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So, I am trying to understand transistors.

I start with the very basics, the part which says that an NPN transistor kinda look like two diodes joined at the hip (at the anode, in fact)

I build the "floating collector" circuit below, which - according to what I've read - should be the functional equivalent of a diode in series with a resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

I then try to calculate by hand the intensity going through the loop, starting at the emitter and walking my way back towards the source.

  • Voltage at emitter is 0 (grounded)
  • Voltage drop in a silicon diode is 0.7V (from textbook)
  • Therefore, voltage at base has to be 0.7V
  • Voltage left of R1 is 10V (source)
  • Therefore voltage drop across R1 is (10-0.7) = 9.3V
  • Therefore (Ohm's law), intensity across R1 is 9.3/1000 = 9.3 milliAmps
  • Current has nowhere to go but through emitter and back to source
  • Therefore (KCL) : intensity is everywhere 9.3 mA

To verify that I've got this right, I build the circuit in LTSpice, and lo and behold, the darn thing disagrees :-).

When I run an LTSpice simulation, it tells me that:

  • Ib = 9.1581 mA
  • Vb = 0.8418 V

Retracing my reasoning backwards, this basically means that my assumption that a voltage drop across a Si diode is 0.7v is wrong.

Researching this further, I find that the V/I characteristic of a Si diode has a "round knee" around 0.7v and that therefore, the rule that says: "Si diode always drops 0.7v when forwards biased" is in fact an approximation and that the characteristic curve is in fact some sort of exponential.

Ok, fine.

But now, I want to be able to derive the actual value of Vb myself, by hand, and I am stuck: in the above reasoning, I was relying on a fixed drop across the diode to derive Vb and go from there to Ib.

Now that Vbe and Ib are tied up in some sort of equation, I'm not sure how to get to Vb from the knowledge that Ve=0.

I've essentially got two unknowns (Vb, Ib) and only one equation (the diode characteristic curve) ... how do I compute Vb ?

Am I missing something glaringly obvious ?

Help appreciated.

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    \$\begingroup\$ The second "equation" is the value of the resistor, which gives you a second V-I curve. The solution is where the two curves intersect. \$\endgroup\$ – Dave Tweed Oct 30 '13 at 13:55
  • \$\begingroup\$ You might find this answer helpful: electronics.stackexchange.com/questions/82508/… \$\endgroup\$ – The Photon Oct 30 '13 at 16:08
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You can use the Shockley equation, which is another idealization, but a better one than a fixed diode drop. To apply it, you have to know, for the given diode, all the parameters:

$$I=I_\mathrm{S}\left( e^{V_\mathrm{D}/(n V_\mathrm{T})}-1 \right)$$

I would suggest to replace the transistor with an actual diode model, from which you can pull out the parameters and calculate by hand.

In this answer to a question, I was working with the formula, in relation to the behavior of a diode model in CircuitLab. See the question's "appendix" at the bottom.

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OK, I think I have it figured out.

@Dave Tweed : thank you very much for your comment, and in fact, doh! : indeed, I do have two equations.

@Kaz : yup, Shockley's equation is what I needed.

From an intuition perspective, here's what I was missing: there are two "constraints", one imposed on the left by voltage source going through resistor, and one on the right imposed by GND going through Shockley's equation.

These constraints "meet" or "clash" at the base of the transistor, where they form a non-linear system of two equations in two unknowns which can easily be solved numerically once you have all the pieces.

It's just the node method, but with some non-linear bits thrown in. Cool !

Although Mathematica can't seem to be able pull off a symbolic solution (not a surprise, there's an ax+b == Exp[cx+d] type headache in there), it can actually pull off a numerical solution effortlessly:

Vs = 10;          (* Voltage at source                                          *)
R1 = 1000;        (* 1k resistor                                                *)
Is = 1/10^12;     (* Typical value, as per Wikipedia on Shockley's equation     *)
n = 3/2;          (* Typical value, something between 1 and 2, as per Widipedia *)           
VT = 2585/100000; (* Typical value = 25.85 mV at 300 Kelvin, as per Wikipedia   *)
N[
    Solve[
        {
            Element[{VN2, Intensity}, Reals],
            Vs - VN2 == R1*Intensity,
            Intensity == Is*(Exp[VN2/(n*VT)] - 1)
        },
        {Intensity, VN2}
    ]
]

$$ \{\{\text{Intensity}\to 0.00911078,\text{VN2}\to 0.889216\}\} $$

And, yes, although I kind of threw random constants at Shockley's equation instead of the 2N2222 values, the numerical solutions cranked out by Mathematica are fairly close to what LTSpice simulates.

Nice !

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