Eber-Moll Equation --- Transistors

[A.J.]

Hello All,

I have been reading an electronics book that shows the following
circuit.


+20V
-----
|
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin /
--------| Q1
\
|
|
|
|
|
|
_____
___
_

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?

Thanks so much for any assistance offered?

Regards

A.J.

 

[Graham Holloway]

"A.J." <itwasme3133@hotmail.com> wrote in message
news:14de980b.0405091855.2404dc7b@posting.google.com...

Hello All,

I have been reading an electronics book that shows the following
circuit.


+20V
-----
|
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin /
--------| Q1
\
|
|
|
|
|
|
_____
___
_

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?

Thanks so much for any assistance offered?

Regards

A.J.

It looks as though the base bias is achieved by applying a voltage across
the base-emitter junction. Not terribly stable.


Graham Holloway

 

[Winfield Hill]

A.J. wrote...

I have been reading an electronics book that shows the following
circuit.

+20V
-----
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin |/
-------| Q1
|\
|
|
_____
___
_

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?

A transistor would not be biased with a fixed temperature-invariant
base voltage in practice, but it is an interesting student problem.

Did your book present the Ebers-Moll equation for you to study? You
can arraign this equation in several ways, but Vbe = kT/q ln (Ic/Is)
is common (dropping a small term that doesn't matter). We often
substitute VT = kT/q, where VT = 25.3mV at 20C, for simplification.

Using this to find the difference of two base voltages, Vbe1 - Vbe2,
call this dVbe, you'll get dVbe = VT ln (Ic2/Is) - VT ln (Ic1/Is),
simplifying to dVbe = VT ln (Ic2/Ic1), which can be converted to
I2 = I1 e^(dVbe / VT). This says if you change dVbe by 8x 2.1mV or
16.8mV, then I2/I1 = e^(16.8/25.3) = 1.94 Which means I2 = 1.943mA
in your problem, which creates a new collector voltage of 0.57 volts.

This analysis may be what your book had in mind, but there's a big
problem, namely that it assumes that Is remains constant. If this
was true, then the Ebers-Moll formula Vbe = kT/q ln (Ic/Is) would
predict about +2.3mV/deg C Vbe tempco. But Is actually has a rather
large positive temperature coefficient, about +18%/deg, so in end we
get a negative Vbe tempco! So while the above analysis is perfectly
fine for a 16.8mV change in Vbe at constant temperature, it cannot be
correct for changing temperatures. But surprisingly, if we painfully
do the analysis the right way, we'll get roughly the same answer.

Here's that analysis. The Ebers-Moll equation can also be expressed
Ic = Is e^(Vbe/VT), which we can use to analyze the circuit with a
8-degree rise, if we know how Is changes with temperature. Using a
spreadsheet to experiment with Is, and assuming a -2.1mV/deg C tempco,
we find that Is must increase by about +18%/deg, as I mentioned above,
and by about 3.77x for an 8 degree change.

Noting that VT increases by about 2.7% (from 25.25 to 25.94mV) for an
8-degree rise (which by itself would lower Ic in the equation), we can
calculate (using Is => 3.77x higher) that Ic increases to 1.91mA for a
fixed Vbe bias. Coincidentally, that's pretty close to 1.94mA for the
"incorrect" method above.

Hmm, neither matches the book's 1.98mA. If we were to also consider
the transistor's Early-voltage effect, the discrepancy would be a bit
larger, and if we were to consider saturation effects, given that Vce
is nearly zero, it would be larger yet. Oh well. :&gt

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

 

[Active8]

On 10 May 2004 07:20:07 -0700, Winfield Hill wrote:

A.J. wrote...

I have been reading an electronics book that shows the following
circuit.

+20V
-----
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin |/
-------| Q1
|\
|
|
_____
___
_

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?

A transistor would not be biased with a fixed temperature-invariant
base voltage in practice, but it is an interesting student problem.

