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This is class B Push-Pull circuit for an amplifier.

enter image description here

The question says:

Calculate the maximum Icmax current of the resistance RL if you knew Rl = 8Ω

The book solved it this way:

Vcc - (Rl * Icmax) = 0

Then by solving the equation, we find Icmax = 3A

The thing is, how they used Kirchoff's law if they didn't count VCE's voltage of the transistor?

When you make a loop, you're supposed to get Vcc - (Rl * Icmax) - VCE = 0, right?

Then how the book here didn't consider it nor included it into the equation?

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    \$\begingroup\$ They approximated \$V_{CE}\approx 0\$. If \$V_{CC}\$ is 24 V and \$V_{CE}\$ is 0.2 V, this is often "close enough". \$\endgroup\$
    – The Photon
    Commented May 11, 2020 at 16:52
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    \$\begingroup\$ @ThePhoton How did you know VCE is 0.2V? And are we supposed to neglect VCE in every equation or circuit or this is just one-off? \$\endgroup\$
    – TechnoKnight
    Commented May 11, 2020 at 16:57
  • \$\begingroup\$ because the question is about the "maximum Icmax current" and 0.2 V is a typical guess for what Vce will be if you drive a BJT to its maximum current. \$\endgroup\$
    – The Photon
    Commented May 11, 2020 at 17:00
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    \$\begingroup\$ @ThePhoton They didn't teach us this. \$\endgroup\$
    – TechnoKnight
    Commented May 11, 2020 at 17:03
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    \$\begingroup\$ Vce=0.2V can only be achieved if the driver stage operates at a higher voltage. It doesn't so the limiting factor is Ib * R6 + Vbe. The book is over-simplifying in not mentioning either Vce or Vbe and Ib. \$\endgroup\$
    – user16324
    Commented May 11, 2020 at 17:18

2 Answers 2

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Then how the book here didn't consider it nor included it into the equation?

Because it's trivial in many cases. After all if the supply is +/- 12 volts then the load current peak is still going to be approximately 12 volts / 8 ohms = 1.5 amps compared to 11.3 volts / 8 ohms = 1.41 amps.

If the supply was +/- 50 volts then it's even less of an error.

Calculations for BJT amplifiers are always a tad inaccurate due to the simplifications we make.

BTW - I chose a VCE volt drop of 0.7 volts because each transistor in the push-pull arrangement is an emitter follower.

But, given that the input source isn't shown in the OP's diagram then who is to say that its amplitude doesn't slightly exceed the power rails by a couple of volts.

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    \$\begingroup\$ Are we supposed to neglect VCE in every equation or calculation no mater the circuit or the voltage though? \$\endgroup\$
    – TechnoKnight
    Commented May 11, 2020 at 16:58
  • \$\begingroup\$ @TechnoKnight no, you use judgement. BJT and MOSFET circuits require judgement on what is important to include and what is unimportant so that it can be ignored. If you want to include stuff that another person might think as trivial then there's nothing wrong with that of course. \$\endgroup\$
    – Andy aka
    Commented May 11, 2020 at 16:59
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    \$\begingroup\$ wait, what? I can't depend on something so inaccurate in an important exam, you know. \$\endgroup\$
    – TechnoKnight
    Commented May 11, 2020 at 17:02
  • \$\begingroup\$ Are you still asking that same question given my late comment amendment? \$\endgroup\$
    – Andy aka
    Commented May 11, 2020 at 17:06
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    \$\begingroup\$ this is an exam for electronic/electricity. I can't depend on a judgement or something inaccurate like that to decide what voltage matters and what voltage doesn't matter. I either don't include it every time or I include it every time. \$\endgroup\$
    – TechnoKnight
    Commented May 11, 2020 at 17:10
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The answer is a quick approximation. Your schematic shows that the supply voltage is ±24 V from batteries. The battery voltage won't be that precise and will vary significantly with state of charge. Similarly, a mains power supply will usually have ripple on the DC side at twice the mains frequency so an average value is used.

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