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The Ebers-Moll equation is given by:$$I_C = I_S(T) * (e^{V_{BE}/V_T})$$ where \$V_T\$ = \$kT/q\$ = 25.3mV at room temperature. In TAOE (2.3.1), it mentioned

\$I_S(T)\$ is the saturation current of the particular transistor (which depends strongly on temperature, T ... \$I_S(T)\$ approximates the reverse leakage current (roughly \$10^{-15}\$A for a small signal transistor like the 2N3904.

I ran a simple simulation just to test the Ebers-Moll equation, where \$I_C\$ is just approximated to \$10^{-15}\$A, and taking \$V_{BE}\$'s value directly from the simulation: $$I_C = 10^{-15} * e^{0.727330 \div 0.0253} = 0.003A \ \ or \ \ 3mA$$

The value is over two orders to one simulated by the program iCircuit (427.267mA). If I just use KVL and multiply the default \$\beta\$ of 100, I get \$I_C = (5-0.727330) \ \div \ 1000 * \beta = 0.427262A \ \ or \ \ 427.267mA\$ matching that in the simulation. How am I supposed to use Ebers-Moll equation and why is it more accurate to determine collector current by base-to-emitter voltage. I just plug in some basic values in and the result did not even turn out close.

BJT simulation

KMC
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    The solution must start with solving for the base current. There are a number of effects not accounted in your model, which is a level 1 Ebers-Moll and doesn't include level 3 effects (such as the Early Effect, not terribly important here) nor the emission co-efficient, to start. But you are really making a huge mistake by using a simulator's VBE value for the formula. You don't fully understand the model used in simulation, and then you apply a different model in your calculations. There's no way that will work out for you. – jonk Apr 20 '20 at 05:12
  • Do you know how to apply KVL to the base-emitter path and solve for the base current ***without*** referring to a simulator, which is likely using different values for the saturation current? – jonk Apr 20 '20 at 05:23
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    You want to solve: \$5\:\text{V}-1\:\text{k}\Omega\cdot I_\text{B} - V_T\,\ln\left(1+\frac{\beta \,I_\text{B}}{I_\text{SAT}}\right)=0\:\text{V}\$ for \$I_\text{B}\$. Once you have that, you can compute \$I_\text{C}=\beta\,I_\text{B}\$. Once you have that, you can work out \$V_\text{BE}=V_T\,\ln\left(1+\frac{I_\text{C}}{I_\text{SAT}}\right)\$. Those calculations will be consistent with the model you have. The simulator is using a different \$I_\text{SAT}\$ and probably a different \$V_T\$ and, as a result, you are getting confused. There are 2nd order effects. But they don't explain two orders. – jonk Apr 20 '20 at 05:48
  • @jonk in the question I did calculate \$I_B\$ which multiplies \$\beta\$ to estimate \$I_C\$. I just started with Ebers-Moll model to see if it gets me more or less the same result - I actually thought I understand the model and assume the equation will indeed gives me the same value until I used an simulation to confirm that it does. So there has to be something really wrong in my understanding fo the equation but I don't know where my misunderstanding is.... – KMC Apr 20 '20 at 06:12
  • @jonk in my reading of TAOE I understand the Ebers-Moll equation gives a more accurate result than \$\beta\$ in calculating \$I_C\$. With Ebers-Moll equation I do not need to use \$\beta\$ but calculated based off \$I_S\$ and \$V_{BE}\$ assuming constant room temperature. – KMC Apr 20 '20 at 06:15
  • @jonk "But they don't explain two orders" - exactly my point here. Even if the simulator use a slightly different \$I_S\$ or different room temperature other than 20C, it's still not supposed to give me two order difference. If I don't use a simulator, I won't even notice I had a problem with my understanding of Ebers-Molls. I've spend nearly two months on BJT and even go down reading solid-state physics and still don't get this model. – KMC Apr 20 '20 at 06:20
  • The Ebers-Moll equation you are starting with assumes that you know the base-emitter voltage. But you don't. And since the collector current varies by a factor of **10** for each \$60\:\text{mV}\$ error in your assumptions about the base-emitter voltage, it doesn't take much of an error on your part in guessing wildly about it to create orders of magnitude errors in your collector current estimate. If you are off by two orders of magnitude then I'd guess that you are wrong about the base-emitter voltage by about 120 mV. – jonk Apr 20 '20 at 06:22
  • I've provided you with the KVL equation above. Solve it for the base current. The value of \$\beta\$ is *not* something you get from Ebers-Moll. It's a parameter to the model, not a result of the model. When I solve the KVL above, I find \$I_\text{B}\approx 4.14843\:\text{mA}\$ using your values for the thermal voltage and the saturation current. See if you can get a similar value. – jonk Apr 20 '20 at 06:22
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    I also find, then, that \$V_\text{BE}\approx 851.571\:\text{mV}\$. Which, you may note, is about a \$124\:\text{mV}\$ difference. And with **10** times the collector current for each \$60\:\text{mV}\$ difference, that just about completely explains the two orders of magnitude you found. – jonk Apr 20 '20 at 06:34
  • @jonk, thank you and I understand your point. But my original thought was because "a circuit that depends on a particular value for \$\beta\$ is a bad circuit", and hence Ebers-Moll uses current and voltage on one terminal (BE) to predict those on the other terminals (CE) without using the value of \$\beta\$ at all. So I'm not supposed to use \$\beta\$ in Ebers-Moll. The E-M equation can calculate the changes on \$V_{BE}\$ from changes in \$I_C\$ without using \$\beta\$ but that's only the delta or change, not a value if a circuit is given. – KMC Apr 20 '20 at 06:51
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    Here are the [Ebers-Moll equations](https://electronics.stackexchange.com/a/252199/38098) (level 1, as they do ***not*** include junction capacitance or the Early Effect [base-width modulation.]) I wrote these down here, some time back. There are three entirely equivalent versions and I provide all three. But jump to the non-linear hybrid-\$\pi\$ version. You will see the beta factors there. They are ***not*** derived from some theoretical model. They are supplied into the model. They are required. – jonk Apr 20 '20 at 06:56
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    It's true, though, that you do not want to depend too highly on their values. That said, you can and ***must*** rely on some range. You can use sensitivity equations to work out a reasoned range of behavior vs any particular parameter variation. And with these, you can work out an "optimal" operating point given some inputs. For example, the operating temperature of the device may vary over some known range. And the beta value for a specific family of devices may be guaranteed to be no less than some particular value when operating at some temperature and collector current. Etc. – jonk Apr 20 '20 at 06:58
  • [Here's an example](https://electronics.stackexchange.com/a/481317/38098) of how you might take into account BJT and other variations in generating a final design for a circuit using BJTs. I cover a lot of ground in it and this should give a flavor, anyway. – jonk Apr 20 '20 at 07:07
  • Something else I forgot to mention is that negative feedback (NFB) is often used (okay, almost always used) in order to stabilize a circuit. This can be local NFB (such as an emitter resistor in a CE amplifier stage) or it can be global NFB, which takes the output and feeds back part of it to counter against the input. (Or it can be semi-global NFB, I suppose, though I don't see that as often.) Just another FYI to be aware of. – jonk Apr 20 '20 at 07:13

