bjt current, hfe, and R.G.'s transistor test

[gaussmarkov]

To do one using real transistors you want a lot of emitter degeneration.  Maybe a volt.  Make the resistors match very well and you will get a very accurate mirror. 

o.k.  if you will bear with me for a moment (remember--i just figured out how to use a zener diode as a voltage regulator ), i need to check on a few items:

• emitter degeneration = 1V -- does this refer to the voltage across an emitter resistor?
• mirror -- current mirror, right?  would you mind pointing to a schem? i have looked at the ones on Wikipedia (bjt, Widlar, and Wilson).  only the Widlar shows an emitter resistor, although i can see that you could add emitter resistors to both xstrs, or give them a common resistor to ground.
• matching resistors -- which ones?  i have some guesses, but there's not much point in displaying additional ignorance 

"no good deed goes unpunished" -- i hope you don't mind these follow-up questions.  i promise to run with any answers.

[Sir H C]

The simplist current mirror is two transistors sharing base and emitter terminals so the base-emitter voltages across both are identical.  Then one has the collector tied to the common base node.  This is the input terminal, the other device is the output terminal.  So assume two NPN transistors here, you have a resistor tied to V+ and the device with the collector tied to the base (this is called a "Diode Connected Device" by the way).  The resistor will pull up the base voltage until it is high enough to allow the collector (and the two, much lower, base) currents to keep the circuit happy.  There is only one steady operating point for this.  Now the second transistor, since the base-emitter is at the same voltage as the other device will have the same output current.  Well within reason because of output impedance as seen Early-er (ha ha ha). 

The Wilson, Cascode, and other more complex current mirrors are used to increase the output impedance of the devices and make the current matching better.  The use of degeneration resistors is great as a way to make the matching of the transistors less important.  You can use transistors that don't match nearly as well with a good amount of degen and still get good current matching.  The difference between current sources and mirrors is that the source is just using the mirror with a resistor to get a well defined current.  The degeneration resistors are the ones to match, one off each emitter on the bottom devices so that these will control accuracy of the current.

Wikipedea seems to be getting all esoteric on this when there are easy thought experiments that can make it pretty clear.

[mac]

and, because you were able to report your dmm's hfe measurement, i went back again to the manual for mine and found another section with the info.  guess what?  same as yours:  10uA at the base and a Vce equal to 2.8V.  maybe it's a standard specification?

thanks for following up, gm

More on DDM. A guy I know at an electronic store unpacked two equal cheap ddms to measure a 2n3904 hfe. Guess what? yes, very different readings.
I wonder how inaccurate my germaniums hfe values are.

More, Vce=Vcb+Vbe. But not to my DDM.


mac

[gaussmarkov]

More on DDM. A guy I know at an electronic store unpacked two equal cheap ddms to measure a 2n3904 hfe. Guess what? yes, very different readings.
I wonder how inaccurate my germaniums hfe values are.

interesting.  and i'm surprised.  do you think it might have been temperature differences?  i'm convinced that you have to keep your fingers off the transistor, Si or Ge.

More, Vce=Vcb+Vbe. But not to my DDM.

are you sure about this?    i remember  Ie = Ic + Ib, but not the voltage relationship.  unless you have a particular circuit in mind?

[Sir H C]

Yes Vce = Vcb + Vbe.  Pretty much if you do it as geometry, you get the line segment CE equals the addition of the other two line segments as the B point gets nulled out.

[gaussmarkov]

Yes Vce = Vcb + Vbe.  Pretty much if you do it as geometry, you get the line segment CE equals the addition of the other two line segments as the B point gets nulled out.

o.k.  thanks.  i guess i have never looked at Vcb, so it was never an issue.    i guess it's "obvious" when you think about it in terms of just using a multimeter to measure voltages.  how could it be otherwise?     funny how one can let things get more complicated than they really are when learning something.

[mac]

interesting.  and i'm surprised.  do you think it might have been temperature differences?  i'm convinced that you have to keep your fingers off the transistor, Si or Ge.

No finger heating or dead batteries, just "chinese" differences 

I guess the resistor of the voltimeter is too low, so when measuring voltage drops, the current flowing through the part of the circuit under test is of the order of the current flowing through the DDM. That is my explanation of why I can not get vce = vcb + vbe with my cheap chinese DDM. Also, I do not expect to have good readings with my RG hfe circuit.


mac