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What is it in an npn transistor that determines the voltage "maximum" for the base?

because in a typical npn transistor it is 5V, meaning if you feed 5 volt into the base it will flow 100%, 1 volt = 20%(if it is a very linear transistor).

Is it possible to somehow change the 5V to something lower, much lower, like 10 milivolts? (Because if so, then this would be a very cheap way to make it act as an amplifier - for sensors)

  • I am talking about the S9014C, and I am sorry but I thought that the voltage that goes to the base was always between 0 and 5 on any transistor.

  • If it now is as the comments say, that I am wrong, then how do I make the 0-10 milivolt range(at the base) the new 0-100% range?

  • The sensor that I would attatch would be a temperature sensor. (Type K Thermocouple)

  • Still, the only thing I'm wondering is how I can "increase" the "sensitivity" range from 0-5 volt to 0-10 milivolt. But maybe that's something that got nothing to do about transistors. What do I know.

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closed as not a real question by Olin Lathrop, placeholder, Anindo Ghosh, Leon Heller, jippie Jan 11 '13 at 17:35

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ Your assumption about typical 5V for baseis wrong, please add a circuit diagram and part number so we can explain where your assumption is off. Transistors can be perfect fit for amplifying sensor signals. \$\endgroup\$ – jippie Jan 11 '13 at 16:44
  • \$\begingroup\$ I guess that you are referring to the maximum reverse voltage that the BE junction can safely handle. That is not a parameter you'd need often in a typical sensor amplifier. \$\endgroup\$ – Wouter van Ooijen Jan 11 '13 at 16:48
  • \$\begingroup\$ The moment the very first transistor was made it was realized that it could be used as an amplifier. Your basic assumptions are wrong, I'd suggest that you search around for how BJT's actually work on wikipedia, and on this site. Repost your question once you've got a simple understanding. Voting to close. \$\endgroup\$ – placeholder Jan 11 '13 at 16:52
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    \$\begingroup\$ This question can't be answered because it is based on incorrect assumptions. Put another way, it makes no sense. It's like "Since tires are made from wood, could I wave a dead fish over it to make my tunafish sandwich get better milage?" \$\endgroup\$ – Olin Lathrop Jan 11 '13 at 17:18
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    \$\begingroup\$ I think there's at least 4 different questions here: 1. How to read the ratings on the datasheet; 2. How to use a BJT as a voltage amplifier; 3. How to make an amplifier to translate a 0-10 mV input to 0-5 V output; 4. How to deal with a thermocouple (which has some special issues). It might help you out most if you work on these from the top down (look at the big picture first). \$\endgroup\$ – The Photon Jan 11 '13 at 17:51
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It looks like the 5 V figure you're talking about is from the line in the datasheet for absolute maximum "emitter-base voltage".

When we talk about an "emitter-base voltage" or a "collector-emitter voltage" the order of the words is important. The first terminal mentioned is the positive terminal for the measurement being described. So if the "emitter-base voltage" is +5 V it means the emitter is 5 V above the base. If we say the emitter-base voltage is -0.5 V it means the emitter is 0.5 V below the base.

So the 5 V emitter-base voltage limit means the base-emitter junction should not be reverse-biased more than 5 V or you risk damaging the device.

I thought that the voltage that goes to the base was always between 0 and 5 on any transistor.

This is not right at all. 5 V applied to the base of most BJTs will destroy them. Typically if you forward bias the base-emitter junction, you need to apply about 0.6 or 0.7 V to turn on the BJT. After that, the device is current-controlled, not voltage controlled, so you need additional components to determine the voltage gain.

Try a search for common-emitter amplifier to see the most common way to make a voltage amplifier with a single BJT.

As I said in comments, an op-amp or in-amp is more appropriate for the application you are describing than a discrete transistor.

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