how to get "h11e" of Ge transistor

[yeeshkul]

I found in my archaic university notes that we can approximately get B-E resistance of a Si transistor (aka H11e) like:

H11e =  Hfe/(40 x Icbias)

Is there anything like that for Ge transistors?

[R.G.]

My (archaic!) textbooks called it hie - hybrid parameter for input impedance with emitter grounded.

Yes, germanium devices have  a base input resistance. I'll have to look up how it's calculated. I ... think ... it's the same way the Shockley resistance is done for silicon: R = 25mV/Ic

I don't recall the equation of hie = hfe/40*Ic, but I could have missed that lecture.

[yeeshkul]

Thank you again R.G.

[R.G.]

Well, I don't know that I told you anything other than I think it exists too and don't know exactly what the numbers are.

[yeeshkul]

actually: 0.025/Ic = 1/(40*Ic) so here we go
hie = hfe /(40*Ic)

it looks like it is valid for Ge transistors too.
I was just trying to sort out the input impedance of my own FF and needed that hie because the rest are just resistors . Just to make my brain a bit bussy ...

[Sir H C]

Remember that the 25mV is for room temperature.

[R.G.]

actually: 0.025/Ic = 1/(40*Ic) so here we go Smiley
hie = hfe /(40*Ic)
it looks like it is valid for Ge transistors too.

... duh... I'm having a bad math day! 

Remember that the 25mV is for room temperature.

It is indeed. So if you're designing a Fuzz Face for use outdoors in Antarctica or Phoenix AZ, you ought to be sure to correct it.   

Actually, the 25mv/Ic is not specifically identical to hie, now that I think about it. The 25mv/Ic value is called the Shockley resistance, after the guy who was part of the team that invented the transistor. He came up with this as an empirical approximation. I'll have to go dig for how the Shockley resistance and hie are related.

[Sir H C]

The 25mV is Vt.  KT/Q IIRC (it has been quite a while since I have had to know this).

[yeeshkul]

It is indeed. So if you're designing a Fuzz Face for use outdoors in Antarctica or Phoenix AZ, you ought to be sure to correct it.   

then i will have to add a little sharp bits on the bottom of it so it doesn't slip on the icy floor

I'll have to go dig for how the Shockley resistance and hie are related.

that would be absolutely great! I thought that 250mV is a voltage drop on Ge transistor B-E/B-C diode, while the drop is about 0.6V on Si. But on the other hand i haven't seen these things for many years by now so i tend to say plenty of rubbish 

[R.G.]

The 25mV is Vt.  KT/Q IIRC

You recall correctly that Vt = kT/Q = 25mV.

But something nags at me that this isn't the same kT/Q. I thought "kT/Q" when I first saw this, but I believe that it's purely empirically derived and without a good explanation of why it is so. I did a bit of a search for the info and came up with this, from http://webtools.delmarlearning.com/sample_chapters/04.pdf

Let us review the concept of dynamic resistance. In Chapter 1,we discussed the detailed diode model that shows how forward-biased current increases steeply with changes in forward-biased voltage.We stated that the dynamic resistance of a diode junction can be calculated by taking a small change in voltage and dividing it by the change in current caused by the change in voltage (r = Δv/Δi). However, these changes in voltage and current are difficult to obtain. Dr. Shockley developed an approximation formula that simplifies solving for dynamic resistance. Shockley's formula states that the dynamic resistance of a diode junction is equal to 25 mV divided by the
forward-biased current through the diode junction (r = 25 mV/I). So, once the bias current is known, the dynamic resistance can be easily calculated.
In a simple diode, there is only one current flowing; however, in a transistor, there are three currents: emitter current, base current, and collector current. The formula for calculating the dynamic resistance of the forward-biased base/emitter junction is r'e = 25 mV/IE. The emitter bias current (IE) is used to calculate the dynamic resistance (r'e) of the base/emitter junction because the emitter current is the current flowing through the base/emitter junction. The fact that only a small percentage of emitter current exits the base has no effect on the dynamic resistance of the junction.

To calculate the input impedance, we now have two parallel paths for signal current flow: one through the base resistance (rb) and another through the forward-biased  base/emitter junction, which has dynamic resistance in the emitter leg. The resistance in the emitter leg will be seen as beta times larger from the perspective of the base.

It might actually turn out that kT/Q is hiding down there somewhere in the math-noise, but as far as I know, this was made up by Shockley to fit observed data, not derived.

[yeeshkul]

my textbook says this:

uT(temperature voltage) = m*k*T/q

where m is a constant which follows technology of making PN junction and it is 1<m<2

for temperature ~ 300K -> uT~20-50mV
we choose uT 25-30mV for the rought approach

but this is ment to be for Si transistors/diodes

[mac]

uT(temperature voltage) = m*k*T/q

The omnipresent term K*T in every eq. derived from quantum statistical mechanics... I guess that this resistance is the partial derivative of the partition function with respect to some variable.
As far as I remember, this was 20yrs ago and big chances that I was sleeping at the back of the classroom, the calculation of this type of systems is like distributing balls into holes, taking into account potential barriers, etc.
What I was trying to figure out what the m is.

mac