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I'm reading a text on BJT common collector amplifier biasing.

Text says:

The voltage divider (R1 and R2) is used to give the input signal (after passing through the capacitor) a positive dc level or operating point (known as the quiescent point).

The actual sizes of R2 and R1 should be such that their parallel resistance is less than or equal to one-tenth the dc (quiescent) input resistance at the base (this prevents the voltage divider’s output voltage from lowering under loading conditions):

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What does that mean? I don't understand why R2 and R1 parallel resistance should be less than or equal to one-tenth the dc? What does loading mean here?


The voltage divider R1 and R2, disconnected from the transistor, would define a voltage (5V) which you want to use as the base voltage.

Loading that voltage divider by connecting another resistor in parallel with R2 will reduce that voltage.

In the days when resistors were 10% tolerance, that voltage would only be accurate to within typically 10% anyway, so a 10% error due to the additional load was (and still usually is) considered acceptable.

So, what additional resistance, connected across R2, will change the voltage by 10%? (Thevenin circuit, Ohm's law). (You do not add this load explicitly, this is the load added by the transistor's base.)

Does the input resistance of the transistor circuit equal or exceed that value? (Re, and the tranistor's current gain)

  • \$\begingroup\$ I also thought that's what the question was asking but then I realized that the equations were comparing Rin to R1||R2, which is a different issue. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 7 '16 at 22:45
  • \$\begingroup\$ additional load to a biasing part?? why would one do that? \$\endgroup\$ – user16307 Jan 8 '16 at 7:17
  • \$\begingroup\$ @user16307 : edited. clearer? \$\endgroup\$ – Brian Drummond Jan 8 '16 at 11:32
  • \$\begingroup\$ but why base's dynamic resistance would change if it was already fixed? can u give an ecample? \$\endgroup\$ – user16307 Jan 8 '16 at 13:37
  • \$\begingroup\$ I cannot understand that last question, sorry. Do you understand that the base connection draws current from the voltage divider, and can be (approximately) modelled by a resistance? \$\endgroup\$ – Brian Drummond Jan 8 '16 at 14:12

I don't understand why R2 and R1 parallel resistance should be less than or equal to one-tenth the dc?

The base always takes "some" current in order to produce an amplified current through the collector. Because the base takes current it in effect places a resistor in parallel with R2 and, the effect of this is seen to lower R2 a little bit. Therefore, if the base takes 0.01 mA then one rule of thumb suggests that the current flowing into R2 should be 0.1 mA.

This prevents the set-point voltage at the base being too much lower than the theoretical set-point determined soley by R1 and R2.

But of course, this is just a rule of thumb and some people use a different rule of thumb that says that "the parallel resistance of R1 and R2 should be 10x lower than the effective resistance looking into the base".

This slightly different rule of thumb has a slightly different result for the nominal values of R1 and R2 but, that doesn't matter because, there can be a wide range of R1 and R2 values that suit a particular amplifier circuit. It's not an exact piece of physics or science.

Given that the first rule of thumb says R1 and R2 should consume 0.1 mA and given the supply voltage is 10V, the series resistance of R1 + R2 should be 100k ohm i.e. 50k each (and not 100k as per your circuit). That of course is not a big-deal.

The 2nd rule of thumb would work like this; the base is at approximately 5V and consumes 0.01 mA therefore its resistance is effectively 500k ohms therefore, the combined parallel resistance of R1 and R2 should be no greater than 50kohm and of course R1 = R2 = 100k satisfies that but, given the "no greater than" clause, so does R1 = R2 = 50k.

  • \$\begingroup\$ why would R2 be loaded too much if we set up the right Q point? and why does it matter? \$\endgroup\$ – user16307 Jan 7 '16 at 22:28
  • \$\begingroup\$ If the Q point is correctly setup then no worries but before that can happen you need to estimate values for R1 and R2. If that estimation does not consider current into the base then the Q point may be significantly different to what is desired. \$\endgroup\$ – Andy aka Jan 7 '16 at 22:40
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    \$\begingroup\$ Of course you could just choose 500 ohm resistors for R1 and R2 and not even bother thinking about the trivial amount of base current but that would be 10mA wasted in current and, on a battery circuit that would be crap. \$\endgroup\$ – Andy aka Jan 7 '16 at 22:44
  • \$\begingroup\$ if u read my question the author says "under loading conditions" what does he mean by loading? \$\endgroup\$ – user16307 Jan 9 '16 at 14:15
  • \$\begingroup\$ The loading condition IS the loading by the base current i.e. base current is seen as an effective resistor in parallel with R2. \$\endgroup\$ – Andy aka Jan 9 '16 at 16:33

Without the transistor in place R1||R2 is the input impedance of the circuit (since either supply is "AC ground" and both resistors lead to them). Once we put the transistor in place we have a voltage divider made of both the parallel impedance and the transistor impedance. In order to supply most of the power to the transistor we need its impedance to be much greater than the impedance from the resistors.

  • \$\begingroup\$ im sorry but i still dont understand what u mean \$\endgroup\$ – user16307 Jan 7 '16 at 22:16
  • \$\begingroup\$ Which part are you having trouble with? \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 7 '16 at 22:17
  • \$\begingroup\$ i need such logic: "if a happens then b would happen" \$\endgroup\$ – user16307 Jan 7 '16 at 22:18
  • \$\begingroup\$ Okay then: "If you have a voltage divider with a large Ra and a small Rb then you will have a small output voltage." \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 7 '16 at 22:19
  • \$\begingroup\$ i mean lets say R1||R2 is less than 2, what would happen and why, \$\endgroup\$ – user16307 Jan 7 '16 at 22:19

For selecting the impedance niveau of this divider we have two conflicting requirements:

  • The resistors should be as large as possible in order (a) to provide a signal input resistance as large as possible and (b) to allow a small DC power cnsumption.

  • The resistors should be as small as possible because, in this case, the base DC voltage would be as "stiff" as possible ( a kind of "good" voltage source). This is desired because only in this case, the negative feedback provided by the emitter resistor can work efficiently (base voltage fixed and emitter voltage regulated).

  • The mentioned factor of "10" (rule of thumb) is trade-off between both requirements. In most cases, this factors is selected in the range 8...15.


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