Did your book present the Ebers-Moll equation for you to study? You
can arraign this equation in several ways, but Vbe = kT/q ln (Ic/Is)
is common (dropping a small term that doesn't matter). We often
substitute VT = kT/q, where VT = 25.3mV at 20C, for simplification.

Using this to find the difference of two base voltages, Vbe1 - Vbe2,
call this dVbe, you'll get dVbe = VT ln (Ic2/Is) - VT ln (Ic1/Is),
simplifying to dVbe = VT ln (Ic2/Ic1), which can be converted to
I2 = I1 e^(dVbe / VT). This says if you change dVbe by 8x 2.1mV or
16.8mV, then I2/I1 = e^(16.8/25.3) = 1.94 Which means I2 = 1.943mA
in your problem, which creates a new collector voltage of 0.57 volts.

This analysis may be what your book had in mind, but there's a big
problem, namely that it assumes that Is remains constant. If this
was true, then the Ebers-Moll formula Vbe = kT/q ln (Ic/Is) would
predict about +2.3mV/deg C Vbe tempco. But Is actually has a rather
large positive temperature coefficient, about +18%/deg, so in end we
get a negative Vbe tempco! So while the above analysis is perfectly
fine for a 16.8mV change in Vbe at constant temperature, it cannot be
correct for changing temperatures. But surprisingly, if we painfully
do the analysis the right way, we'll get roughly the same answer.

Here's that analysis. The Ebers-Moll equation can also be expressed
Ic = Is e^(Vbe/VT), which we can use to analyze the circuit with a
8-degree rise, if we know how Is changes with temperature. Using a
spreadsheet to experiment with Is, and assuming a -2.1mV/deg C tempco,
we find that Is must increase by about +18%/deg, as I mentioned above,
and by about 3.77x for an 8 degree change.

Noting that VT increases by about 2.7% (from 25.25 to 25.94mV) for an
8-degree rise (which by itself would lower Ic in the equation), we can
calculate (using Is => 3.77x higher) that Ic increases to 1.91mA for a
fixed Vbe bias. Coincidentally, that's pretty close to 1.94mA for the
"incorrect" method above.

Hmm, neither matches the book's 1.98mA. If we were to also consider
the transistor's Early-voltage effect, the discrepancy would be a bit
larger, and if we were to consider saturation effects, given that Vce
is nearly zero, it would be larger yet. Oh well. :&gt

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

Isn't this the same circuit from AoE where as an excersize you ask
the reader to show that for an 8 deg C rise in temp the transistor
saturates? p. 94?
--
Best Regards,
Mike

 

[Winfield Hill]

Active8 wrote...

On 10 May 2004 07:20:07 -0700, Winfield Hill wrote:

A.J. wrote...

I have been reading an electronics book that shows the following
circuit.

+20V
-----
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin |/
-------| Q1
|\
|
|
_____
___
_

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?

A transistor would not be biased with a fixed temperature-invariant
base voltage in practice, but it is an interesting student problem.

Did your book present the Ebers-Moll equation for you to study? You
can arraign this equation in several ways, but Vbe = kT/q ln (Ic/Is)
is common (dropping a small term that doesn't matter). We often
substitute VT = kT/q, where VT = 25.3mV at 20C, for simplification.

Using this to find the difference of two base voltages, Vbe1 - Vbe2,
call this dVbe, you'll get dVbe = VT ln (Ic2/Is) - VT ln (Ic1/Is),
simplifying to dVbe = VT ln (Ic2/Ic1), which can be converted to
I2 = I1 e^(dVbe / VT). This says if you change dVbe by 8x 2.1mV or
16.8mV, then I2/I1 = e^(16.8/25.3) = 1.94 Which means I2 = 1.943mA
in your problem, which creates a new collector voltage of 0.57 volts.

This analysis may be what your book had in mind, but there's a big
problem, namely that it assumes that Is remains constant. If this
was true, then the Ebers-Moll formula Vbe = kT/q ln (Ic/Is) would
predict about +2.3mV/deg C Vbe tempco. But Is actually has a rather
large positive temperature coefficient, about +18%/deg, so in end we
get a negative Vbe tempco! So while the above analysis is perfectly
fine for a 16.8mV change in Vbe at constant temperature, it cannot be
correct for changing temperatures. But surprisingly, if we painfully
do the analysis the right way, we'll get roughly the same answer.

