hey,

in the art of electronics there are two formulations for how bjt's react to temperature changes, they are...

(1) Vbe falls at 2mV/degree C and
(2) Ic grows at 9%/ degree C

the text uses an example (page 85) where it suggests that a 20 degree C temp increase leads to a 25% increase in collector current. Using the first formulation its easy to calculate that Vbe decreases by 40mV and therefore Ve increases by 40mV (because Vb is fixed). The 40mV increase of Ve increases Ie and therefore Ic by 25% from 1mA to 1.25mA.

Can anybody tell me how the second formulation (Ic grows at 9%/ degree C) is consistent with these calculations.

 

(2) Ic grows at 9%/ degree C

On what page did you find this?

 

oops...

(2) Ic grows at 9%/ degree C

that bit isnt explicitly mentioned in the book itself, however, it is mentioned in the student manual that accompanies the book. I should have made that clear.

Its true because you can use Ebers Moll to show they are equivalent statements. Ebers Moll says VTln1.09 = 2mV which in turn is saying 1.09 is the ratio of the collector currents before and after a 2mV change in Vbe i.e if we change Vbe by 2mV then the collector current will be 1.09 times what it was, a 9% increase

The thing that was bothering me was if the temperature climbed by 20 degree C then by the second formulation the transistor (in a common emitter amp) would saturate.

 

should I have put my reply to you here ?

 

Yes. You are doing ok.

The 'Ic grows at 9%/ degree C' is not in the Art of Elecronics, but in the student manual?

 

Well, the first condition is very well known. But you have a misconception: Vbe actually drops while the temperature rises, so the correct rate of change is -2mV/oC.
Doing our math we verify, as you said, the conclusion the author reaches at page 85 of the mentioned book, about the 25% increase of the collector current.

Now, about the second formula, I haven't met it anywhere before. But if we wanted to express the increase of the Ic with a percentage, that would be close to 1.1%. You can verify that by substituting the arithmetic conclusions we came to earlier, in the formula Ic20=Ic0 (1+x)^20. Indeed, a change of 9% increases the collector current to a whole lot more than expected.

Could you please revise what you have read?

 

My book (2nd edition) says
1) An 8°C rise on a grounded emitter amplifier with fixed base voltage biasing will cause it to saturate (p.84, exercise 2.9).
2) A 20°C rise in fig. 2.38 will cause the collector current to rise by nearly 25%.
Are you confusing the two statements, or do you have a different edition?

I verified (1), and can show you how if you want. I didn't try to do (2) mathematically, but a simulation showed a increase of Ic of 17% from 25°C to 50°C.

 

Well, the first condition is very well known. But you have a misconception: Vbe actually drops while the temperature rises, so the correct rate of change is -2mV/oC.

He said that

Vbe falls at 2mV/degree C

 

I do understand that Vbe diminishes with rising temperature

the second formula is just a restatement of the first, by using the Ebers Moll equation you can prove they are equivalent.

 

first of all thanks for the input.

The circuit I was asking about is the fig 2.38 that you mention. The problem ive got is that there are two ways to see how Ic is affected by temp. First, we could use

Vbe falls at 2mV/degree C

this would tell us that Vbe would fall by 40mV for a 20 degree C increase in temp. We could then say that Ve grows by 40mV because Vb is fixed by the divider. Ohms law would then tell us that the emitter current and therefore the collector current must increase from 1mA to 1.25mA, a 25% increase.

However...

if we use the other formulation...

Ic grows at 9%/ degree C

then we should see 20*9% = 180% increase = 2.8mA

the transistor would saturate well before it was able to supply 2.8mA

cheers

 

And maybe even earlier, depending on how you interpret the 9% increase for every oC. If you think of it as +9% of a fixed Vbe (@ 25\(^{\tiny{o_C}}\) for example), then it is as you say. If, however, you interpret that as, for every increase of 1\(^{\tiny{o_C}}\) the new Vbe is 1.09 times the previous Vbe, then for a 20\(^{\tiny{o_C}}\) increase, Vbe would rise 5.604 times, at 3.36 volts, which of course is farfetched. This formula is trully problematic...

 

It almost seems like it was refering to a particular case.

Like, in this case, Ic rose at 9%/degC.

I dont have the student manual its just a guess.

 

first of all thanks for the input.

The circuit I was asking about is the fig 2.38 that you mention. The problem ive got is that there are two ways to see how Ic is affected by temp. First, we could use

Vbe falls at 2mV/degree C

this would tell us that Vbe would fall by 40mV for a 20 degree C increase in temp. We could then say that Ve grows by 40mV because Vb is fixed by the divider. Ohms law would then tell us that the emitter current and therefore the collector current must increase from 1mA to 1.25mA, a 25% increase.

However...

if we use the other formulation...

Ic grows at 9%/ degree C

then we should see 20*9% = 180% increase = 2.8mA

the transistor would saturate well before it was able to supply 2.8mA

cheers

The 9%/°C only happens if the emitter degeneration resistor (175Ω in this case) is zero. That's the whole point of the resistor. It provides negative feedback. The increase in emitter current increases the emitter voltage, which tends to cut off the transistor. Not that it actually cuts it off, but the rate of current increase vs temperature is reduced.

 

I had wondered about how to apply the formula too

either

1mA + (180%*1mA) =2.8mA

or

((1.09)^20)* 1mA =5.6mA

I think the latter is the the way to apply it. Either way produces an answer that is wrong. Anyway, Ron H and I have come to same conclusion that the two temperature formulations only strictly apply to the transistor itself and not the circuit in which the transistor is placed.

 

After thinking about this problem all yesterday I have come to the same conclusion as you Ron H, that strictly speaking the temperature formulations apply to the transistor itself and not to the overall circuit.

As you mentioned, the emitter resistor applies feedback by reducing Vbe when Ic grows, this of course would reduce Ic.

I am still left with the feeling that the formulation...

Ic grows at 9%/ degree C

should still work when applied to the transistor in an emitter degenerated circuit as long as the feedback scenario is taken account of.

Thanks again for the help