2N3904 input voltage range

hello,

i have question of whether the input to 2n3904 can be like 2V? is there some input voltage range for 2n3904?

thank you

okbro said:

hello,

i have question of whether the input to 2n3904 can be like 2V? is there some input voltage range for 2n3904?

thank you

Click to expand...

The question really makes no sense, it's the circuit that controls the operation, not the transistor itself - as a silicon transistor the turn-on threshold is around 0.7V, as with all single bipolar silicon transistors.

To add to Nigel's answer:
The voltage can be anything above around 0.7V, but the base current must be limited to a reasonable value by a resistor in series with the base.

If you are switching something with the transistor, the base current needs to be at least (load current / worst case hFE).
eg. If the transistor has a worst case gain of 50 and you want to switch 200mA, the base current must be at least 4mA;
10mA is still a safe, low value and would be OK.

For 10mA with a 2V signal, there would be around 1.3V across the resistor, as the base conducts at about 0.7V, so 1.3V / 0.01A = 130 Ohms.
Either 120 or 150 Ohms would be near enough for that.

If the controlling signal does not actively switch to 0V when it's off, also add a higher value resistor between the base and emitter of the transistor to eliminate any leakage and ensure it turns fully off.
10K should generally be OK for that.

As a bipolar is a V controlled device, Ic = f(Vbe), and log at that, you are asking
for roasted plastic in less time it takes to microwave a hummingbird
for dinner....unless external, as mentioned prior, external circuit limits Ic.
That of course applies to V driving the base....

Shockley's equation







Reply # 6 - https://www.eevblog.com/forum/beginners/current-controlled-device-vs-voltage-controlled/



Regards, Dana.

danadak said:

As a bipolar is a V controlled device,

Click to expand...

Controversial

Transistors are generally considered current controlled devices, and FET's are voltage controlled.

Simple enough - transistors take current in the base to work, FET's don't take current in the gate.

okbro said:

i have question of whether the input to 2n3904 can be like 2V?

Click to expand...

We need a schematic. To get 2V at the input there must be a resistor on the emitter. Need more infornation.

It depends on how much current you want to control across the C-E junction..... 2m3904's typically have a maximum collector current of 200mA. The B-E junction saturation voltage is typically 0.65V but can be as high as 0.95V. The Gain (Hfe) in combination with the maximum C-E Current is what you want to look at. unfortunately the Hfe can typically range from 100 to 400 and vary wildly from transistor to transistor.

What this means basically ... Suppose you need to deliver 150mA of current across the C-E junction.

Taking the worst case Hfe scenario of 100, we would need at least 1.5 mA of current at the B-E junction of the transistor .. 150mA / 100Hfe = 1.5mA

To calculate the current limiting resistor required at the B-E junction of the transistor would be...

( Vsupply - Vsat ) / Current

Assuming a 5V supply....
( 5V - 0.65V ) / 1.5mA = 2.9k

Assuming a 3.3V supply....
( 3.3V - 0.65V ) / 1.5mA = 1.8k

Beau Schwabe said:

The B-E junction can supposedly tolerate 6V.

Click to expand...

That is the max allowed reverse voltage where the junction begins avalanche breakdown.

Beau Schwabe said:

The Gain (Hfe) in combination with the maximum C-E Current is what you want to look at. unfortunately the Hfe can typically range from 100 to 400 and vary wildly from transistor to transistor.

Click to expand...

hFE is used when the transistor is an amplifier with plenty of Vce. Then it is not saturated.
When a transistor is saturated then the datasheet shows a base current that is 1/10th the emitter current.

audioguru said:

That is the max allowed reverse voltage where the junction begins avalanche breakdown.

hFE is used when the transistor is an amplifier with plenty of Vce. Then it is not saturated.
When a transistor is saturated then the datasheet shows a base current that is 1/10th the emitter current.

Click to expand...

A transistor whose base current is 1/10 Ie occurs because user forces 1/10 the current into base.
This practice is called forced beta, and is an approximation to make sure the transistor goes
into saturation.

Its a condition user forces on transistor, not vice versa.

Regards, Dana.

[about voltage-input controlling transistor current]

Nigel Goodwin said:

Controversial

Transistors are generally considered current controlled devices, and FET's are voltage controlled.

Click to expand...

Not very controversial when you do measurements.

The input currents of a transistor aren't negligible, but aren't accurately predictable, either.
Shockley's equation for transistor action accurately relates input base-emitter voltage
to emitter current (and thus, collector current), and that voltage/current relationship
is called transresistance. The name 'transistor' derives from that word.

I'm sure all would agree the original question is invalid as you need a reference for voltage like 0V or emitter voltage.
When the emitter is grounded (Ve=0) or resistor Re=0 then both base voltages, Vb are the same.

The only way you can make the Vb >1 V without exceeding the base current is to add Re (or a current sink) to limit the current. When Re is the only limiting factor and then Vb becomes more linear above Vbe=0.7 and thus the exponential Shockley equation turns into the linear Ic= hFE*Ib . This is due to Re being greater than the internal resistance for Vbe.

As Ib = (Vb-Vbe)/Re you can assume Vbe is compressed and choose constant value between 0.6 to 0.8 for collector currents from Ic= 1mA to 100mA which varies somewhat with transistor bulk size and power capacity at the higher levels.

Thus all BJT's are logarithmic voltage-controlled collector=current or inversely resistance by themselves but in circuits like Common Emitter with \( R_E > > r_{\pi} \) (base-emitter bulk resistance) .

- When above is true, they become more linear current amplifiers as long as Vce >> 1V which is roughly the edge of saturation and the threshold depends on Ic value. Unfortunately, hFE is never constant but usually peaks around 10% of the maximum rated current and drops off below Ic=1mA. It also has a large process variance with > 3:1 max:min as commonly seen in datasheets. Thus, the use of impedance ratios Av= Zc/Ze is preferred when you want a more stable voltage-gain in a config. such as a Common Emitter, CE amplifier.