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I have a circuit where a 5.055V 2.17A (MAX) DC circuit is supposed to be switched using a 3.404V 0.99A(MAX) DC source.

Theoretically, 3.404V 0.99A between base and emitter should allow quite a large current to pass from collector to emitter, but when I am using BC548 B type (Min theoretical gain 200X) as my transistor, the Collector-emitter current that I am actually getting is only about 1.34A.

Of course, BC548 is not designed to handle that high level of current. So I replaced it with TIP31C O Type (Min gain 20X), but something strange hapened with the TIP31C. That transistor is passing at least 1.63A between collector and emitter even when there is nothing connected to base and therefore no potential difference is going on between base and emitter.

schematic

simulate this circuit – Schematic created using CircuitLab

Admittedly, I do not know how to read all the parameters on a transistor datasheet, but even the BC548 did not break down at that voltage, and there is no OFF characteristic mentioned on the TIP31C datasheet at all.

Not only that, even after applying the 3.404V 0.99A potential difference between base and emitter, the collector-emitter current is only going up to 1.72A.

Is there any theoretical explanation behind this strange behaviour?

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    \$\begingroup\$ Check the pins you assumed were E, B and C. \$\endgroup\$ – Andy aka Feb 21 '18 at 12:00
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    \$\begingroup\$ You might be putting 3.4V on the base of the transistor, but if 1A is flowing base to emitter you won't have a transistor for very long. It is likely to pop and emit smoke if you push that much current through the base of a BC548. \$\endgroup\$ – JRE Feb 21 '18 at 12:01
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    \$\begingroup\$ May I suggest that you have a look around on this site and the rest of the internet to look for information on how you should use a transistor to operate a relay. Your schematic will not work because you're using the transistor wrong. There are many ways to do this wrong and only few to do it right. So investigate how others do this and learn from that. Trying to figure it out yourself is almost guaranteed to end in tears. Also, you will learn that you are going to need a flyback diode, without it the transistor will break. \$\endgroup\$ – Bimpelrekkie Feb 21 '18 at 12:34
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    \$\begingroup\$ Fritzing cartoons won't be greated favourably in all quarters. Show a proper circuit diagram and a picture of what you think you did. \$\endgroup\$ – Andy aka Feb 21 '18 at 14:00
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    \$\begingroup\$ Even Fritzing can produce a schematic diagram. The wiring diagrams don't help you understand the circuit. All they show you is where the wires go. To see what the circuit is doing, you have to look up the function of each chip and each pin and try to make sense of it. Or, you could use a schematic diagram where all the needed info is readily visible. You might not have gotten beyond Fritzing's limitations yet, but when you do you will have to start all over and learn a new program. Or, save yourself the hassle and learn a good program first. \$\endgroup\$ – JRE Feb 21 '18 at 17:37
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In that configuration, the transistor is an emitter follower/ common collector. As such, the maximum voltage the load will ever see is your 3.4V - 0.7V = 2.7V. Since that's about half what the load expects, it will only take half the current.

Your circuit needs to be common emitter, more like this.

schematic

simulate this circuit – Schematic created using CircuitLab

HOWEVER: With the load currents you are talking about the base current will be a lot with a simple transistor, likely more than whatever you have driving this can supply. You would be wise to replace Q1 with a suitable N-Channel MOSFET instead.

Obviously, whichever part you chose needs to be able to carry the load current indefinitely without burning out.

Note D1 is a fly-back diode. It is required so the transistor does not kill itself when it turns off.

Leaving the base open

Not a good idea for any transistor circuit. Leakage current from collector becomes base current. That's why good designs include some form of biasing resistor to hold the base at the right "parked" voltage.

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    \$\begingroup\$ Thanks for explaining the theory in a very approachable way. Now I see why the transistor is acting like it is blown. That should be enough hints for me to search up the rest. \$\endgroup\$ – Spero Feb 21 '18 at 16:37
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    \$\begingroup\$ I just redesigned the circuit based on the theory you presented, and now it is working flawlessly. Turns out I did not need the TIP31C. BC548 did the job fine. I express my gratitude to all who contributed. \$\endgroup\$ – Spero Feb 22 '18 at 14:58
  • \$\begingroup\$ @Spero thanks for letting us know. It's nice when folks get back, usually we are left hanging. \$\endgroup\$ – Trevor_G Feb 22 '18 at 15:19

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