Here's that analysis. The Ebers-Moll equation can also be expressed
Ic = Is e^(Vbe/VT), which we can use to analyze the circuit with a
8-degree rise, if we know how Is changes with temperature. Using a
spreadsheet to experiment with Is, and assuming a -2.1mV/deg C tempco,
we find that Is must increase by about +18%/deg, as I mentioned above,
and by about 3.77x for an 8 degree change.

Noting that VT increases by about 2.7% (from 25.25 to 25.94mV) for an
8-degree rise (which by itself would lower Ic in the equation), we can
calculate (using Is => 3.77x higher) that Ic increases to 1.91mA for a
fixed Vbe bias. Coincidentally, that's pretty close to 1.94mA for the
"incorrect" method above.

Hmm, neither matches the book's 1.98mA. If we were to also consider
the transistor's Early-voltage effect, the discrepancy would be a bit
larger, and if we were to consider saturation effects, given that Vce
is nearly zero, it would be larger yet. Oh well. :&gt

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

Isn't this the same circuit from AoE where as an excersize you ask
the reader to show that for an 8 deg C rise in temp the transistor
saturates? p. 94?

Aha, you mean Exercise 2.9 in our book, page 84? Yes, perhaps so.
Apparently, according to my calculation above, it almost saturates,
but not quite. But I certainly didn't mean to work out one of the
problems in the book! <sigh>

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

 

[Fred Bloggs]

Winfield Hill wrote:

Active8 wrote...

On 10 May 2004 07:20:07 -0700, Winfield Hill wrote:


A.J. wrote...

I have been reading an electronics book that shows the following
circuit.

+20V
-----
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin |/
-------| Q1
|\
|
|
_____
___
_

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?

A transistor would not be biased with a fixed temperature-invariant
base voltage in practice, but it is an interesting student problem.

Did your book present the Ebers-Moll equation for you to study? You
can arraign this equation in several ways, but Vbe = kT/q ln (Ic/Is)
is common (dropping a small term that doesn't matter). We often
substitute VT = kT/q, where VT = 25.3mV at 20C, for simplification.

Using this to find the difference of two base voltages, Vbe1 - Vbe2,
call this dVbe, you'll get dVbe = VT ln (Ic2/Is) - VT ln (Ic1/Is),
simplifying to dVbe = VT ln (Ic2/Ic1), which can be converted to
I2 = I1 e^(dVbe / VT). This says if you change dVbe by 8x 2.1mV or
16.8mV, then I2/I1 = e^(16.8/25.3) = 1.94 Which means I2 = 1.943mA
in your problem, which creates a new collector voltage of 0.57 volts.

This analysis may be what your book had in mind, but there's a big
problem, namely that it assumes that Is remains constant. If this
was true, then the Ebers-Moll formula Vbe = kT/q ln (Ic/Is) would
predict about +2.3mV/deg C Vbe tempco. But Is actually has a rather
large positive temperature coefficient, about +18%/deg, so in end we
get a negative Vbe tempco! So while the above analysis is perfectly
fine for a 16.8mV change in Vbe at constant temperature, it cannot be
correct for changing temperatures. But surprisingly, if we painfully
do the analysis the right way, we'll get roughly the same answer.

Here's that analysis. The Ebers-Moll equation can also be expressed
Ic = Is e^(Vbe/VT), which we can use to analyze the circuit with a
8-degree rise, if we know how Is changes with temperature. Using a
spreadsheet to experiment with Is, and assuming a -2.1mV/deg C tempco,
we find that Is must increase by about +18%/deg, as I mentioned above,
and by about 3.77x for an 8 degree change.

Noting that VT increases by about 2.7% (from 25.25 to 25.94mV) for an
8-degree rise (which by itself would lower Ic in the equation), we can
calculate (using Is => 3.77x higher) that Ic increases to 1.91mA for a
fixed Vbe bias. Coincidentally, that's pretty close to 1.94mA for the
"incorrect" method above.

