Timeline for How to use Ebers-Moll equation to calculate collector current?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 24, 2020 at 5:57 | vote | accept | KMC | ||
| Apr 21, 2020 at 14:28 | answer | added | G36 | timeline score: 1 | |
| Apr 20, 2020 at 7:13 | comment | added | jonk | 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. | |
| Apr 20, 2020 at 7:07 | comment | added | jonk | Here's an example 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. | |
| Apr 20, 2020 at 6:58 | comment | added | jonk | 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. | |
| Apr 20, 2020 at 6:56 | comment | added | jonk | Here are the Ebers-Moll equations (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. | |
| Apr 20, 2020 at 6:51 | comment | added | KMC | @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. | |
| Apr 20, 2020 at 6:34 | comment | added | jonk | 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. | |
| Apr 20, 2020 at 6:22 | comment | added | jonk | 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. | |
| Apr 20, 2020 at 6:22 | comment | added | jonk | 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. | |
| Apr 20, 2020 at 6:20 | comment | added | KMC | @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. | |
| Apr 20, 2020 at 6:15 | comment | added | KMC | @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. | |
| Apr 20, 2020 at 6:12 | comment | added | KMC | @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.... | |
| Apr 20, 2020 at 5:48 | comment | added | jonk | 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. | |
| Apr 20, 2020 at 5:23 | comment | added | jonk | 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? | |
| Apr 20, 2020 at 5:14 | answer | added | Leoman12 | timeline score: 1 | |
| Apr 20, 2020 at 5:12 | comment | added | jonk | 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. | |
| Apr 20, 2020 at 3:56 | history | asked | KMC | CC BY-SA 4.0 |