I encountered a text which says that the leakage current Ico (or called Icbo) adds up to the signal current hence causes thermal runaway because it causes temperature to increase and it also increases with temperature. Below is the related illustration circuit(I guess we are not in control of Vbe anymore):

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As a remedy the text shows these alternatives:

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What I understand from above alternatives is that it seems if we are in total control of Vbe(by using a voltage source at BE terminals), then the Ico will not affect the biasing circuit.

If my understanding is correct, how does the circuit on the right(with voltage divider) is equivalent to the circuit on the left(with voltage source)? How is this source transformation done in electric circuit theory? Somehow the author equalizes the voltage source to a voltage divider but how?


enter image description here

  • \$\begingroup\$ Thevenin equivalent circuit is the answer \$\endgroup\$ – G36 Jun 11 '18 at 18:07
  • \$\begingroup\$ What can we say about the voltage divider resistor values? For Thevenin to produce a battery equivalent should they be very low value or high? \$\endgroup\$ – pnatk Jun 11 '18 at 18:10

A voltage divider produces a voltage at the junction between the two resistors and the value of that voltage is easy to calculate, given the source voltage and resistor values. Now, if you connect something else to the junction between the two resistors (such as the base of your BJT) then you need to worry about how much current will be drawn by the thing you have connected. As long as the current drawn by the transistor's base is much less than the current that would flow through the resistors without the base connection, then the resistor divider is a reasonable approximation of an ideal voltage source.

So, in this case you just need to select resistor values for the voltage divider so that

  • the resulting voltage is equivalent to the battery in the left-hand circuit, and
  • the current through the resistors is much greater than the base current.
  • \$\begingroup\$ Can you a bit clarify the reason for the condition "current through the resistors is much greater than the base current" ? \$\endgroup\$ – pnatk Jun 11 '18 at 18:22
  • \$\begingroup\$ @panicattack If the base current is significant compared to the current through the resistors, then it will significantly affect the voltage at the base. That means the divider is not acting like a voltage source. Try a sample calculation with the base current equal to half the divider current and with the base current 0.1X the divider current. \$\endgroup\$ – John D Jun 11 '18 at 18:25
  • \$\begingroup\$ @JohnD How can this be shown with an equivalent circuit or algebraically? \$\endgroup\$ – pnatk Jun 11 '18 at 18:29
  • \$\begingroup\$ Please see my edit. I modelled BE junction as a resistor and called it Zbe. And I failed the Zbe from zero to 100k. As Zbe increased current through Zbe decreases and the Voltage divider approaches to 5V. Can this be used to demonstrate the relation between the amount of current drawn and the voltage stability? \$\endgroup\$ – pnatk Jun 11 '18 at 18:42
  • \$\begingroup\$ So should we say that the R2 must be much smaller(10 times at least) than the input impedance of the BE junction? \$\endgroup\$ – pnatk Jun 11 '18 at 18:44

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