Hmm, neither matches the book's 1.98mA. If we were to also consider
the transistor's Early-voltage effect, the discrepancy would be a bit
larger, and if we were to consider saturation effects, given that Vce
is nearly zero, it would be larger yet. Oh well. :&gt

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

Isn't this the same circuit from AoE where as an excersize you ask
the reader to show that for an 8 deg C rise in temp the transistor
saturates? p. 94?


Aha, you mean Exercise 2.9 in our book, page 84? Yes, perhaps so.
Apparently, according to my calculation above, it almost saturates,
but not quite. But I certainly didn't mean to work out one of the
problems in the book! <sigh

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

I read this a decrease in V(Rc) drop of 0.2V - this does work out at a
constant Vbe bias of 0.7V to have obtained Ic=1mA, and Ieo nicely
doubles for each 8oC- so that assuming T=300oC then VT=25.9mV and
elevated VT=26.6mV- so that Ieo@300oC must be 1.83E-15 making the Ie at
308oC 2*1.83E-15*exp(0.7/26.6m)=0.98mA for an exact -0.2V change across
the collector resistor.

 

[Fred Bloggs]

Fred Bloggs wrote:

Winfield Hill wrote:

Active8 wrote...

On 10 May 2004 07:20:07 -0700, Winfield Hill wrote:


A.J. wrote...

I have been reading an electronics book that shows the following
circuit.

+20V
-----
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin |/
-------| Q1
|\ | |
_____ ___ _

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?


A transistor would not be biased with a fixed temperature-invariant
base voltage in practice, but it is an interesting student problem.

Did your book present the Ebers-Moll equation for you to study? You
can arraign this equation in several ways, but Vbe = kT/q ln (Ic/Is)
is common (dropping a small term that doesn't matter). We often
substitute VT = kT/q, where VT = 25.3mV at 20C, for simplification.
Using this to find the difference of two base voltages, Vbe1 - Vbe2,
call this dVbe, you'll get dVbe = VT ln (Ic2/Is) - VT ln (Ic1/Is),
simplifying to dVbe = VT ln (Ic2/Ic1), which can be converted to
I2 = I1 e^(dVbe / VT). This says if you change dVbe by 8x 2.1mV or
16.8mV, then I2/I1 = e^(16.8/25.3) = 1.94 Which means I2 = 1.943mA
in your problem, which creates a new collector voltage of 0.57 volts.
This analysis may be what your book had in mind, but there's a big
problem, namely that it assumes that Is remains constant. If this
was true, then the Ebers-Moll formula Vbe = kT/q ln (Ic/Is) would
predict about +2.3mV/deg C Vbe tempco. But Is actually has a rather
large positive temperature coefficient, about +18%/deg, so in end we
get a negative Vbe tempco! So while the above analysis is perfectly
fine for a 16.8mV change in Vbe at constant temperature, it cannot be
correct for changing temperatures. But surprisingly, if we painfully
do the analysis the right way, we'll get roughly the same answer.

Here's that analysis. The Ebers-Moll equation can also be expressed
Ic = Is e^(Vbe/VT), which we can use to analyze the circuit with a
8-degree rise, if we know how Is changes with temperature. Using a
spreadsheet to experiment with Is, and assuming a -2.1mV/deg C tempco,
we find that Is must increase by about +18%/deg, as I mentioned above,
and by about 3.77x for an 8 degree change.

Noting that VT increases by about 2.7% (from 25.25 to 25.94mV) for an
8-degree rise (which by itself would lower Ic in the equation), we can
calculate (using Is => 3.77x higher) that Ic increases to 1.91mA for a
fixed Vbe bias. Coincidentally, that's pretty close to 1.94mA for the
"incorrect" method above.
Hmm, neither matches the book's 1.98mA. If we were to also consider
the transistor's Early-voltage effect, the discrepancy would be a bit
larger, and if we were to consider saturation effects, given that Vce
is nearly zero, it would be larger yet. Oh well. :&gt

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)


Isn't this the same circuit from AoE where as an excersize you ask
the reader to show that for an 8 deg C rise in temp the transistor
saturates? p. 94?



