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Our professor in college was making us understand the stiff voltage-divider bias concept of a BJT.

schematic

simulate this circuit – Schematic created using CircuitLab

(Ignore the resistor and voltage source values) Here he says, "The voltage-divider can be called/assumed a "stiff" divider if the current through R1 and R2 is quite large compared to the current going into the base (which makes sense ofc). Such an assumption can be made if the resistance seen looking into the base is much larger compared to (R1+R2), which ensures very less current flowing into the base". How do we measure this resistance looking into the base at bias conditions? I can understand and calculate small signal impedances no sweat, but this confused me idk why. Or for that matter, how would we calculate the impedance looking into any of the terminals at DC? Any kind of hint will help. Thanks!

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    \$\begingroup\$ V1 is too low (if this is a real circuit); it should be at least a few volts (e.g., 10 V). Q1 acts as an emitter follower that copies VR2 on R4. Its input (the base) consumes (1 + beta) times smaller current than if R4 was directly connected to the output of the voltage divider. So, "looking" into the base, the voltage divider sees resistance (1 + beta).R4. \$\endgroup\$ Dec 10 '19 at 18:28
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It’s all about Ohm’s law. The voltage at the base is

$$V_b = R_e I_e + V_{be} = I_b (\beta + 1) R_e + V_{be}.$$

Hence

$$R_b = \frac{V_b}{I_b} = (\beta + 1) R_e + \frac{V_{be}}{I_c / (\beta + 1)} = (\beta + 1)(R_e + R_{be}).$$

Since \$\beta\$ is large, the resistance looking into the base will be large as well.

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  • \$\begingroup\$ Besides Ohm's law, we should see here such important concepts as negative feedback, emitter follower, emitter degeneration... \$\endgroup\$ Dec 10 '19 at 18:24
  • \$\begingroup\$ @user 110971...According to your expression we have re=Vbe/Ic. Why you have chosen a small symbol (re) for a pure DC resistance? This is confusing because this symbol is used very often for the inverse transconductance (re=1/gm). \$\endgroup\$
    – LvW
    Dec 11 '19 at 9:33
  • \$\begingroup\$ @LvW You are right. I was thinking about AC signals for some reason. \$\endgroup\$
    – user110971
    Dec 11 '19 at 14:57
  • \$\begingroup\$ @user110971...Putting some numbers into your formulas (beta=100, Ic=1mA) we arrive - even for moderate values - at at a DC input resistance of app. Rb=(100...200)kohms. Normally, for R1 and R2 we choose some lower kOhm values and it becomes clear that in many (if not for most) cases we can neglect this large resistance Rb in parallel to R2 and - with it - the DC base current. It is even not necessary to know the actual Ib value. A good indication for voltage control Ic=f(Vbe). \$\endgroup\$
    – LvW
    Dec 11 '19 at 15:51
  • \$\begingroup\$ @LvW Indeed. However the question was how to actually calculate it, so that’s what I’ve answered. \$\endgroup\$
    – user110971
    Dec 11 '19 at 15:56

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