My questions pertains to the \$ \beta\$ value of a transistor. On page 162 of this book, it says that \$ \beta\$ increases as temperature of the transistor increases. But it doesn't explain why. Can anyone explain that bit more thoroughly?
Transistor Behavior – How Temperature Variations Affect Current
currenttemperaturetransistors
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2nd Edit! Modified my answer about semi-conductors based on jk's answer below, read the history if you want to see the wrong bits I modified!
Everything gets weird within certain limits. I mean, sure, the resistance improves in conductors but it increases in semi-conductors, and that change effects how the IC works. Remember that the way that transistors work on the basis that you can modify their resistance, and if the temperature drops so low that you can no longer decrease their resistance, you've got an issue! Imagine that suddenly your semi-conductor essentially became a resistor... how do you control it? It no longer behaves the same way! Now I'm a bit confused at where you're getting the -25°C, as the industrial/military spec should put it at -40°C for the minimum operating temp.
But for the space question, I can answer that as I work in a space lab! In general you have three thermal concerns in space:
1) In space, you only radiate heat. Radiation is a terrible way to get rid of heat. In the atmosphere, you conduct heat into the air around you which makes cooling a lot easier. So in space, you have to put big heatsinks on to get the heat into larger radiative surfaces.
2) If you have a component which doesn't generate heat, then space is happy to let you get really friggin' cold! In general, what you do is you have active heating elements to keep components which don't generate more heat than they radiate but have thermal limits.
3) Heat swings are common because you will exit and re-enter the sun's rays. Thus you need to have active thermal management where you have a big heatsink which can radiate heat when it's hot, and a heater for when it's not.
You can also get extended temperature range devices which go lower and higher, but there's pretty much always a limit. Some of them are for where the cold temperature will crack the die because the metal will shrink more than the plastic (or vice versa) which is why they list limits for storage as well!
The limit is mostly in materials. You also tend to get space-rated chips made out of ceramic for the packaging, which can also raise or lower the thermal limits.
Anyway, I hope that explains it for you. I can try and answer any other questions, but I'll admit the physics of low-temperature semiconductors is not my forte!
1st Edit:
Here's a link to a wikipedia entry about the idea that at lower temperatures there are fewer electrons which are excited enough to generate a current flow through a semiconductor lattice. This should give you a good idea of why the resistance becomes higher, and why 0 Kelvin would have never been an option.
If you look at your top circuit you will notice that there are shorts where the input, emitter and output capacitors were - this is the first step to doing an AC analysis. The caps are assumed to pass AC without hinderance so thay are shorted. Resistors are presumed to to attenuate so these are left in.
Whether a resistor is connected to Vsupply or ground is irrelevant - The power supply is assumed to be one big capacitor and so this is also shorted to ground (note the ground symbols on R1, Rc and Rout1).
You then use the Hfe of the transistor (current gain of the transistor) to calculate the signal-gain as it goes from base to collector.
So, on the base is a fraction of Vin and this fraction is initially determined by the attenuation produced by Rs, R1 and R2. BUT importantly it is mainly dictated by the B-E of the transistor - it can be assumed to be forward-conducting and sometimes it is easier to assume it is a short circuit hence the current into the base is purely Vin / Rs.
This current is amplified by Hfe (current gain) and produces a voltage on the collector that is dictated by the parallel combination of Rc, Rout1 and Rout2.
Thus you can determine the approximate signal gain for AC.
Me, I use a simulator - it's quicker especially if you are trying different values out AND takes all the transistor details into account and can give a fairly accurate frequency response too.
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Best Answer
If you look at the Ebers -Moll model, one can see that:
$$I_C = \dfrac{\beta_F}{\beta_F +1}I_E$$
which can be rewritten as
$$\beta_F +1 = \beta_F \dfrac{I_E}{I_C}$$
$$1+\dfrac{1}{\beta_F}=\dfrac{I_E}{I_C}$$
$$ \beta_F = \dfrac{1}{\frac{I_E}{I_C}-1}$$
Where
$$I_E = I_{ES}\{e^{\frac{V_{BE}}{V_T}}-1\}$$
where $$V_T=\dfrac{k_BT}{q}$$ is the thermal voltage.
Here it should be clear that the temperature cause a change in the thermal voltage which leads to a change in beta.
The equation is basically a statement about the distribution of energy per unit of charge, as the temperature increases, the exponential in the emitter current equation decreases, which causes the ratio of collector current to emitter current to increase, and since that ratio is in the denominator of the equation for beta, it has an inverse relationship, which causes beta to increase. So as temperature increases, beta increases.