Aha, you mean Exercise 2.9 in our book, page 84? Yes, perhaps so.
Apparently, according to my calculation above, it almost saturates,
but not quite. But I certainly didn't mean to work out one of the
problems in the book! <sigh

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)


I read this a decrease in V(Rc) drop of 0.2V - this does work out at a
constant Vbe bias of 0.7V to have obtained Ic=1mA, and Ieo nicely
doubles for each 8oC- so that assuming T=300oC then VT=25.9mV and
elevated VT=26.6mV- so that Ieo@300oC must be 1.83E-15 making the Ie at
308oC 2*1.83E-15*exp(0.7/26.6m)=0.98mA for an exact -0.2V change across
the collector resistor.

VOOPS- 300oK

 

[Ric]

Winfield Hill <Winfield_member@newsguy.com> wrote:

A.J. wrote...

I have been reading an electronics book that shows the following
circuit.

+20V
-----
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin |/
-------| Q1
|\
|
|
_____
___
_

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?

A transistor would not be biased with a fixed temperature-invariant
base voltage in practice, but it is an interesting student problem.

Did your book present the Ebers-Moll equation for you to study? You
can arraign this equation in several ways, but Vbe = kT/q ln (Ic/Is)
is common (dropping a small term that doesn't matter). We often
substitute VT = kT/q, where VT = 25.3mV at 20C, for simplification.

Using this to find the difference of two base voltages, Vbe1 - Vbe2,
call this dVbe, you'll get dVbe = VT ln (Ic2/Is) - VT ln (Ic1/Is),
simplifying to dVbe = VT ln (Ic2/Ic1), which can be converted to
I2 = I1 e^(dVbe / VT). This says if you change dVbe by 8x 2.1mV or
16.8mV, then I2/I1 = e^(16.8/25.3) = 1.94 Which means I2 = 1.943mA
in your problem, which creates a new collector voltage of 0.57 volts.

This analysis may be what your book had in mind, but there's a big
problem, namely that it assumes that Is remains constant. If this
was true, then the Ebers-Moll formula Vbe = kT/q ln (Ic/Is) would
predict about +2.3mV/deg C Vbe tempco. But Is actually has a rather
large positive temperature coefficient, about +18%/deg, so in end we
get a negative Vbe tempco! So while the above analysis is perfectly
fine for a 16.8mV change in Vbe at constant temperature, it cannot be
correct for changing temperatures. But surprisingly, if we painfully
do the analysis the right way, we'll get roughly the same answer.

Here's that analysis. The Ebers-Moll equation can also be expressed
Ic = Is e^(Vbe/VT), which we can use to analyze the circuit with a
8-degree rise, if we know how Is changes with temperature. Using a
spreadsheet to experiment with Is, and assuming a -2.1mV/deg C tempco,
we find that Is must increase by about +18%/deg, as I mentioned above,
and by about 3.77x for an 8 degree change.

The dependence of Is on temperature is well-known, so there's no need
to compute a value that gives you the dVbe/dT you want.

Is(T) = Is0*((T/T0)^XTI)*exp(q*EG*(T-T0)/(k*T*T0))
(where Is0 = saturation current measured at T0)

plug that into the diode equation:
Vbe = k*T/q * ln(Ic/Is(T))
= k*T/q * ln(Ic*(T/T0)^(-XTI)*exp(q*EG*(-T+T0)/(k*T*T0))/Is0)

Then differentiate with respect to T:
dVbe/dT = (k*T*ln(Ic*(T/T0)^(-XTI)*exp(-q*EG*(T-T0)/(k*T*T0))/Is0)-XTI*k*T-q*EG)/(T*q)

substitute Vbe back into that lot:
dVbe/dT = 1/T * (Vbe - XTI*Vt - EG)

Job done. XTI = 3 for Silicon, EG=1.11eV or 1.21eV, depending on who you believe.
This gives about -2mV/C for a junction potential of 0.7V at 300K.

