1
\$\begingroup\$

I can't figure out how the DC current through the diodes is calculated in eq. 8.22 on this PDF from University of North Carolina.

$$ I_D = \frac{V_{CC}/2-V_{BE}}{R_f+R_2} $$

\$I_D\$ should be equal to the current through the resistor \$R_2\$ minus the current going through the base of the transistor \$Q_1\$. I am able to obtain this equation only supposing \$I_B = 0\$, which is not true, specially because they are matching the diode to the transistor to avoid thermal problems.

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ I am able to obtain this equation only supposing IB=0 So they're assuming that the transistors have a large \$\beta\$? That is quite common. \$\endgroup\$ – Bimpelrekkie Feb 14 at 10:12
  • \$\begingroup\$ which is not true, specially because they are matching the diode to the transistor to avoid thermal problems. How does this prevent us from assuming that \$\beta\$ is high enough to assume that \$I_B\$ = 0. The diodes and transistors aren't matched but thermally coupled and that makes the DC biasing current through the transistors more constant over temperature. This works because the forward voltage of a diode and the \$V_{BE}\$ of a transistor have a similar temperature coefficient. This is about the voltages. How is the base current relevant to that? \$\endgroup\$ – Bimpelrekkie Feb 14 at 10:20
  • 1
    \$\begingroup\$ @Bimpelrekkie The base current is relevant because it affects the voltage drop over \$R_2\$ \$\endgroup\$ – user253751 Feb 14 at 10:53
  • \$\begingroup\$ @user253751 That is true however in practice in a proper design the base current is highly unpredictable this is because \$\beta\$ is also unpredictable. Look up the datasheet of any transistor and note how the range over which \$\beta\$ varies is quite large. Values like min/typ/max = 50/100/200 are not uncommon. So there can be a factor 4 difference in base current. You don't want that to affect the behavior of your circuit. So that's why designers choose the current through R2 to be much larger than the larges base current (smallest \$\beta\$). \$\endgroup\$ – Bimpelrekkie Feb 14 at 11:08
  • 1
    \$\begingroup\$ @Bimpelrekkie We can't selectively ignore numbers because they are unpredictable. "We can't predict who will get cancer, therefore we can ignore cancer for the purpose of mortality statistics" \$\endgroup\$ – user253751 Feb 14 at 11:23
2
\$\begingroup\$

Yes, that equation assumes Ib=0. We all know that can't be true IRL, but if the transistors aren't in saturation, it's likely that Ib is far lower than the bias currents used in the resistor/diode network. So the estimate isn't a perfect solution in the real world, but it's close enough...and engineering is never about perfection, it's all about being close enough.

\$\endgroup\$
  • \$\begingroup\$ Thank you. I agree with you, only I would have stated before in the document that \$\beta\$ was large. I did a simulation of the circuit with \$\beta = 40\$ and \$I_B\$ was 3 mA, while \$I_{R2}\$ was 16 mA. Also I would have used this symbol \$\approx\$ instead of equal. \$\endgroup\$ – stenio Feb 14 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.