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I have the following scheme.

enter image description here

The goal to drive 2 A current through resistor R1. For example purposes I chose BD743 bipolar transistor which should be half opened and limit current up to 2 A.

Using datasheet I've took hfe as 35.

enter image description here

Using this hfe I've calculated base current:

I = 2 A / 35 = 0.05714... A

Knowing base current I've found base-emitter voltage from datasheet (link above), which is 0.9V

enter image description here

Knowing base-emitter voltage drop I've calculated transistor R2:

R2 = (5.5 - 0.9) / 0.0571 = 80.560

The question: what am I missing? Am I calculating transistor correctly? With another models of transistors I've had either higher or lower resulting values, but still wrong values.

Update 1

I've made changes in the scheme according to all comments collected. Also I've changed source voltage to mach Vce=4V on the charts for hfe value. Scheme still showing wrong values. New R2 calculated the same way

R2 = (4 - 0.9) / 0.0571 = 54.290

What's wrong now?

enter image description here

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    \$\begingroup\$ You have your transistor connected backwards, so it will work in reverse active rather than forward active mode, and the \$\beta\$ for reverse operation is generally lower than for forward operation. \$\endgroup\$
    – The Photon
    Commented Nov 14, 2019 at 17:59
  • \$\begingroup\$ But there is more wrong with your circuit than that, so I am not posting this as an answer. \$\endgroup\$
    – The Photon
    Commented Nov 14, 2019 at 17:59
  • \$\begingroup\$ @ThePhoton Is it usual practice to connect load to emitter chain instead of collector? \$\endgroup\$
    – Artsiom Miksiuk
    Commented Nov 14, 2019 at 18:01
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    \$\begingroup\$ @QuestionAndAnswer A simple rule of thumb is to have the arrow of a BJT (pnp or npn) point to the ground path. \$\endgroup\$
    – Aaron
    Commented Nov 14, 2019 at 18:06
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    \$\begingroup\$ The load may be connected to the emitter or to the collector, but it is common practice to connect the emitter to a lower voltage than the collector. Your schematic makes my brain hurt. \$\endgroup\$
    – Elliot Alderson
    Commented Nov 14, 2019 at 19:40

1 Answer 1

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Using this hfe I've calculated base current:

There's your mistake. That chart only applies when Collector to Emitter voltage is +4V. You have the Collector and Emitter swapped around, so VCE is negative and HFE will be very low.

Also bear in mind that those curves only show typical values. Actual current gain will vary between individual devices, as well as being temperature dependent. This circuit is OK if you only want to limit current to about 2A, but not if you want a more precisely defined value.

Update 1

I've made changes in the scheme according to all comments collected. Also I've changed source voltage to mach Vce=4V on the charts for hfe value. Scheme still showing wrong values. New R2 calculated the same way

R2 = (4 - 0.9) / 0.0571 = 54.290

What's wrong now?

After swapping Collector and Emitter connections the current through R1 is now much closer to your calculation, but still off by 0.92A. Calculating HFE in the simulation we get (1.08-0.0505)/0.0505 = 20. The datasheet says minimum HFE at 4V and 1A is 40. So the model is 'wrong'.

But models are only an approximation of the real world, so they are always 'wrong'. This particular model may be too simple to track the actual performance closely, or it may have been derived from another manufacturer's datasheet which had different specs.

Your calculations also have a significant error - you have not accounted for the voltage drop across R1. At 2A this will be 1V, so the voltage across R2 is 4 - (1 + 0.9) = 2.1V, and the value required for 57.1mA Base current is 2.1/0.0571 = 36.8Ω. You should also account for Base current going through R1, though as this is 35 times less than the Collector current it may be small enough to ignore.

If your only concern is getting an 'accurate' 2A through R1 then a single resistor by itself could do a better job than your circuit. It will still be sensitive to supply voltage, but not as much as the transistor. It would probably also be less temperature dependent and have much more tightly controlled specs.

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  • \$\begingroup\$ So, is there any way to calculate transistor for 2A current 5% error rate? Where can I get hfe values for my Vce? \$\endgroup\$
    – Artsiom Miksiuk
    Commented Nov 14, 2019 at 20:30
  • \$\begingroup\$ According to the datasheet Hfe can be anywhere from 40 to 150 at 1A and 4V. That's way too variable to guarantee 5% tolerance in this simple circuit. To get greater accuracy you must create a 'constant current' circuit which closely regulates current independently of Hfe en.wikipedia.org/wiki/Current_source \$\endgroup\$
    – Bruce Abbott
    Commented Nov 15, 2019 at 1:16
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    \$\begingroup\$ "hFE is not a 'good' transistor parameter; for instance, its value can vary from 50 to 250 for different specimens of a given transistor type. It also dependes upon the collector current, collector to emiter voltage, and temperature. A circuit that depends on a particular value for hFE is a bad circuit" (Horowitz & Hill) \$\endgroup\$
    – mguima
    Commented Nov 16, 2019 at 18:15
  • \$\begingroup\$ @mguima, I can imagine how to calculate circuits in which transistors are working in switch mode, but what about amplifying or sensors mode? What are the appropriate calculation technique for such schemes? \$\endgroup\$
    – Artsiom Miksiuk
    Commented Nov 18, 2019 at 8:43

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