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I wanted to plot a Vb Ic curve of an NPN transistor in LTspice as follows: Vbe is swept from 0 to 2V.

enter image description here

But all the tutorials I encounter shows this curve as the following:

enter image description here

Why in this characteristics they always omit plotting the settled part(I guess the saturation part) or am I doing something wrong?

Edit:

Here when I set series source resistance to zero:

enter image description here

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    \$\begingroup\$ With a \$V_{BE}\$ much larger than 0.8V, your 2N2222 transistor would burn. \$\endgroup\$ Mar 30, 2017 at 15:22

2 Answers 2

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This is the standard 2N2222 model in LTSpice:

* Copyright © 2000 Linear Technology Corporation.   All rights reserved.
*
*
.model 2N2222 NPN(IS=1E-14 VAF=100
+   BF=200 IKF=0.3 XTB=1.5 BR=3
+   CJC=8E-12 CJE=25E-12 TR=100E-9 TF=400E-12
+   ITF=1 VTF=2 XTF=3 RB=10 RC=.3 RE=.2 Vceo=30 Icrating=800m  mfg=Philips)

As you can see, there will still be about 0.5 Ohm (RC+RE) to that will limit the current through the transistor.

Large signal BJT model

Furthermore, there is a base resistor of 10 Ohm in that model. This will limit the base current to some 100-120 mA when your input source is 2 V. On top of that, the DC current gain will not keep constant at BF = 200, but will fall at high collector currents due to parameter IKF. You can expect BF to fall down to 20 or so (something similar can be seen in the actual device datasheet).

2N2222 hfe

Hence you'll have 20 x 120 mA = 2.4 A collector current @ 2 V input. However, a 2N2222 will blow well before reaching that current.

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It looks like you have 100 ohm in series with your 24V voltage source. 24V/100ohm=240mA. Even if you replace the transistor by a short, more current will not flow from the source.

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  • \$\begingroup\$ Please see my edit. It still doesn't represent the generic plots. I plotted log log. \$\endgroup\$
    – user16307
    Mar 30, 2017 at 14:56
  • \$\begingroup\$ I see. Can you plot the base current too? In general simulation models are more complex then first order equations, but I wonder if something else is going on here. \$\endgroup\$ Mar 30, 2017 at 15:06
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    \$\begingroup\$ The 2N2222 model in LTSpice has Rb=10, Rc=0.3, Re=0.2, so there's still series resistances present even after you remove the source resistance. From the model values I roughly calculate the current should limit at about 8.4 A. That's a bit higher than what your graph shows, but in the ballpark. \$\endgroup\$
    – The Photon
    Mar 30, 2017 at 15:47
  • \$\begingroup\$ Since 2N2222 (at least one version I found) has maximum Ic of 800 mA, there's not much point in worrying about whether the model is accurate with Ic > 5 A. \$\endgroup\$
    – The Photon
    Mar 30, 2017 at 15:49
  • \$\begingroup\$ @ThePhoton I agree. For all we know whoever designed the model just didn't care what it output after a certain point. On the other hand, maybe the model is more or less ideal and this is some weird simulation artifact. A third possibility is that if you could somehow prevent the transistor from catching on fire, maybe you would observe this behavior, and there is some physical model for it. It would be interesting to know for sure which category this falls into, but maybe not worth the effort. \$\endgroup\$ Mar 30, 2017 at 15:57

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