Cheers,

--
Rick

 

[Ric]

Fred Bloggs <nospam@nospam.com> wrote:

I read this a decrease in V(Rc) drop of 0.2V - this does work out at a
constant Vbe bias of 0.7V to have obtained Ic=1mA, and Ieo nicely
doubles for each 8oC- so that assuming T=300oC then VT=25.9mV and
elevated VT=26.6mV- so that Ieo@300oC must be 1.83E-15 making the Ie at
308oC 2*1.83E-15*exp(0.7/26.6m)=0.98mA for an exact -0.2V change across
the collector resistor.

Your 0.2V drop is due to rounding of your numbers.
The exact answer by your method boils down to:

2 * 1mA * exp (0.7 * q/k * (300-308)/(300*308))

which is .990mA, giving a drop of 0.1V

--
Rick

 

[Winfield Hill]

Rick wrote...

Fred Bloggs <nospam@nospam.com> wrote:
I read this a decrease in V(Rc) drop of 0.2V - this does work out at
a constant Vbe bias of 0.7V to have obtained Ic=1mA, and Ieo nicely
doubles for each 8oC- so that assuming T=300oC then VT=25.9mV and
elevated VT=26.6mV- so that Ieo@300oC must be 1.83E-15 making the Ie
at 308oC 2*1.83E-15*exp(0.7/26.6m)=0.98mA for an exact -0.2V change
across the collector resistor.

Your 0.2V drop is due to rounding of your numbers.
The exact answer by your method boils down to:

2 * 1mA * exp (0.7 * q/k * (300-308)/(300*308))

which is .990mA, giving a drop of 0.1V

Shheeessh, you guys! Super-accurate wrong answers!

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

 

[Fred Bloggs]

Rick wrote:

Fred Bloggs <nospam@nospam.com> wrote:

I read this a decrease in V(Rc) drop of 0.2V - this does work out at a
constant Vbe bias of 0.7V to have obtained Ic=1mA, and Ieo nicely
doubles for each 8oC- so that assuming T=300oC then VT=25.9mV and
elevated VT=26.6mV- so that Ieo@300oC must be 1.83E-15 making the Ie at
308oC 2*1.83E-15*exp(0.7/26.6m)=0.98mA for an exact -0.2V change across
the collector resistor.


Your 0.2V drop is due to rounding of your numbers.
The exact answer by your method boils down to:

2 * 1mA * exp (0.7 * q/k * (300-308)/(300*308))

which is .990mA, giving a drop of 0.1V

Oookay- well just keep adjusting that Vbe until it comes out right.

 

[Winfield Hill]

Rick wrote...

Winfield Hill wrote:
A.J. wrote...

I have been reading an electronics book that shows the following
circuit.

+20V
-----
|
| +10V
R1 (10k) O
|_______________|
|
|
Vin |/
-------| Q1
|\
|
|
_____
___
_

If The base Bias is set so 1mA flows through R1 it will have 10V
dropped accross. It then goes on to say that an 8 Degree rise in
temperature will cause the voltage accross R1 to drop to 0.2V. I
can't say I really unbderstand how this is working! I know that Vbe
will drop -2.1mV/Degree but how does this affect the current flowing
through the collector? Does someone with a better undersatnding of
the Ebers-Moll equation know exactly what is happening here and why
the 8 degree rise will cause this problem?

A transistor would not be biased with a fixed temperature-invariant
base voltage in practice, but it is an interesting student problem.

Did your book present the Ebers-Moll equation for you to study? You
can arraign this equation in several ways, but Vbe = kT/q ln (Ic/Is)
is common (dropping a small term that doesn't matter). We often
substitute VT = kT/q, where VT = 25.3mV at 20C, for simplification.

