Need help with the process for finding input resistance of a transistor ??
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Posted by jalbers@bsu.edu on September 19, 2008, 3:30 pm
Please log in for more thread options I have been trying to follow along with a books calculation for the input resistance of a transistor which they define as Rin = delta (Vbe) / delta (Ib) or Rin would be 1 / (slope of the diode curve) or 1 / (derivitive of the diode curve) . Doing the math and assuming that KT/Q = 26 mv and assuming that (Is) the reverse saturation current is negligible, I get the desired Rin = 26 mv / Ib. My question is why is Rin = delta (Vbe) / delta (Ib) and not just Vbe / Ib . My some how flawed reasoning is that if you look at ohms law E = IR, then R = I/E and therfore Rin should be Vbe / Ib . Why is Rin equat to the change in current divided by the change in voltage? Any help would be greatly appreciated. Thanks | |||||||||||||
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Posted by Kevin Aylward on September 19, 2008, 5:06 pm
Please log in for more thread options jalbers@bsu.edu wrote: It may be better to think of this as: rin = hie = rbb' + (1 + hfe).re where rbb' is the base resistace, usually of the order 10-500 ohms, and re=1/(40.IC) = 1/gm > My question is why is Rin = delta (Vbe) / delta (Ib) and not just
> Vbe / Ib . My some how flawed reasoning is that if you look at ohms > law E = IR, then R = I/E and therfore Rin should be Vbe / Ib . Why > is Rin equat to the change in current divided by the change in > voltage? > > Any help would be greatly appreciated. Thanks The fundamental reason is: del_y = f'(x).del_x if del_y, and del_x are small, as a general math result, from differential calculus. For small signals, i.e., when an input signal only changes a small amout about a fixed bias point, what is the resulting change in the output, about a fixed bias. rin, is a small signal resistance. It is not equal to VINDC/INDC. Since the inputs and outputs that are wanted are the small signal changes, than one needs the small signal resistance, not the large signal resistance. If one actually wanted large signal values, than one would use such values throughout. Kevin Aylward kevin@kevinaylward.co.uk www.kevinaylward.co.uk | |||||||||||||
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Posted by John Popelish on September 19, 2008, 6:21 pm
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jalbers@bsu.edu wrote: > My question is why is Rin = delta (Vbe) / delta (Ib) and not just
> Vbe / Ib . My some how flawed reasoning is that if you look at ohms > law E = IR, then R = I/E and therfore Rin should be Vbe / Ib . Why > is Rin equat to the change in current divided by the change in > voltage? Because the resistance is not ohmic. Ohmic resistances hold a constant ratio of voltage to current, regardless of the magnitude of voltage or current. Diode junctions do not. So resistance can be defined for such devices only incrementally (over tiny ranges of voltage and current). -- Regards, John Popelish | |||||||||||||
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Posted by Kevin Aylward on September 20, 2008, 6:54 am
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John Popelish wrote: > jalbers@bsu.edu wrote:
> >> My question is why is Rin = delta (Vbe) / delta (Ib) and not just
>> Vbe / Ib . My some how flawed reasoning is that if you look at ohms >> law E = IR, then R = I/E and therfore Rin should be Vbe / Ib . Why >> is Rin equat to the change in current divided by the change in >> voltage? >
> Because the resistance is not ohmic. Interesting point. The resistance does not follow ohm's law, so in that sense is not ohmic. However, it does have characteristic similar to ohmic resistors but not similar inductors and capacitors. That is, the current and voltage are independent of time, i.e not +/-j For DC conditions, the actual value of V/I may be important, especially for values of Vdc of around 0.5 to 1V. >Ohmic resistances hold
> a constant ratio of voltage to current, regardless of the > magnitude of voltage or current. Diode junctions do not. > So resistance can be defined for such devices only > incrementally (over tiny ranges of voltage and current). Well... technically one can actually define the large signal resistance as V/I. So that: R = Vd/I = (Vd/Io).exp(-Vd/Vt) And using the results of http://www.kevinaylward.co.uk/ee/widlarlambert/widlarlambert.html, for the Lambert W() function One can obtain the inverse relation; Vd = -Vt.W(-R.Io/Vt) I will leave it as an exercise for the reader to deduce formulas expressing R in terms of I, and I in terms of R:-) Kevin Aylward www.blonddee.co.uk www.anasoft.co.uk | |||||||||||||
> input resistance of a transistor which they define as Rin = delta
> (Vbe) / delta (Ib) or Rin would be 1 / (slope of the diode curve) or
> 1 / (derivitive of the diode curve) . Doing the math and assuming
> that KT/Q = 26 mv and assuming that (Is) the reverse saturation
> current is negligible, I get the desired Rin = 26 mv / Ib.
>