really have no clue about Ic, esp over temp range. If I told you Ib = 1.0 mA, you can estimate Ic as 50 to 500 mA. Quite loose indeed, but in the ball park.

Bit if I said Ie = 100 mA, you know what Ic is, don't you? Say 99 mA, and you are very close, over the range of super cold to hot, any device, and any current value within the part limits.

3) Ie = alpha*Ie is the best predictor of Ic. Comments/questions welcome. Best regards.

Claude

#### joeyd999

Joined Jun 6, 2011
4,425
The circuit above is not a good practice, the bjt likely dies. With bjt, one must remember it is current driven. Attempting to voltage drive a base-emitter junction is certain doom for the device. The people who insist that Vbe is what "controls" Ie/Ib/Ic are simply assumimg that current exists because of a "driving voltage". Voltage does not in general drive current nor vice versa.
There are 3 equations that describe Ic in terms of input quantities Ib, Vbe, & Ic as follows:
1) Ic = beta*Ib
2) Ic = alpha*Ies(T)*exp((Vbe/Vt) - 1)
3) Ic = alpha*Ie

The most accurate at predicting Ic is clearly equation 3). Alpha describes Ic in terms of Ie. This relation holds over a huge range of currents from nanoamps up to 100's of mA, even amps for power devices. Also, alpha shows very little temperature variation. However equations 1) & 2) deserve some attention.

Equation 1) describes Ic in terms of Ib, with parameter "beta". Devices are available where beta is fixed over 3 to 5 decades of Ic. But there are variations from part tp part, known as "specimen" variations". Also beta can vary from low to high temp by a factor of 2 to 3 or so. Even if flat beta parts are used, and devices hand selected after beta testing, temp variation remains.

So if a general purpose bjt, like a 2N2222A, or any other, has 1.0 mA of base current, what is Ic. At low temp, it is around 50 to 200 mA. At high temp, 100 to 300 mA. So it can vary by a factor of 6 or so. Good texts show students how to implement networks that set emitter current Ie to a known value, then depend on alpha for Ic. Alpha is very reliable, more so than beta.

Alpha = beta/(beta+1). In the previous example, with beta at 50 min, alpha = 0.98 min. At beta of 300 max, alpha = 0.997 max. Beta varied 500% (from 50 to 600), yet alpha changed by a mere 1.73%. Clearly equation 3) is more predictable an indicator of Ic than equation 2). I am a practicing EE, and since the 70s I exploit this property.

How about equation 2)? If I know that Vbe = 0.65 volts, can I safely estimate Ic with good accuracy? The Ies current (base-emitter junction saturation value), is a strong function of temp, and varies with specimen. Over the temp range, Ies can vary from femto amps to nanoamps or even microamps for power parts. Also, Vt changes with absolute temp, i.e. Vt = nkT/q. For a 0.65V value of Vbe, Ic can range from microamps to amps. Beta is known to be variable by a factor of up to 6 or so, maybe 10 if you vary current in the collector. But the Ic as a function of Vbe can literally vary by a million (not a typo, I said a million).

Equation 3) is the basis for bjt applications. If we know Ie, then Ic is extremely predictable. If I told you Vbe = 0.65 V, you really have no clue about Ic, esp over temp range. If I told you Ib = 1.0 mA, you can estimate Ic as 50 to 500 mA. Quite loose indeed, but in the ball park.

Bit if I said Ie = 100 mA, you know what Ic is, don't you? Say 99 mA, and you are very close, over the range of super cold to hot, any device, and any current value within the part limits.

3) Ie = alpha*Ie is the best predictor of Ic. Comments/questions welcome. Best regards.

Claude
Yet Eber-Molls (eq. 2 -- assumedly without parsing your text) precisely accounts for temperature. Alpha does not.