Using this to find the difference of two base voltages, Vbe1 - Vbe2,
call this dVbe, you'll get dVbe = VT ln (Ic2/Is) - VT ln (Ic1/Is),
simplifying to dVbe = VT ln (Ic2/Ic1), which can be converted to
I2 = I1 e^(dVbe / VT). This says if you change dVbe by 8x 2.1mV or
16.8mV, then I2/I1 = e^(16.8/25.3) = 1.94 Which means I2 = 1.943mA
in your problem, which creates a new collector voltage of 0.57 volts.

This analysis may be what your book had in mind, but there's a big
problem, namely that it assumes that Is remains constant. If this
was true, then the Ebers-Moll formula Vbe = kT/q ln (Ic/Is) would
predict about +2.3mV/deg C Vbe tempco. But Is actually has a rather
large positive temperature coefficient, about +18%/deg, so in end we
get a negative Vbe tempco! So while the above analysis is perfectly
fine for a 16.8mV change in Vbe at constant temperature, it cannot be
correct for changing temperatures. But surprisingly, if we painfully
do the analysis the right way, we'll get roughly the same answer.

Here's that analysis. The Ebers-Moll equation can also be expressed
Ic = Is e^(Vbe/VT), which we can use to analyze the circuit with a
8-degree rise, if we know how Is changes with temperature. Using a
spreadsheet to experiment with Is, and assuming a -2.1mV/deg C tempco,
we find that Is must increase by about +18%/deg, as I mentioned above,
and by about 3.77x for an 8 degree change.

The dependence of Is on temperature is well-known, so there's no need
to compute a value that gives you the dVbe/dT you want.

Is(T) = Is0*((T/T0)^XTI)*exp(q*EG*(T-T0)/(k*T*T0))
(where Is0 = saturation current measured at T0)

plug that into the diode equation:
Vbe = k*T/q * ln(Ic/Is(T))
= k*T/q * ln(Ic*(T/T0)^(-XTI)*exp(q*EG*(-T+T0)/(k*T*T0))/Is0)

Then differentiate with respect to T:
dVbe/dT =
(k*T*ln(Ic*(T/T0)^(-XTI)*exp(-q*EG*(T-T0)/(k*T*T0))/Is0)-XTI*k*T-q*EG)/(T*q)

substitute Vbe back into that lot:
dVbe/dT = 1/T * (Vbe - XTI*Vt - EG)

Job done. XTI = 3 for Silicon, EG=1.11eV or 1.21eV, depending on who
you believe. This gives about -2mV/C for a junction potential of 0.7V
at 300K.

Excellent, Rick, that nicely fills out the subject.

Thanks,
- Win

(email: use hill_at_rowland-dot-org for now)

 

[Rick]

Winfield Hill <Winfield_member@newsguy.com> wrote:

Rick wrote...
The dependence of Is on temperature is well-known, so there's no need
to compute a value that gives you the dVbe/dT you want.

Is(T) = Is0*((T/T0)^XTI)*exp(q*EG*(T-T0)/(k*T*T0))
(where Is0 = saturation current measured at T0)

plug that into the diode equation:
Vbe = k*T/q * ln(Ic/Is(T))
= k*T/q * ln(Ic*(T/T0)^(-XTI)*exp(q*EG*(-T+T0)/(k*T*T0))/Is0)

Then differentiate with respect to T:
dVbe/dT =
(k*T*ln(Ic*(T/T0)^(-XTI)*exp(-q*EG*(T-T0)/(k*T*T0))/Is0)-XTI*k*T-q*EG)/(T*q)

substitute Vbe back into that lot:
dVbe/dT = 1/T * (Vbe - XTI*Vt - EG)

Job done. XTI = 3 for Silicon, EG=1.11eV or 1.21eV, depending on who
you believe. This gives about -2mV/C for a junction potential of 0.7V
at 300K.

Excellent, Rick, that nicely fills out the subject.

You're welcome, Win.
Note that the equation shows that if you make Vbe large enough, say by
pumping 100,000A through a diode-connected transistor, then the tempco
goes positive. Spice confirms my equation, but I don't know if reality
would agree - I'm having trouble finding a transistor beefy enough!

--
Rick