0
\$\begingroup\$

I tried to solve this problem and I could find the answer but i try to look for more attention to the circuit and finding some doubts. Let me try to explain myself.

enter image description here

the question ask about the current through the \$R_L\$ resistor.

I compute this way:

Current though \$R_1\$ and \$R_2\$ resistors:

\$\require{cancel} \dfrac{0-5}{\cancel{10k}} = \dfrac{5-V_E}{\cancel{10k}} \Rightarrow V_E = 10V\$

Current thought \$R_L\$ is:

\$I_L = \dfrac{V_E-0}{200}\Rightarrow \boxed{I_L = 50mA}\$

Ok. Too easy.

My some questions are:

What the operational mode NPN is?

If it would be on saturation region \$V_{ce} = 0V\$, how could it possible if \$V_b\$ (output of ampop) was 0V, so the BJT would be in reversed bias mode due to \$V_{be}\$. And if the \$V_{bc}\$ would be reversed mode the transistor was on Cutt-off region?

If it would be cut-off region where the emitter current had cames from? Because

\$ I_E -\dfrac{5-10}{10k} = 20mA\Rightarrow I_E = 20.5mA\$

I hope my English was a minimum clear.

\$\endgroup\$
9
  • \$\begingroup\$ Why do you think the op amp output is 0 V? \$\endgroup\$
    – Hearth
    Commented Apr 23, 2020 at 18:16
  • \$\begingroup\$ @Hearth, because V- = V+ (virtually zero ground) or do i need consider more than that? \$\endgroup\$
    – miguel747
    Commented Apr 23, 2020 at 18:24
  • 2
    \$\begingroup\$ Yes, V- = V+, but that doesn't mean that the output is zero. The output will be whatever it has to be to make V- = V+. \$\endgroup\$
    – Hearth
    Commented Apr 23, 2020 at 18:28
  • \$\begingroup\$ RIght ^^^ The opamp will drive it's output to make it's inputs the same voltage. In the example above, it will drive the transistor into saturation, putting 10V at point "B", and thus 5V at point "A" (because R1 = R2, and they are a voltage-divider). \$\endgroup\$
    – Kyle B
    Commented Apr 23, 2020 at 18:33
  • \$\begingroup\$ Makes sense, but every books (e.g Sedra) says that the ideal ampop with the same input signal the output will be zero. \$\endgroup\$
    – miguel747
    Commented Apr 23, 2020 at 18:34

1 Answer 1

Reset to default
0
\$\begingroup\$

You need to know what the opamp is doing first. It looks like the opamp is performing two tasks:

  • Translating the voltage V1 from 5v to 10v. (by comparating inputs and saturating)
  • It's also buffering V1.

All this assuming that the VCC of the opamp is 10v. And it seems like the opamp is saturating to VCC. This means the output of the opamp is at 10v volts (assuming full swing opamp).

Back to your question, about the transistor. It's not switching as you think, it's not completely saturated nor in cutoff, it's on the "Active Region", the current is coming from the transistor (that's the point of the circuit).

transistor regions

Source of image: https://www.electronics-tutorials.ws/transistor/tran_4.html

To fully understand your circuit: Ok, why would I make this circuit? What's the purpose?

Three reasons:

  • Voltage
  • Current
  • Control

The voltage of your apparent signal is 5V. But the circuit is translating it to 10v.

Opamps usually can't supply too much current (usually 20mA max). But your load consumes 50mA approx. That's why you need the BJT.

You may ask, why not connect the load directly to the 10v rail? The answer is "Control": With this circuit you can control your load with the 5v signal, it may be turned on or off. Yes, this is a driver. And the good thing about it is that you can control 10v with 5v.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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