2 Answers2

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You are supplying 5V on the 1k resistor will give a base current of about 4.273mA. However you say you should get 3mA as the collector current. However this would violate the idea that the current gain is 100.

If beta is truly 100 then the base current should be 30uA to give a collector current of 3mA when Vbe is 727mV. I suggest changing either the base resistor or voltage at end of base resistor to give a base current of 30uA and see if things improve.

Leoman12
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  • Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/109075/discussion-on-answer-by-leoman12-how-to-use-ebers-moll-equation-to-calculate-col). – Voltage Spike Jun 08 '20 at 22:36
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Let me examine this simple circuit

enter image description here

We know that for an ideal BJT the collector current will follow this equation:

$$I_C = I_S \times \left(e^{\frac{V_{BE}}{V_T}}-1 \right)$$

As you can see in this case we have \$I_S = 1E-14 = 10\text{fA}\$ and by default the ambient temperature is equal to \$ 27°C\$ thus \$V_T = 25.8649\text{mV}\$ and \$\beta = 100\$. And the emission coefficient/ideality factor \$NF = 1\$ by default.

And we can try to solve this circuit using the old method known as an iterative method.

First, we as usually assume some \$V_{BE}\$ value and solve for \$I_B\$ current.

$$I_B(1) = \frac{10V - 0.6V}{100k\Omega} = 94\mu A$$

Now I will use this equation to solve for the new \$V_{BE}\$ value.

$$ V_{BE} = V_T \ln \left(\frac{I_C}{I_S}+1\right)$$

But because you want to find the base current we must modify the equation to:

$$I_{SB} = \frac{I_S}{\beta} = 1E-16 = 0.1\text{fA}$$

$$ V_{BE} = V_T \ln \left(\frac{I_B}{I_{SB}}+1\right)$$

So, we finally can find the new \$V_{BE}\$

$$ V_{BE}(2) = V_T \ln \left(\frac{I_B}{I_{SB}}+1\right) = 0.7131V$$

Now I will use this new \$V_{BE}\$ value to find the new base current value.

$$ I_B(2) = \frac{10V - 0.7131V}{100k\Omega} = 92.869 \mu A$$

And we continue and find new \$V_{BE}\$ value and base current value.

$$ V_{BE}(3) = 25.8649\text{mV} \ln \left(\frac{ 92.869 \mu A}{0.1\text{fA}}+1\right) = 0.71276V$$

$$ I_B(3) = \frac{10V - 0.7128V}{100k\Omega} = 92.872 \mu A$$

$$ V_{BE}(4) = 25.8649\text{mV} \ln \left(\frac{ 92.872 \mu A}{0.1\text{fA}}+1\right) = 0.71276V$$

AS you can see because we are getting almost the same numbers we can conclude that.

\$V_{BE} = 0.7128V\$ and \$I_B = 92.872 \mu A\$ and \$I_C = \beta I_B = 9.2872mA\$

Of course, sometimes the equation does not converge, then we need to use for example the average value of the previous estimate and the calculated value.

G36
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