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Transistor battery question Answered

Hi, I want to know if I can hook up an 18v 3ah battery to a npn transistor with a max of 1a. Will the transistor only let through 1a or will it get burnt out? Thanks.

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user
iceng (author)2018-02-28

So i'm very interested to learn how you changed non to npn ?

Is there an edit function on a mobile platform ?

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seandogue (author)2018-02-28

A transistor in full on mode is effectively a short circuit (sans some internal resistance). If connected so the emitter is tied to ground and the collector is tied to the source, and the source is 18VDC, and if it is only capable of transferring a maximum of 1A, it will almost certaintly smoke as soon as the base is fed sufficient current to turn it on completelt

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seandogue (author)seandogue2018-02-28

and by "if it is only capable of transferring a maximum of 1A, I mena the transistor, not the battery.

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steveastrouk (author)2018-02-28

Backing up a bit.

Why do you ask ?

;-)

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Jack A Lopez (author)2018-02-27

Speaking vaguely, it is pretty easy to make a naive circuit which will cause a BJT to burn.

For this reason, prudent designs will have resistors, or other circuit elements to limit current, placed in series with both the base-emitter junction (BE) and the collector-emitter junction (CE).

To supplement my crude explanation, you probably also want to look at a more complicated one, like the Wikipedia article for BJT, here,

https://en.wikipedia.org/wiki/Bipolar_junction_tra...


Anyway, this is a tale of two currents: the current through the CE junction, and the current through the BE junction.

Also I am just going to consider only a very simple case of what you can do with a transistor, specifically the case of using a transistor as a switch. More specifically, the thing I am imagining being switched is a relay coil, because that is actually a pretty common task for a NPN transistor.

In this case of a transistor switching a big current to something else, that something else is what is limiting current to the CE junction. For example if it is a transistor turning on a relay, the maximum current that can flow is just the supply voltage divided by the resistance of the relay coil. And this is equivalent to assuming the resistance of the CE junction, when the transistor is turned on, is zero. Or that this will be approximately true, with a sufficient amount of base current.

As an example, suppose I have a 12 volt supply, and a relay whose coil has resistance of Rrelay = 120 ohm. The maximum current I could possibly get to flow through the relay coil is simply(12V)/(120 ohm) = 0.1 A = 100 mA.

By the way, for this circumstance of just switching current through a load, I have the emitter of my NPN transistor tied to ground, or the low side of the supply. (If it is single sided supply, the low side is ground.) And this is typical for NPN transistors. Very often the emitter of NPN transistor is seen tied to ground, and the base and collector will naturally be at voltages higher than the emitter.

The last part of this mojo is picking a resistor to go in series with the base-emitter junction, for to limit the base current.

The promise of the BJT is that it is capable of switching large collector current (Ice) with small base current (Ibe). Moreover, there is a magic number called, "beta". Speaking vaguely, beta is the second letter in the Greek alphabet. More specifically, in the context of BJTs, beta is the ratio of collector current to base current... approximately, in some circumstances.

beta = Ice/Ibe

The "in some circumstances" part is confusing, I admit. More truthfully, the relationship, Ice/Ibe = beta, is true when the transistor is in its so-called, "active mode", and I did not really want to have stop and explain what active mode is.

But I promise that this beta is essentially the largest current gain, Ice/Ibe, I'm going to get while the transistor is turned on.

Also a typical, like rule of thumb, guess for the magnitude of beta, for a typical BJT, is about 100.

Also for now, I ignore the fact that beta is actually temperature dependent too.

So I guestimate that I could make a 100 mA collector current flow, with only a 1 mA base current, and this is equivalent to assuming beta = 100 = Ice/Ibe.

Also, it turns out I can actually pick a resistor to determine the base current very precisely. The B-E junction looks just like a forward biased diode, and picking a resistor to limit current to this is as easy as picking a resistor to limit current to a LED (just another kind of diode).

Ibe = (Vsupply - Vdiode)/Rb

Actually, it gets easier if Vsupply is much larger than Vdiode, which it is when Vsupply=12 V, and Vdiode = 0.6 V.

Then the approximation,

Ibe = Vsupply/Rb

is pretty good.

Anyway, if I pick a 10K resistor, that gives approximately 12/10 = 1.2 ~= 1 mA of base current. And that is base current corresponding to Ice/Ibe = beta = 100.

The final step is to realize, I can make Ibe larger, if I want to, because, as I was saying before, I can basically just choose what I want Ibe to be, via choice of Rb. However making Ibe larger, for example by a factor of 2 or 5, will not at this point make Ice larger by a factor of 2 or 5.

The reason why Ice will get bigger in this way is because Ice is already as big as it can get, namely Vsupply/Rrelay. I started with the assumption that the CE junction was a dead short, or perhaps something like a lump of solid copper or gold, i.e. a near perfect closed switch, with almost exactly zero volts across it.

So what is the reason for pushing more current through the BE junction? Well, the answer is that maybe my guess for beta was too high, and making Ibe larger by a factor of 2 or 5 (by making Rb smaller by the same factor, i.e. from 10K, to 5K or 2K, respectively) should make the same trick happen for a beta as small as 50 or 20.

By the way, it makes sense that this region, where Ibe can be increased, yet Ice remains the same, is called, "saturation".

If I wanted to, perhaps I could increase the excess base current by a stupid huge factor, like 100 or 1000 times, but this would be dumb. Why would this be dumb? The main reason is doing this takes me away from the basic promise of transistor action, which is switching a large current with a small one. There might also be practical reasons not to make Ibe too big. For example, the Ibe junction dissipates heat, in watts, equal to Ibe*Vdiode, just like any other forward biased diode.

As I mentioned previously, the Wikipedia article for BJT, linked above, gives more complete explanation of these various regions of operation, called "active", "saturation", et al. This explanation is under the heading titled, "Regions of operation", here,

https://en.wikipedia.org/wiki/Bipolar_junction_tra...

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CornishKiwi (author)2018-02-27

Hi. Just type transistor part No. into Google and click on data sheet for that part, a BC547 for example will cope with 50v using suitable resistors. Here is a link for BC 547.

https://components101.com/bc547-transistor-pinout-datasheet

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iceng (author)2018-02-27

Welcome to ibles, what is left of a community in decline..

I'm still getting used to kids wanting answers to simple first lab skills..

Your answer is it; depends on the non (NPN) transistor base current.

Discussion, a non transistor is a 3 pin current amplifier which can have a gain of 100..

Example a current of .010 amperes from the base to the emitter will control a current of 1.0 amps from the collector to the emitter..

A base current of .04 amperes will destroy the transistor ! tell me why ?

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