Question:
What controls the current at the base of transistors?
Trevor
2013-08-10 11:53:41 UTC
Ive learned that in order to turn a transistor on you have to apply a positive voltage at the base. But where does this voltage come from and what controls when it turns on or off? Any answer would be greatly appreciated. Thanks!
Five answers:
Jonathan
2013-08-11 10:41:17 UTC
Transistors are controlled by a voltage applied between the emitter and base. The current is a parasitic effect caused by internal recombination of charges. This recombination isn't generally desired, but it is always present in an active BJT. It simply turns out that this recombination current is roughly proportional (over some reasonable range) to the collector current. So a new factor called β is defined to describe this constant of proportionality. People then falsely imagine that it is the current itself that causes the collector current in the first place, but that isn't true. It is the impressed voltage.



The voltage follows the following approximation:



        Vbe = k⋅T ⁄ q ⋅ ln( Ic ⁄ Is )



where Is is a temperature dependent model constant, k is Boltzmann's constant, q is the unit of elementary charge (electron or proton, for example), and T is the absolute temperature (usually expressed in Kelvin since 'k' usually carries that unit.) As a rule of thumb in a small signal BJT, you can broadly speaking assume that Vbe = 0.7V at about Ic = 2mA. For smaller or larger values of Ic, you adjust Vbe by 60mV (at room temps) for every decade change in Ic. So if Ic = 200μA, then Vbe = 640mV. If Ic = 20mA, then Vbe = 760mV. Like that. A designer usually knows (or can estimate) the value of Ic, so they can then roughly guess at the Vbe value. This is then used in calculating the values of resistors or other circuit elements that will be used to ensure that this voltage is met, while ALSO making sure that "at least" a certain amount of recombination current is also present.



If the BJT is used as a switch, with the collector voltage very near the emitter voltage, then the β is usually taken to be approximately 10 or 20 (in small signal BJTs, 20 is usually just fine -- even 30 works often enough -- but in large BJTs able to handle amps the value of 10 is almost certainly necessary as a guess.) If there is a question, the datasheet is used to clarify, adjust, or confirm such estimates. Then the designer makes sure that at least this fraction of the collector current is available as base current drive.



If the BJT is operated as an amplifier, β is often taken to be at least 100 -- if a small signal device. Most typical small signal BJTs have β somewhere between about 150 and 300. If it is a larger BJT, then the data sheet is consulted because the β estimate varies from device to device too much. A class-A amplifier is "always on" the designer makes sure that the base has that fraction of the collector current available as base drive, again. Same thing if the BJT is going to sometimes on and sometimes off as it may be in a class-B amplifier, but in this case there needs to be a way to make sure the Vbe voltage turns off and on, as needed.



How it is all done is a matter for design. There are so many factors to take into account and if you set two designers on the same task they may each take slightly different, or even wildly different, means to get there. There is no one solution fits all problems answer here, as Bill's answer suggests.
Griz806A
2013-08-11 18:35:10 UTC
In the case of a BJT, in some application to turn something on or off, it may come from a sensor, an integrated circuit or a digital chip. It could also be taken from a switch, a relay or another transistor's collector or emitter. The exact type of component the voltage comes in from and when it gets there to turn it on are more a matter of the design of the circuit and its relationship to the purpose of the circuit. For example, a digital IC (microcontroller) may take an input from a keypad, compare the input and find it needs to turn on an LED controlled by an NPN transistor. Its digital output comes into the base, is summed to a DC bias provided by a resistor and a connection to the supply voltage and turns the transistor on. That is, assuming the collector and emitter are each provided a proper voltage / current source from some supply voltage, either the same or another than the IC is using. When the transistor is on, the current will flow from collector to emitter, which in our supposed case will have the LED connected with a protection resistor and should light, assuming we did the design correctly and accounted for Ohm's Law, Kirchoff's Laws and didn't make too many mistakes. When the signal from the IC quits, then the base will drop back to below the level needed to activate it and the LED will shut off.



I hope that this gives you a better general understanding about what happens to turn on a transistor in an everyday circuit. Good luck and stay curious!
Robert
2015-10-30 23:26:27 UTC
The base current increase (decrease) is due to an increase (decrease) in vBE.

The increase (decrease) in vBE increases (decreases) the injection of carriers from the heavily doped emitter.

Most of these carriers cross the thin base region without recombining and are then swept across the base-collector junction into the collector region. A small percentage don't and these form the base current. the collector current is exponential in the base-emitter voltage:

iC=IS^e(vBE/VT)

To get a feel for this, consider this question: to double the collector current, how much would vBE need to increase?

For example, assuming the bias value is VBE=0.7V, increasing this voltage by a mere 17mV (an increase of just under 2.5%) will double the collector current.

According to the collector current equation, if we change the base-emitter voltage from its quiescent value by some small amount, the change in collector current is approximately:

ΔiC=(IC/VT)ΔvBE

As a typical example, let the quiescent collector current IC=1mA. At room temperature, VT=25mV.

Then, for these numbers, the collector current changes by 4% when the base-emitter voltage changes by 1mV.



https://www.electrikals.com/
anonymous
2016-12-30 22:13:56 UTC
8 amp divided by making use of 6ma is 1300. that is modern-day that controls, no longer voltage. Transistor features do no longer pass that best (1300) Even darlingtons have worry, yet you're able to looking one with a blended earnings of the two tiers equivalent to that huge variety, even nevertheless you % a minimum of earnings of 2000 in case you % saturation. Darlingtons are 2 tiers cascaded, so which you % some million.4 volts on the backside. .
billrussell42
2013-08-10 11:57:22 UTC
Depends on the circuit. It may come from a pull up resistor, another transistor, a transducer, a diode, etc.


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