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On my protoboard have a BJT CDIL transistor TIP 112 N2, working in the way that on base I have a Vbb = 5V and Rb = 450 Ohms. On Collector, Vcc 24V and Ic = 0.13A. A gearmotor is working properly on that way, (covered by a diode and a capacitor).

So I tried to measure voltages on BJT just for verifying everything is ok, and then simply do some other calculations. If I assume Vbe is approx 0.7 V, then I should expect Ib have an approx value of (5-0.7)V/450 Ohms = 9.5 mA... However when using the multimeter, a current of 7.8 mA is measured.

The fact is when I measure voltage between base and emitter I notice that Vbe = 1450 mV (!!!)... I was expecting a value near 700 mV as I had learnt, but there's no way to let it down (tried replacing the gearmotor for a cooler, increasing Rb, etc). That value of Vbe reamains pretty constant. Is it ok?? Why is Vbe >> 0.7V?

PD: other measures Vce = 0.71 V and Vcb = -0.75.

Thanks a lot (not an electronic here, please be patient :)

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  • \$\begingroup\$ The datasheet says it loud and clear that Vbe is <2.8V. What's wrong with 1450 mV? \$\endgroup\$
    – Dmitry Grigoryev
    Commented Jun 16, 2020 at 10:44
  • \$\begingroup\$ Hi Dmitry. Dont see any problem. Check answers bellow. I didnt notice I was working w/ a Darlington. So, Im trying to identify a Darlington based on its datasheet. As I see, a common NPN transistor doenst mark you that its Vbe is around 0.7 V. If you check for example the TIP 31C datasheet, at the field "Base Emiiter on Voltage" you are pointing out, the value pointed is 1.8 V. However, the Vbe you get under operation is near 0.7 V. Cheers \$\endgroup\$
    – LeandroP
    Commented Jun 16, 2020 at 13:13
  • \$\begingroup\$ That's probably because 1.8V is a maximum value. 0.7V is less than that, so I don't see a contradiction. \$\endgroup\$
    – Dmitry Grigoryev
    Commented Jun 16, 2020 at 14:53

2 Answers 2

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The TIP112 is a Darlington device and as such will have essentially 2 diode drops between base and emitter when conducting. (Darlington Transistor.

Since the base-emitter voltage is higher than in your calculation this would account for the difference between the measurement of the base current and the calculation.

Furthermore, you will often see differences between calculated and measured values for other reasons such as the tolerance of components and the fact that inserting a meter into the circuit will alter the circuit- a current meter, for example, will typically drop from 200-500mV when at full scale.

TIP 112 datasheet

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  • \$\begingroup\$ Many thanks Kevin! I did not expect such a quick and effective answer. I'll check that link. Cheers \$\endgroup\$
    – LeandroP
    Commented Jun 16, 2020 at 0:12
  • \$\begingroup\$ How can I know if a transistor is Darlington or not? For example, the DCIL TIP 112 Data Sheet doesnt clarify it. Of course w/ an electrical diagram Ill see the couple of transistors connected. But is there other way? Maybe some letter of the codename of transistor, or some characteristic parameter. The fact is I Have a transistor long list in Livewire simulator and need to choose a Darlington. Many thanks, cheers \$\endgroup\$
    – LeandroP
    Commented Jun 16, 2020 at 12:14
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The TIP112 is an NPN Darlington transistor, so it effectively has two base-emitter junctions in series to give the 1.4 volt Vbe you measured.

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  • \$\begingroup\$ Many thanks Peter! This doubt was driving me crazy. Cheers! \$\endgroup\$
    – LeandroP
    Commented Jun 16, 2020 at 0:13

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