Electrical – What happens if we force a voltage on Vbe of a bjt transistor

Firstly: current flows into the base, through the emitter. secondly, current flows through the collector, and out of the emitter. The total current through the emitter is that through the base plus that through the collector.

You will need a datasheet to determine the exact voltage drop. Also bare in mind, however, that no two transistors are identical.

The datasheet will have graphs which you can use to look up the expected values. For some calculations, it is helpful to assume that the Vbe is typically around 0.7v. The base-emitter junction is essentially a diode, so it clamps the voltage across itself to around 0.7v. Using that fact, it is trivial to calculate the current going into the base: the voltage across R is 5-0.7 = 4.3v approximately. So the current going into the base must be approximately:

I = V/R = 4.3 / R

So if you know R, you can approximate the current flowing into the base. This will give you one factor to help you read the graphs from the transistor's datasheet. Say R is 10k ohm, the current flowing into the base would be approximately 0.43mA.

Now with that base current, you can calculate the current flowing through the collector - simply multiply it by hfe, the transistor's current gain. But be aware that these can vary enormously firstly between transistors of one model, as well as under different operating conditions for that model. Say hfe is 50, the current flowing through the collector would be approximately 22.5mA.

Using your diagram above, assume the LED has a voltage drop of 2v at 22.5mA, that would mean Vbe has to be 5-2 = 3v. Again, however, the voltage drop of the LED at a given current will vary slightly between LEDs of the same model, and some LEDs such as white LEDs tend to have a higher voltage drop, for example, 4v.

To try to arrive at the exact Vbe, there is a formula you can use, however given the variation between individual transistors, it is far simpler just to use the graphs. Since you know the approximate voltage Vce, and the approximate base current Ib, you can look up Vbe on the graph.

And given the range of possible hfe values stated on the datasheet, (usually they provide three values: a minimum, typical, and maximum). Using the upper and lower hfe bounds, you can then calculate an upper and lower bound of current that will flow through the collector. From that, and the datasheet of the LED, you can calculate the upper and lower bound of Vce. This value will be useful in refining the possible values of Vbe, since often Vbe significantly depends on Vce and Ice; it could make a difference of +/- 0.2v or thereabouts.

Other considerations which could also be quite significant, is the temperature of the transistor's junction. So how much power is flowing through it, for how long, as well as how well it conducts heat to its environment, and the temperature of that environment will all determine the transistor's junction temperature, which will in turn affect values such as the hfe, Vbe, and so on.

For your circuit above, you could use a transistor such as the BC547 NPN which is a general purpose low power BJT NPN transistor. That datasheet should be enough for you to get a good idea of how it would behave. The hfe values I stated above would be different in the BC547; the datasheet states the minimum as 110, and maximum as 800. So your circuit would give a very wide range of potential Ice values, so be careful not to blow the LED. You can determine the hfe of any individual transistor by putting a small current through the base, and measure the current through the collector; then divide Ic by Ib and that's it's hfe for that situation. (Unless the transistor is "saturated", meaning the LED or whatever in its place in your circuit has nearly 5v across it, meaning the transistor is not able to increase Ic any further, as it is already acting as a short circuit.) So to calculate the hfe of one particular BC547, you could assume it will not have an hfe of less than 110, and then calculate the resistor to replace R and the LED, (let's call it Rled), with. Make R 800 times greater than Rled, then measure the current through Rled. Finally, divide that current by the current through R, and that will give you the hfe (current gain of that particular transistor).

Edit in response to your answer:

I do agree that Vr + Vbe = Vce + Vled = 5, and therefore Vbe = 5 - R * Ib.

However, the substitution into the ebers-moll equation does not seem right. Admittedly I haven't used the equations in a very long time, and my maths isn't what it used to be! But I think firstly the reason is the equation Ie = A(e^(Vbe/kt) -1) does not apply to when the transistor is saturated, since Ie will be limited. (See below). Secondly, the solution involves plotting the intersection of two plots: The voltage across R vs Ir according to ohm's law, and Vbe vs Ib with the Ebers-moll equation, solving for where Ir and Ib are equal. The Ir vs Vr will be a straight line, and Ib vs Vbe will be exponential.

The graph in your answer looks about right, assuming it's taken from the circuit you posted in your question? (I.e. with the LED?) The reason it might stop being linear is because the voltage across the LED becomes close to 5v, which means the transistor is saturating. So more base current results in only slightly more current through the LED, due to slightly reduced saturated values of Vce. This is mirrored by the following:

If you have a look at this datasheet for the 2N3904, it tabulates these two values:

VBE(sat) Base-Emitter Saturation Voltage:

• With IC = 10 mA and IB = 1.0 mA then Vbe = 0.65
• With IC = 50 mA and IB = 5.0 mA then Vbe = 0.85

The voltage divider rule between your two resistors does not work like you think because the base emitter junction of the BJT tends to go up to about 0.7V and then not go much higher whilst the current into the base can increase more and more. In other words the BE junction clamps the voltage level between the two resistors to about 0.7V.

When the R1 value is increased to a certain level the voltage at the BJT base lowers down below the 0.6 to 0.7V level and the transistor starts to shut off. At some point the voltage divider will begin to act like normal as the current into the base approaches zero.

ADDITIONAL INFORMATION

Since the OP is not yet quite getting it let me be specific with the examples that were posted. It is correct that at a voltage in range of 0.6 to 0.7V the transistor will begin to turn on.

Let's look at the 20K//1K case in the left picture. Assume for a moment that the transistor base is not connected to the two resistors. By the voltage divider equations the divider voltage is:

Vb = (Vsupply * R6)/(R5 + R6) = (12V * 1K)/(20K + 1K) = 0.571V

This voltage is less than the voltage needed to turn on a transistor so if you would reconnect the transistor base to the divider there will be virtually no current flowing into the base of the transistor and the voltage divider will remain near this 0.571V value.

Next step is to visualize what happens in the above equation when the R5 value is decreased. The divider voltage will increase slowly as the R5 value is decreased.

As R5 decreases more and more the Vb divider voltage will rise up to to the point where the transistor wants to begin turning on. That will be in the 0.6 to 0.7 voltage range. At this point the transistor base begins allowing some of the current from R5 to flow into the base of the transistor.

Be aware that transistors are current mode devices and are actually turned on when the current into the base starts to flow. Below the Vbe threshold the current is nearly zero. As the divider gets past the Vbe threshold the current into the base increases and the transistor starts to turn on.

Ok lets go back and decrease the value of R5 a little more. The lower resistance of R5 allows more current from the 12V supply to flow to R6 and the base of the transistor. The voltage across R5//R6 divider will no longer follow the above equation because the base of the transistor is placing a load on R5 and stealing current so that R6 does not get as much. The nature of the transistor base-emitter junction is that the current into the base can increase more and more whilst the voltage of the base will change only a little.

As I said before the base of the transistor begins to act like a clamp on the voltage divider not allowing the Vb to increase much above the 0.7V level as R5 is made increasingly smaller and smaller. Instead the base current increases to the point that the collector current starts to flow and the transistor eventually turns full on.

The amount of base current needed to turn the transistor full ON will depend on how much collector current is allowed to flow which is limited by components in the collector circuit. The relationship between the base current and the collector current is called the transistor gain or Beta. If the collector current is limited then the transistor will saturate to a Vce of near zero volts when the base current has reached a sufficient level.

It is possible to keep lowering the value of R5 more and more causing the base current to increase more. But beyond the level that caused saturation (Vce near zero) the Vb will only increase slightly and no additional collector current will flow because it has reached the level limited by the components in the collector circuit.