# Transistor: Getting “incorrect” graph for input voltage to output voltage

I tried drawing the graph of $$\v_o\$$ vs $$\v_i\$$ for a bipolar junction transistor. I used the below setup (common emitter:)

By design of how the base is a small part compared to others, $$\I_b << I_c\$$ and so I take $$\I_c \approx I_e\$$.

I’m taking this transistor to be equivalent to two diodes, one forward biased and another backward, and matching the graphs pointwise against $$\I_c\$$ to get graph between $$\v_o\$$ and $$\v_i\$$:

I’m getting a graph different from what’s given in my textbook.

Mine:

Textbook & circuit given:

I’m not sure where I’m going wrong.

• Very simple: A BJT does not behave like two diodes back to back. Jan 9, 2021 at 18:23
• @ThePhoton Oh oops. Would it be accurate if I said thats because the depletion layers are much more intricately related than just two detached diodes? Jan 9, 2021 at 18:30
• Yes, that's one way to put it. Really it's that the two junctions are close enough together that injected minority carriers from the b-e junction can diffuse to the c-b junction and get carried over to the collector. Jan 9, 2021 at 18:32
• The reverse diode effect only occurs in saturation with Vce difference due to the ratio of Ic/Ib vs Vf of each diode. So @ThePhoton is only half right. Normally in saturation Ic/Ib is 10% of hFE then Rce effects arise with Ic. Yet Vce(sat) is defined for some fixed ratio like 10,20, or 50 for super-Beta types >>500 Jan 9, 2021 at 20:05
• The missing information is Vcc and Rc as Ic is a Vbe-voltage controlled current sink and @ThePhoton is also half wrong Jan 9, 2021 at 20:10

Below is a simplified diagram of the PNP BJT in active mode, with the relative thicknesses exaggerated in order to make it more readable. This image is taken from Jacob Millman's "Microelectronics: Digital and Analog Circuits and Systems" circa 1979 I think:

The middle N layer is actually a lot thinner and it is also the more highly doped of the three layers, as well. In active mode, the base-collector (middle and right regions) are reverse-biased and the base-emitter (middle and left regions) are forward-biased.

As a forward-biased charge-carrying current, $$\I_{pE}\$$, transitions from the left to the middle region, drawn into the middle region by the forward-biased state of affairs of barrier $$\J_E\$$, only a small part of it is collected by the base lead. These charges are moving through a very, very thin layer and most of them readily find themselves crossing the junction barrier, $$\J_C\$$, which despite being reverse-biased, more lightly doped, and a much larger region, is even still more negatively charged than the base they passed through. So they are now simply accelerated towards the collector.

If you try and imagine putting two diodes together, appropriately for demonstration purposes, I think you can see that the thin N layer of the PNP BJT has now been replaced by rather thicker N layers in each diode as well as the bonding points and a length of wire.

The whole idea of the BJT's mechanism of active-mode operation has been completely undone when you try to mimic a BJT by "gluing" two PN diodes together.

## Schottky Barrier Diode

The Schottky barrier diode is a unipolar (uses only an N type semiconductor material but no P type) diode where a metal contact point is applied. Despite the lack of a PN junction, it still functions as a diode, but with a lower forward voltage requirement. But any thoughts of trying use a pair of these to make a BJT should be still more obviously flawed!

There's a funny story I read, many years ago, that's tertiarily related to the idea of making Schottky barrier diodes. It's preserved here and worth a moment's reading. The story may help you well imagine some of the difficulties facing early attempts to make point-contact Schottky diodes.

The modern Schottky barrier diode avoids much of the "fun" illustrated in that story by often (not always) using metalized silicides, which bond more easily than what unfolds in the above story.

If you thought about this intuitively you could tell your graph is backwards because as current through the transistor increases, of course its voltage will drop! A transistor is a "Transfer resistor" , it acts like a resistance that can change based on input. When a resistance goes down, current increases and voltage decreases.

So even though you drew the graph, if you stop to try and analyze it to see if it matches overall behavoir you would have realized your graph is backwards.

• Im not quite sure what you mean. Derive a graph, realize its opposite of what's happening, assume this graph and that graph are just the same thing? How does that work? Jan 10, 2021 at 2:59
• I cant quite tell what your bottom graph axis is, is it V(I) or does it say V(E)? Jan 10, 2021 at 3:10
• First one's $I_e$ vs $V_i$ and second $I_c$ vs $V_o$ Jan 10, 2021 at 3:13