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Here, in the picture attached, I've simulated a constant current source using an NPN transistor, TIP series. My task is to vary the R2 resistor and measuring the current change. This picture holds the info for a certain value, R2=50k. But the current, Ic is still constant. even at that value. The calculation in Ares is quite wrong as the voltage drop is 500V in this scenario while the supply is only 100V! Is there any delicate simulation software that can do the math correctly? Thanks in advance. enter image description here

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  • \$\begingroup\$ Don't assume that the software is wrong or whetever, it is a common trap to fall into if you don't understand how a simulator works. If you do the same with a different simulator, very likely you will get the same results. OK, look at the circuit and voltages: 1) there's only 20 mV across Vce, so the transistor is fully on it does not regulate/control anything because it can't. It tries to do its best but you're giving it no "room" (voltage) to work with. 2) what current is there supposed to flow? The emitter current is 10 mA, calculate the collector current. Is it correct? (No). \$\endgroup\$ – Bimpelrekkie Nov 29 '17 at 15:45
  • \$\begingroup\$ I strongly suggest to draw the circuit on a piece of paper and then calculate by hand what the voltages and current should be. Change resistor values and voltages as needed. When all is as you want, then (and only then) put that in the simulator and compare. If you've done your calculations properly the values will be very similar. \$\endgroup\$ – Bimpelrekkie Nov 29 '17 at 15:47
  • \$\begingroup\$ Yes, the desired current is 10mA. In my circuit, the load is R2, I need to optimize a range of load resistance. So, how can I solve this problem using the software? I understood the math. \$\endgroup\$ – Arnab Paul Nov 29 '17 at 15:50
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    \$\begingroup\$ " R2=50k. But the current, Ic is still constant. even at that value" - WRONG! Most of that 10mA is through the base-emitter junction. Ic is about 1.9mA. \$\endgroup\$ – JIm Dearden Nov 29 '17 at 15:52
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    \$\begingroup\$ Where do you see 500 V drop? \$\endgroup\$ – The Photon Nov 29 '17 at 17:13
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Your simulator is poor in a couple of ways:

  1. The ammeters only show tens of milliamps in the least significant digit. You need finer resolution on their readings.
  2. The ammeters apparently don't round values up, but instead simply truncate. So a value of \$9\:\textrm{mA}\$ will read as \$0.00\:\textrm{A}\$ rather than \$0.01\:\textrm{A}\$, as you might normally expect.
  3. The TIP122 is a Darlington. It is impossible that \$V_{CE}=0.02\:\textrm{V}\$ in the circumstances you show there. As low as 20 times that much, perhaps. But that figure isn't right for a TIP122.
  4. The TIP122 is a Darlington. It is impossible for \$V_{BE}=5\:\textrm{V}-4.39\:\textrm{V}=610\:\textrm{mV}\$. Again, it might be as low as \$1.2\:\textrm{V}\$. But again, that figure isn't right for a TIP122.

As a result, I'm almost certain that the transistor shown in your diagram is NOT a TIP122, at all. Nor is it any existing Darlington. But something else. Probably some regular NPN BJT model is being applied here.

My conclusion is that your transistor is a simple NPN BJT and not a Darlington.

Let's analyze the following (much more likely) case:

schematic

simulate this circuit – Schematic created using CircuitLab

  • Here, the base is nailed to \$5\:\textrm{V}\$. This means the emitter is going to be about \$700\:\textrm{mV}\$ less, or about \$4.3\:\textrm{V}\$.
  • This means that the emitter current will be about \$I_E= \frac{V_E=4.3\:\textrm{V}}{R_1=430\:\Omega}\approx 10\:\textrm{mA}\$. Note that this is very close to what your ammeter says.
  • The collector cannot be lower in voltage than the emitter. It can get close. Your reading says \$0.02\:\textrm{V}\$, which is possible. This means the voltage across \$R_2\$ is \$100\:\textrm{V}-4.3\:\textrm{V}-0.02\:\textrm{V}\approx 95.68\:\textrm{V}\$. Note that this is very close to what your voltmeter says.
  • The collector current will then be \$I_C=\frac{95.68\:\textrm{V}}{R_2=50\:\textrm{k}\Omega}\approx 1.91\:\textrm{mA}\$. This is too low for your ammeter to read it out. So it shows zero, instead.
  • The base current will then be the difference, or in other words: \$I_B=I_E-I_C= 10\:\textrm{mA}-1.91\:\textrm{mA}\approx 8.09\:\textrm{mA}\$. This is also too small for your ammeters and is apparently truncated to zero in your reading.
  • Given the base current is so much more than the collector current, the BJT is in fact very heavily saturated. Which is consistent with the measured \$V_{CE}\$ in your diagram.

The analysis is complete. There are no mysterious \$500\:\textrm{V}\$ voltage drops anywhere in the circuit. The currents make sense, the voltages make sense, the heavily saturated condition of the NPN BJT makes sense.

The problems are: (1) The simulator's inability to round values, or show them with the needed precision here; (2) The fact that you aren't simulating a TIP122, but instead some regular NPN BJT; and (3) Your internal mental state which incorrectly concluded that the measured emitter current must also be taken as close to the collector current. Instead, the base current is quite a bit larger than the collector current, so most of the emitter current is accounted for by examining the base current instead.

That's all there is to it.

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  • \$\begingroup\$ @ArnabPaul I hope it helped. I tried to be as detailed as I could be. \$\endgroup\$ – jonk Dec 3 '17 at 19:08
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enter image description here

@ThePhoton, see. There is nothing to believe here. It's all about facts, needs to be cleared.

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