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Let us consider the following circuit drawn by LTSPICE: enter image description here

  1. Let us assume that we want to find the temperature for which the thermal voltage is VT=KT/q=25 mV
  2. The Boltzmann constant is defined to be exactly K=1.380649×10−23 J⋅K−1
  3. The default temperature in LTSPICE is 27 C which is T=273.15+27=300.15 K
  4. elementary charge =1.60217662 × 10-19 coulombs
  5. Therefore T=q*VT/K-273.15=16.963 C
  6. On the transistor, we hit ctrl +right click and in the "SpiceLine" field we enter: temp=16.963
  7. Let us assume the collector current is IC=1mA
  8. So the Collector voltage is VC=10-6.8=3.2V
  9. The Ebers-Moll formula: IC=IS*(exp(VBE/VT)-1)-(IS/ALPHAR)*(exp(VBC/VT)-1)
  10. Let us assume VBE is around 0.7 V, so VBC=0.7-3.2=-2.5 (AROUND).
  11. exp(VBC/VT)=exp(-2500/25.865)=1.0543e-042
  12. Therefor exp(VBC/VT) can be easily ignored and the EBERSMOLL formula is simplified as below: IC=IS*(exp(VBE/VT)-1+(1/ALPHAR))
  13. The reverse current gain is BR=ALPHAR/(1-ALPHAR)=-1+(1/ALPHAR)
  14. Therefore EBERS-MOLL formula is simplified as below: IC=IS*(exp(VBE/VT)+BR)
  15. For VBE around 0.7 V, exp(VBE/VT)=1.4463e+012
  16. The BR for 2N2222 is BR=3, which is negligible compared to exp(VBE/VT)=1.4463e+012, for our example EBERS-MOLL formula is simplified as below: IC=ISexp(VBE/VT) or VBE=VTLn(IC/IS)
  17. Let us assume that IS=1E-15 A, so VBE=25*Ln(1m/1E-15)=690.78mV 18.We put IS=1E-15 A, VBE=690.78mV, when we run LTSPICE we expect to get IC=1mA but we get IC=.2 mA.

Question: which other parameter of the transistor should also be changed to get IC=1mA in LTSPICE?

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  • \$\begingroup\$ Have you looked in the help under LTspice > Circuit Elements > Q. ... to see what are the default values for the rest of the parameters that you haven't used in the .MODEL card? Even if you are using the 2N2222, it doesn't make use of every parameter. \$\endgroup\$
    – a concerned citizen
    Commented Oct 15, 2021 at 11:32
  • \$\begingroup\$ Yes I did and try to change them too but no change to the results \$\endgroup\$
    – Aria
    Commented Oct 15, 2021 at 13:42
  • \$\begingroup\$ Maybe this helps? Also, I see you're making a few assumptions in there, make sure those do not end up as wrong conditions for the next set of equations. \$\endgroup\$
    – a concerned citizen
    Commented Oct 15, 2021 at 16:18
  • \$\begingroup\$ The assumptions are very reasonable such as 3 is ignored compared to 10 ^12 \$\endgroup\$
    – Aria
    Commented Oct 15, 2021 at 16:42
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    \$\begingroup\$ The assumption at point #17 is wrong. IS is temperature dependent so it needs to be adjusted in your manual calculations. If you're running at 16.963°C, you need to match what SPICE is doing under the hood. Check out the equation involving XTI and EG here: mathworks.com/help/physmod/sps/ref/npnbipolartransistor.html \$\endgroup\$
    – Ste Kulov
    Commented Oct 15, 2021 at 17:20

1 Answer 1

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As @Ste Kulov mentions in the comments below, there is a temperature dependency (which I missed on your point #6). That changes Is to be (see this):

$$I'_S=I_S\cdot\left(\dfrac{T_{\text{meas}}}{T_{\text{NOM}}}\right)^{XTI}\cdot\exp\left[-\dfrac{E_g}{kT_{\text{meas}}}\left(1-\dfrac{T_{\text{meas}}}{T_{\text{NOM}}}\right)\right] \tag{1}$$

Since the .MODEL does not provide any XTI or Eg, their values are set by the defaults to be 3.0 and 1.11, respectively. With these, the new value is \$I_S=2.046\cdot 10^{-16}\$. Now, inserting this value in your step #17 results in \$V_{\text{BE}}=0.73044\;\text{V}\$. LTspice confirms it:

test

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    \$\begingroup\$ But you didn't do Point #6. I think his problem is that he's not compensating for the temperature dependence of Is in Point #17. He needs to manually calculate the new Is via XTI and Eg, as shown here: mathworks.com/help/physmod/sps/ref/npnbipolartransistor.html \$\endgroup\$
    – Ste Kulov
    Commented Oct 15, 2021 at 17:10
  • \$\begingroup\$ @SteKulov You're right, I missed that. However, by accident, both of us haven't applied the temperature, which makes this result the one correct , as seen, Ic ~ 1 mA (as long as temperature is not considered). But with temperature, XTI=3 by default (since it's not present in the .MODEL card), and Is is recalculated to be 1.65684e-16, which results in Vbe=0.7281. Now, LTspice shows 0.926 mA. Some of the other terms in Is would also be influenced by temperature, so I'd say it's closer to the truth. \$\endgroup\$
    – a concerned citizen
    Commented Oct 15, 2021 at 18:15
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    \$\begingroup\$ But that's the whole point of the exercise. Your example is already congruent across temperature, by default. The questioner specifically wants to adjust for temperature and match SPICE's results to the calculations, such that it is congruent again. BTW, your Is calculation is wrong. I got Is=2.0462e-16 which results in Vbe=0.73044 and a collector current very close to 1mA. \$\endgroup\$
    – Ste Kulov
    Commented Oct 15, 2021 at 19:11
  • \$\begingroup\$ @SteKulov Darn it, I reversed the temperatures outside the log, and that 1e-16 abomination was an intermediary step, I actually got 2.509e-15. I read the wrong line. I'll modify accordingly. In the impossible case you're thinking of upvoting, please don't. \$\endgroup\$
    – a concerned citizen
    Commented Oct 15, 2021 at 19:51
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    \$\begingroup\$ @Aria It's Kelvin, and don't forget that \$E_g\$ is \$1.11\cdot q\$, with \$q=1.602\cdot 10^{-19}\$, therefore \$v_T\$ also changes. \$\endgroup\$
    – a concerned citizen
    Commented Oct 16, 2021 at 9:05

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