What does this NPN BJT circuit represent, is it a switch or just a PN diode, and what should this plot look like when we sweep the value of Vbe, and why?

• Is this homework? What have you tried to far? Any simulation or calculation? Oct 16 '20 at 10:38
• This is the fourth homework question you've posted, another straight from the textbook. Please note that this site is not a free homework-answering service or on-line technical encyclopaedia, copied out to you on demand. People will help you take the next step if your question shows you've already done as much as you possibly could - which yours doesn't, I'm afraid. Please edit your question and greatly improve it. Show your own work and own findings in considerable detail, real detail. Please don't post any more unattempted homework questions, it looks very lazy and they get closed. Oct 16 '20 at 19:13
• I’m voting to close this question because it's unattempted homework. Oct 16 '20 at 19:14
• Its not a homework and it is not from a textbook, it's a lab experimental circuit. I am just trying to figure out how it works. Oct 17 '20 at 9:27
• It is not a homework question, I am just asking about the behavior of the circuit. Oct 17 '20 at 9:29

1. It is simply a test circuit for you to test and observe. It is not a "complete" circuit in the sense you expect - it is not a "switch."
2. The point of the exercise is for you to carry out the experiment and develop an understanding of how to assemble the circuit and test setup, and then to gain an understanding of how the base current of a transistor behaves. It does you no good for someone to describe it to you - you need to carry out the experiment.

If you have difficulty carrying out the experiment, then you should explain the problem you are having and ask what is going wrong.

Once you carry out the experiment, you could post your completed setup and the results and ask if you have done it correctly and if the results are reasonable.

Since there seems to be some dispute about what $$\V_{bias}\$$ plotted against $$\I_B\$$ would look like, I whipped up a DC sweep generator and generated the plot.

I used an Arduino Nano I had laying around handy to make the $$\V_{bias}\$$ generator. The Nano puts out a 10kHz PWM signal that is filtered down to DC. By varying the duty cycle of the PWM signal, I can change the output voltage. The PWM generator has 1023 steps.

This is the circuit:

The red circled area generates the $$\V_{bias}\$$ from the PWM signal.

R2 takes the place of $$\R_b\$$ in the drawing in the question. I used a 2N3904 transistor because I had one close to hand.

The Nano can only measure voltage, but $$\I_b = I_{R_2} = (V_1-V_2)/R_2\$$.

I wrote a simple Python program to drive the Nano, collect the data, and crunch the numbers. It dumps the finished numbers into CSV file, and I used LibreOffice to make the plot itself.

This is the plot of $$\I_B\$$ against $$\V_{bias}\$$:

As you can see, the statement "$$\I_b\$$ vs $$\V_{bias}\$$ does not make much sense because practically, it is the IV curve of the resistor $$\R_b\$$" is incorrect.

The plot of $$\I_B\$$ against $$\V_{bias}\$$ clearly shows the typical response of a diode junction.

For comparison, here's the plot of $$\V_{B}\$$ against $$\I_B\$$:

That shows the diode response of the junction as well.

• JRE, Perfectly prepared and conducted computerized experiment! For comparison, I would only use one potentiometer for V1 and a digital oscilloscope in XY mode:) But what I see is exactly what I said. Indeed, the first IV curve is slightly shifted to right because of Vbe but it is negligible; Rb dominates in the curve. The second IV curve is what OP needs to realize about the base-emitter junction properties. Oct 17 '20 at 22:10
• There is no "shift" in the first curve. It clearly shows the influence of the diode junction on the current.
– JRE
Oct 17 '20 at 22:13
• JRE, The situation is the following: You apply a varying input voltage Vin (Vbias) through a forward-biased diode (base-emitter junction) to the resistor Rb. The diode "eats" 0.5 V from Vin so the IV curve begins changing from this point. What you see after that, is the IV curve of Rb (Vbe variation is negligible; "the influence of the diode junction on the current" also). There is no point of this picture; the next is important for OP. Oct 17 '20 at 23:06
• The first picture is the one the OP was instructed to generate.
– JRE
Oct 17 '20 at 23:31

I think the real purpose of this setup is to measure and draw the input IV curve of the transistor - the function Vbe = f(Ib), which is actually the IV curve of a diode (p-n junction, as you noted).

"Ib vs Vbias" does not make much sense because practically, it is the IV curve of the resistor Rb.

"Sweep the value of Vbe" is not so correct since, in this arrangement, we sweep the base current Ib = Vbias/Rb as an input quantity and should measure Vbe as an output quantity.

As you probably understand, I have neglected Vbe. OK, here is my philosophy about the graphical representation of the transistor input circuit.

For the purposes of this technique (we can name it "superimposed IV curves"), we have to reduce the input circuit of three elements (Vbias, Rb and Vbe) to a network of only two elements connected to each other - a source and load. Each of them will have an IV curve and since the voltage across them and current through them are the same, we can impose their IV curves on the same coordinate system. The intersection (operating) point will represent the voltage and current at this moment.

Usually, we combine Vbias and Rb into the first element (a real voltage source); the base-emitter junction will serve as the second element (a non-linear load).

But to impose the IV curves, we must first obtain them. The most common idea for this is: we change (move, rotate) the one curve to obtain the other. We can name the variable curve "scanning" and the other - "scanned". To do that, we have to change something of the "scanning element".

In our case, to obtain the base-emitter IV curve, we can change either Vbias or Rb. As a result, the IV curve of the Vbias-Rb composed element will either translate horizontally or rotate around the coordinate origin… and the intersection point will picture the famous diode exponent.

If we want to obtain the Vbias-Rb IV curve (aka "load line" or, more correctly, "generator line"), we should change something of the base-emitter junction (e.g., its threshold voltage)... but here it is impossible since it is fixed. It is possible, for example, when investigating the transistor output curve.

If we want to obtain the Rb IV curve, we should group Vbias and Vbe as the first element and Rb as the second one. Then, we have to vary Vbias to scan the Rb IV curve (line).

Let's finally consider again the "Ib vs Vbias" IV curve. If Vbias is the voltage of the input voltage source and Ib is the current through it, it should be the IV curve of an ideal voltage source (without internal resistance)... but it is not interesting for us here.

• What happens when $V_{Bias}$ is below 0.7 V? What happens to $I_B$ when $V_{Bias}$ is in the range 0-0.7V? What happens as $V_{Bias}$ continues rising? I think you will find that the plot is not the IV plot of $R_b$
– JRE
Oct 16 '20 at 12:50
• @Circuit fantasist (your last line): I think, the relation must be Ib=(Vbias-Vbe)/Rb. And we can identify Vbe=Vbias-Ib*Rb,
– LvW
Oct 16 '20 at 13:00
• @JRE and LvW, Exactly... I just neglected Vbe... See my edit above. Oct 16 '20 at 14:27
• I suppose, the questioner means "sweep Vbias" (use triangular or sawtooth) ...and create the corresponding plot. I can see absolutely no cause or sense to "neglect" the influence of the main component (test object)
– LvW
Oct 16 '20 at 15:05
• @LvW, I neglected Vbe since OP had to plot "Ib vs Vbias" IV curve... But this "bias" is also confusing and misleading... Oct 16 '20 at 19:36