Transistor Characteristics

Yes, Just look at the Spice model ;-) Or measure your own... just turn the equation around and force a current and measure a voltage... account for Rs as well... measure quickly to avoid heating. ...Jim Thompson

n~=1 for a diode-connected transistor (typically about 1.004 for a

There's also some series resistance (maybe < 1 ohm for a chunky small signal transistor).

I'm sure if JT felt like it he could expound on this at length..

Best regards, Spehro Pefhany

The tempco of I_s is large and negative, so V_BE isn't PTAT but drop around 2.1 mV/K at ordinary currents, rising to around -3 mV/K at very low currents (picoamps to nanoamps).

Grey and Meyer has the equation for I_s but not much in the way of physics. IIRC the second edition of Sze has more. (The third edition, edited by some other random guy, is not nearly as good on the fundamentals.)

Cheers

Phil Hobbs

The transport version (Ebers-Moll, version 1) is

I_cc(V_BE,T) = I_s(T)*[e^(q*V_BE/(k*T)-1] I_ec(V_BC,T) = I_s(T)*[e^(q*V_BC/(k*T)-1] I_s(T) = I_s(T_nom)*(T/T_nom)^3*e^[(-E_g/k)*(1/T-1/T_nom)] I_c = I_cc + (-1/alpha_R) *I_ec I_b = (1/alpha_F-1)*I_cc + (1/alpha_R-1)*I_ec I_e = (-1/alpha_F)*I_cc + I_ec

Something like that. I think EM1 is usually written without the emission coefficient that is taken as 1.

If you want, Tim, I do have a copy of an excellent book I can send or give you on the topic. It is entirely on the subject of BJTs, nothing else, and half of it covers methods, procedures, and caveats for extracting various parameter values, as well. The author is a local (to us) resident expert on BJTs and used to work at Tektronix in that capacity. I could put you in contact, if you like.

The value of E_g represents the effective energy gap (in eV) and is by default usually taken to be 1.11eV. T and T_nom are in Kelvin, of course. You already know what q and k are, I'm sure.

The usual equation you mention for Ie (do not confuse it with my use of I_e, above) replaces kT/q with Vt. Expanding that back out and reordering gives:

V(T) = (k*T/(q*n)) * ln( 1 + Ie/Iss )

Assuming that Iss were fixed and not a function of T, the derivative is then trivially:

d V(T) = (k/(q*n)) * ln( 1 + Ie/Iss ) dT

which is clearly a positive trend and very nearly +2mV/K for modest Ic... but __positive__. Which of course isn't right and doesn't correspond to experience. The reason is that Iss isn't fixed, but is itself a function of T -- per the I_s(T) I mentioned above.

If you take the derivative of that, as well, and factor it back in you will find that for an example case where Ie=10uA, Iss=1E-15 near T=20C, the figure comes out to about

-2.07mV/K. Which is more what you'd experience.

Not sure what you are looking for, though.

Jon

Only for sufficiently large currents. As I recently had to relearn, a diode connected transistor is actually the world's simplest feedback amp, and its GBW is roughly proportional to collector current.

Cheers

Phil Hobbs

Sorry, tempco of I_s is large and _positive_, so V_BE can be less at higher temperature for the same I_C.

Cheers

Phil Hobbs

I think at this point that the book would be too much information -- but don't be surprised if I ask for information at some point in the future.

I think I'm set. I just realized that this thing is sitting in close proximity to a microprocessor, that the temperature dependencies are very predictable, and that there's going to be at least one temperature sensor on the board -- which means that I don't even have to temperature compensate the circuit: I can just provide some equations and let the software guy worry!

Hmm... HK Gummel & HC Poon say that the base current is non-ideal at low currents, but the collector current pretty much follows the ideal diode equation, so the total current should be close to ideal provided Ic/Ib is large, no?

Best regards, Spehro Pefhany

At DC, it works as long as the beta holds up, that's true. However, the closed-loop bandwidth goes into the tank at low current, so at any given frequency, it stops working for some I_C.

Cheers

Phil Hobbs

That's not the way on-board temperature sensors work. They take a voltage measurement at I, and some multiple, say 10*I, then do the math, or read some back issues of Pease ;-) ...Jim Thompson

That sounds like a recipy for a dissaster. I'd go for a dual diode or dual transistor (preferably single die) and measure one leg with a known signal or incorporate some calibration scheme which switches between a known signal and the sigal to be measured.

Nico Coesel has said it already, but the temperature sensor on the board isn't sensing the temperature of your junction, which is going to be warmer than ambient by whatever power you are dissipating in the junction times the thermal resistance to the ambient environment around your temperature sensor.

A minor point, but sometimes crucial. Monolithic dual transistors allow you to measure the temperature of a point a lot closer to the junction of interest.

Hey, does the (n) go as 1 +1/beta (where beta is the current gain) for diode connected transistors?

(My first post via eternal-sept, since google groups seems to have $hit the bed.)

George H.

What's "sufficiently large"? It seems to still be pretty good at 10 nA.

George H.

I was making a sub-Poissonian current sink for a low noise front end amplifier, and my string of diode-connected BFT25As stopped working properly about 10-15 nA. They were still okay at DC, but got noisy and weird at 1 MHz. I wound up using just the CB junctions.

Cheers

Phil Hobbs

I'm not sure what you meant. What I meant was that if there's a temperature sensor in close proximity to the transistor I use for the log amp, then the software is just a hop, skip and a jump (or a look-up, add and multiply) away from knowing the characteristics of the amplifier at that particular moment.

I'm also considering a dual transistor chip, using one with a fixed current to establish zero, and another for the log amp. That would still leave me with a gain error as the temperature changes, however.

The circuit that I found calls out one transistor for the log function, one to correct for offset, and a thermistor to correct for gain changes with temperature. It adds up.

If that thermistor is going to work, why not just plumb it to the processor, and let the processor do the math?

OK. There are some MAT-something-or-other-number that are multiple matched transistors (Analog Devices ??). ...Jim Thompson

Yup. It's Anal-og Devices. They have duals & quads. Art