simulate this circuit – Schematic created using CircuitLab
Why is the base current into Q1 so high according to the simulator? It is higher than I expect.
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1\$\begingroup\$ How much current do you expect in the base? Can you provide your calculation? \$\endgroup\$– saiMar 30 at 2:05
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\$\begingroup\$ But, most importantly, could a TL081 provide that current? \$\endgroup\$– Sredni VashtarMar 30 at 4:01
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3\$\begingroup\$ 1. Change simulator: use LTSpice or Microcap, both free and no time limits. 2. Use a model for a real op amp 3. Add an intermediate driver to pump up the current supplied from the op amp (a couple of BJTs will do) \$\endgroup\$– Sredni VashtarMar 30 at 4:26
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\$\begingroup\$ It appears that D1 would have a forward voltage higher than 5V at such current levels (can you confirm that?) so the current is having to be made up from the base instead of the collector. \$\endgroup\$– user253751Mar 30 at 7:09
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2\$\begingroup\$ microcap is now at web.archive.org/web/20200114114919/http://… \$\endgroup\$– Antonio51Mar 30 at 7:32
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3 Answers
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Firstly, you are trying to get the voltage across R2 to be 50mV, which would result in 5A of current through it.
If somehow that happened, the voltage across the LED would be many many volts, far more than the 5V available. If you check the voltage across the LED in your simulation, it's 4.9V, the transistor is completely saturated, and its collector potential can't fall any further. So to make this circuit work, you need a load that has less than 5V across it with 5A through it. Try replacing the LED with a 0.5Ω resistor, which will drop 2.5V when passing 5A. That will produce collector potential of +2.5V, and keep the transistor out of saturation.
When the transistor is saturated like that, the op-amp doesn't know, and it still tries to raise its output to achieve 50mV at the emitter, an impossible task with that LED there. [Edit: user253751 pointed out correctly that whatever current in R2 that can't be sourced from the collector, can come via the base, making it technically possible to get 50mV across (5A through) R2. In the simulation that seems to be what's happening, but no real-life "normal" op-amp could possibly source that much current.] The op-amp output will rise as much as it can, which you see as +14V. The reason it's able to reach +14V is because you haven't explicitly set its limits (as far as the simulator is concerned), which you should do using an opamp with power rails connected to 0V and +5V.
Secondly, the TL081 is a bad choice for this design, since its output cannot drop below 1.5V above its negative supply. If you did as I suggested, and used an explicitly powered op-amp, with 0V and +5V supplies, the TL081's inability to produce a controlled output under +1.5V means that you can't guarantee good control of base potential near 0.7V.
While that's not necessarily going to cause the circuit to fail (since in this configuration the op-amp output is always sourcing current, rather than sinking), a wiser decision would be to use an op-amp that can officially get its output near to ground, like the LM358.
Lastly, for 5A collector current, the required base current would be unreasonably large. If you're unlucky, your particular 2N3055 could have current gain β as low as 20, requiring base current of \$I_B = \frac{5A}{20} = 250mA\$. There is no way any real regular op-amp can source anything even approaching that. Typically you should draw no more than 10mA from a TL081 or LM358 output. You really need to increase the current gain of the transistor stage a lot, perhaps by using a darlington pair arrangement.
Putting all that together, you have something like this:
simulate this circuit – Schematic created using CircuitLab
As you can see, the voltage across the load, R4, is the expected 2.5V, meaning that the transistor is operating in the linear region, not saturated.
Importantly, current demand from the op-amp output is a much more reasonable 560μA, and the voltage at Q2's base is well within the op-amp's output range, between +0.1V and +3.5V.
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1\$\begingroup\$ It is not impossible to achieve 50mV at the emitter as long as the rest of the 5A is made up through the base - which is what happens. \$\endgroup\$ Mar 30 at 10:44
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\$\begingroup\$ @user253751 that's a great point. I'll think of a way to shoe-horn it into my answer. \$\endgroup\$ Mar 30 at 11:02
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- You used an op-amp without voltage rails, so OA1's voltage output is unlimited (does not saturate at the +5 V rail as you might have intended) and VM1 shows it as 29.1 V.
- If R2 is 10 mΩ and you have set feedback V2 to get 100 mV across it, you're asking for about 10 A through R2, which is about what you are seeing. These 10 A have to come through Q1's base or collector. In this case, it can't get more current from the collector no matter how high the op-amp goes. So the extra current comes through the base.
In real life, this would certainly burn out Q1. But in real life, OA1's output would saturate.
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\$\begingroup\$ I exceeded my allowed time simulating the OP's circuit, but I fail to realize why there even is an op-amp model without power rails, and even then, I see no reason why the output would not provide the 500 mA or so at about 1 volt to cause the transistor to conduct the LED current through the collector. \$\endgroup\$ Mar 29 at 23:22
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\$\begingroup\$ Find the current gain of Q1 at the desired current. Divide the current consumed by the load by that current gain. That is (approximately) the amount of current supplied into the base. This is, of course, assuming ideal components. \$\endgroup\$ Mar 29 at 23:47
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The older TL081 has Jfet inputs that do not work properly if they are 4V or less higher than the negative supply (its input common-mode voltage limit). Then the opamp output goes as high as it can. The minimum supply for a TL081 is plus +4V and -4V when the inputs are 0V or a positive voltage. The input common-mode voltage limit on a newer TL081H is 2V above the negative supply.
