0
\$\begingroup\$

Does the base voltage has to always be (0.7v) in order for a bipolar NPN transistor to work? what happens if the base voltage increases or decreases? I'm really not into the technical stuff I'm just curious about how electrons exactly work in transistors. Like what if the base voltage is more than the collector's one, wouldn't electrons just flow through the base current and completely ignore the collector?

\$\endgroup\$
1
  • \$\begingroup\$ Welcome. When you have many questions please list them on separate lines so they are easy to see and correlate. \$\endgroup\$
    – user105652
    Aug 13, 2020 at 2:27

3 Answers 3

2
\$\begingroup\$

Does the base voltage has to always be (0.7v) in order for a bipolar NPN transistor to work?

No. But a silicon small-signal BJT transistor usually has a base to emitter voltage of around 0.6 to 0.7V when it's operating.

(Note that you say "base voltage", and I'm changing that to base-emitter voltage -- the voltage of the base vs. some arbitrary point in the circuit is meaningless, and the emitter isn't always grounded. It's the voltage from the base to the emitter that matters).

What happens if the base voltage increases or decreases?

Roughly, the emitter current of a BJT increases exponentially as the base-emitter voltage increases. More base-emitter voltage -- regardless of the collector voltage -- means more emitter current.

what if the base voltage is more than the collector's one, wouldn't electrons just flow through the base current and completely ignore the collector?

This is where things get fun. The way that a BJT is constructed, if there are carriers flowing out of the emitter into the base, and if the collector voltage is right, those carriers mostly fly through the base without contributing to the base current. In an NPN transistor (which is what you appear to be talking about) the carriers are electrons, so they are attracted to positive voltages. So if the collector voltage is above the base voltage, electrons from the emitter, in around a 50:1 or 100:1 ratio, fly through the base and into the collector.

That ratio is pretty close to a transistor's "beta", or amplification ratio.

This happens even if the base and collector are at the same voltage (!). Basically, one way to model a transistor is as the base-emitter diode, which is forward biased in normal operation, and the base-collector diode, which is reverse biased in normal operation, and a magic current source that causes 99% of the emitter current to flow from collector to emitter.

If the collector voltage drops below the base, the transistor will still amplify current -- at this point, the base-collector diode is forward biased, with some current flowing in it, but that action can still be overwhelmed by the "magic current source". In fact, this is a useful mode of operation, where in a typical 1970's-era transistor like the 2N3904, you can have a current amplification of 10:1 or so when the collector-emitter voltage is 0.2V. In new "super-beta" transistors you can get useful amplification with this voltage less than 0.05V.

\$\endgroup\$
0
\$\begingroup\$

0.7V is not an absolute value. It varies depending on the transistor part number and even across any given lot of "identical" transistors. It's an approximation.

It's also not a hard-cut-off. You don't get 0.69999V it's "off", 0.7000001V it's "on". It's a gradual range across maybe 0.5V-0.8V or so (again, transistor dependent).

The 0.7 comes from the result of the base being a diode. You feed it MORE than 0.7V, and the diode 'clamps' the voltage you see there at ~0.7V, just like any normal silicon diode.

BJT transistors are CURRENT controlled devices. You turn them on/off by controlling the amount of current passed into them, not by the voltage you apply. (COntrast to MOSFET's which ARE voltage controlled devices)

I know you don't wanna get 'technical', but you also said you wanna understand how they work. These are technical devices and to understand how they work, you have to understand the technical details.

I'm sure there's plenty YouTube vids showing qualitatively how they work. Just search a bit. Keep in mind that any qualitative understanding doesn't put you in a position to design with them. You really do have to "get it" to apply them successfully. You have to be able to do the math or you'll melt parts and not know why.

\$\endgroup\$
6
  • \$\begingroup\$ So what happens when you increase the base current/voltage, can there be a way for electrons to flow through the base if the voltage/current of the base is greater than that of the collector? \$\endgroup\$
    – Moe
    Aug 12, 2020 at 18:43
  • \$\begingroup\$ The voltage at the base will be higher than the voltage at the collector if the transistor is fully turned on (saturation). How they work is really way too complicated to just describe in text. You need diagrams and maybe some animations. Seriously, look up "How does a BJT transistor work" on YouTube, you'll probably get 100 hits. MOSFETs, to me, are much simpiler to understand and apply \$\endgroup\$
    – Kyle B
    Aug 12, 2020 at 18:46
  • \$\begingroup\$ I did look up on YouTube and Google before I came here, the only thing that I don't understand is at 5:29 in this YT video youtube.com/watch?v=7ukDKVHnac4 It explained everything perfectly except when he said "If you increase the base current the collector current will increase proporionally" can you please explain that to me in a way that it makes sense \$\endgroup\$
    – Moe
    Aug 12, 2020 at 18:51
  • \$\begingroup\$ It just dosen't seem right that when you increase the base current the collector current will ALSO increase, shouldn't one decrease if the other increased? \$\endgroup\$
    – Moe
    Aug 12, 2020 at 18:52
  • \$\begingroup\$ Base current is like a water valve- You can control the collector current by controlling the base current. The more base current you drive, the more collector current can flow. One doesn't fight the other. FUrther, you have to keep in mind, the collector current flowing is a result of the EXTERNAL CIRCUIT as well as the BJT itself, so you can't analyze it in isolation. \$\endgroup\$
    – Kyle B
    Aug 12, 2020 at 19:07
0
\$\begingroup\$

If the base voltage is higher than the collector voltage, and the base voltage is higher than the emitter voltage, the transistor is in the "saturation mode" of operation. Current will be flowing into the base (or out of base for PNP). This is something you can enter into a search engine for better understanding.

If the base voltage is higher than the collector voltage, and the base voltage is LOWER than the emitter voltage, then the transistor is in the "reverse active" mode of operation. Current will be flowing into the base (or out for PNP). That is also a viable search term.

The base emitter voltage does change depending on base current. It generally goes UP in magnitude with higher current and down with lower current. But the relationship is not linear. A small voltage increase leads to a large current increase.

Also, the base emitter voltage changes with temperature. Generally it goes DOWN in magnitude with higher temp, and UP with lower temp.

0.7V is just a typical number used to help do rough analysis. Some people use 0.6V instead of 0.7. If you need a more accurate number, you can start by using 0.7, then solve for the base current, then use the diode equation to solve for the base voltage, then go back and solve for the base current again. After you do that a few times you will probably get very close to the true base current. Or you can use a simulator.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.