There is a formula in Malvino's book (page 210 of the 6th edition) which states that the input base resistance of a transistor is beta times the emitter resistance.

To quote: "When seen from the base side of the transistor, the emitter resistance RE appears to be 100 times larger."(assuming beta = 100).

I don't get how it arrives to this conclusion. It implies that base and emitter voltage are the same. So no VBE.

Could someone please explain more?

P.S. I would provide a screenshot of the above but the book is in Greek and also in printed format. (Edited to include the relevant quote from the 8th edition.)

enter image description here

  • \$\begingroup\$ Base and emitter voltages can't be the same, so what implies they are? And could it have something to do with Ie being roughly beta*Ib? \$\endgroup\$
    – Justme
    Commented May 6, 2022 at 17:46
  • \$\begingroup\$ @Justme assuming IB = VB/RIN and IE = VE/RE it follows that VE*RIN = beta * RE * VB. If RIN = beta * RE then VE = VB. \$\endgroup\$
    – kosgian90
    Commented May 6, 2022 at 17:51
  • 1
    \$\begingroup\$ @kosgian90 In editing your question, did I correctly identify the part you are looking at? \$\endgroup\$
    – jonk
    Commented May 6, 2022 at 19:02
  • 1
    \$\begingroup\$ @jonk yes you did. \$\endgroup\$
    – kosgian90
    Commented May 6, 2022 at 19:06
  • 1
    \$\begingroup\$ @kosgian90 Have you already learned about and feel comfortable with the idea that a resistor divider can be replaced by its Thevenin equivalent that includes a new ideal voltage source with a Thevenin series resistance? (So, replacing two resistors and a voltage source with one new voltage source and one new resistor, thereby simplifying things a bit?) Are you comfortable with that? \$\endgroup\$
    – jonk
    Commented May 6, 2022 at 19:44

1 Answer 1


I understand it like this: If you were to measure the base's input resistance, you would apply a certain base voltage and measure the base current. Then you would calculate the resistance Rin=Ub/Ib. As an example I just assume arbitrary values: Rin=1V/1mA=1000 Ohm.

But as the base current in practice causes 100 times the amount of emitter current, the actual current through the resistor is 100 as high, too. So the resistance is actually Re=1V/100mA = 10 Ohm = Rin/100.

The point is that you only have a fraction of the actual current available to do the measurement.


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