# Transistor Base

#### gerases

Joined Oct 29, 2012
177
Hello everybody,

In the attached circuit, how do I calculate the base current?

Also, if I move the LED to the collector, how will that affect the calculations?

Thanks!

#### Ron H

Joined Apr 14, 2005
7,014
The base voltage divider has to be higher than the sum of the LED forward voltage plus the Vbe forward voltage. Otherwise, the base current will be essentially zero.
Your voltage divider puts 1.5V on the base, which is too low.

#### gerases

Joined Oct 29, 2012
177
I took the schematic out of Charles Platt's Encyclopedia of Electronic Components. Specifically, this one is supposed to be able to test a transistor.

I tried it with a 2N2222 transistor yesterday and unless I did something wrong, the LED did light up. I might have put the LED on the collector though.

And how can the voltage on the base be 1.5V if it shouldn't go higher than 0.6-0.7? Though it's true, I checked it with a multimeter.

#### John P

Joined Oct 14, 2008
1,899
Moving the LED to the collector should cause it to light. (The LED, not the transistor!)

With that circuit, you will probably see some significant voltage drop across the transistor, because the current through the LED with no transistor would be 15mA, but your base current will be less than .1mA. For that to work, you'd need a gain of 150 in the transistor, which isn't very likely. In reality, the current through the LED will be limited by the actual transistor gain, and that's highly dependent on the item that you pick, and will vary with temperature.

Having a resistor where that 2K is placed isn't a bad idea, but it will steal too much base current for no good purpose. Use 100K instead.

Hmm--it's not possible that you have the resistors in the wrong order, is it?

#### gerases

Joined Oct 29, 2012
177
John, how did you get to that 1mA current on the base? I.e., please tell me how you calculated it.

#### Ron H

Joined Apr 14, 2005
7,014
I took the schematic out of Charles Platt's Encyclopedia of Electronic Components. Specifically, this one is supposed to be able to test a transistor.

I tried it with a 2N2222 transistor yesterday and unless I did something wrong, the LED did light up. I might have put the LED on the collector though.

And how can the voltage on the base be 1.5V if it shouldn't go higher than 0.6-0.7? Though it's true, I checked it with a multimeter.
A red LED's forward voltage is typically 2V at rated current.

If you put the LED in the collector circuit, the base current will be about 0.5mA.

#### crutschow

Joined Mar 14, 2008
27,402
Moving the LED to the collector should cause it to light. (The LED, not the transistor!)

With that circuit, you will probably see some significant voltage drop across the transistor, because the current through the LED with no transistor would be 15mA, but your base current will be less than .1mA. For that to work, you'd need a gain of 150 in the transistor, which isn't very likely. In reality, the current through the LED will be limited by the actual transistor gain, and that's highly dependent on the item that you pick, and will vary with temperature.

...............................
I believe your base current calculation is off. The equivalent voltage from the resistive divider at the base is 1.5V. The equivalent resistance at that point is 1.67KΩ. The base current is thus about (1.5 -.65) / 1.67K = 0.51mA.

#### Ron H

Joined Apr 14, 2005
7,014
I believe your base current calculation is off. The equivalent voltage from the resistive divider at the base is 1.5V. The equivalent resistance at that point is 1.67KΩ. The base current is thus about (1.5 -.65) / 1.67K = 0.51mA.
Here is another way of looking at it. Same results, given the fact that you assumed Vbe=0.65V, and I assumed 0.7V.

#### gerases

Joined Oct 29, 2012
177
And how can the voltage on the base be 1.5V if it shouldn't go higher than 0.6-0.7? Though it's true, I checked it with a multimeter.
Answering my own question. Ignore it. The BE voltage can't be more than 0.6, but voltage divider/base can be whatever.

#### gerases

Joined Oct 29, 2012
177
Here is another way of looking at it. Same results, given the fact that you assumed Vbe=0.65V, and I assumed 0.7V.
Can you tell me how you calculated each of those currents?

#### Ron H

Joined Apr 14, 2005
7,014
Can you tell me how you calculated each of those currents?
Do you know Ohm's law? If not, it is one of the first things you should learn when you start to play with electronics.

In a resistor, V=I*R
solving for I,
I=V/R

The voltage across the 10k resistor is (9V-0.7V)=8.3V. I=8.3V/10k=830uA.
The current through the 2k resistor is (0.7V/2k)=350uA.

The difference of these two currents flows through the base.

Ib=830uA-350uA=480uA.

• gerases

#### gerases

Joined Oct 29, 2012
177
OK, got it:

I1 = (9 - 0.7) / 10K
I2 = (1.5 - 0.7) / ((R1*R2)/(R1+R2))
I3 = I1 - I2.

Cool.

Now, if we don't move the LED to the collector, what's going to change in the calculations?

Finally, and I hope I don't cause anybody's fury by this question, but... Do the 470 resistor and the LED on one side AND the 10K + 2K on the other create 2 equivalent resistors connected in parallel?

#### Ron H

Joined Apr 14, 2005
7,014
OK, got it:

I1 = (9 - 0.7) / 10K
I2 = (1.5 - 0.7) / ((R1*R2)/(R1+R2))
This is wrong. I2=0.7/2k.
I3 = I1 - I2.

Cool.

Now, if we don't move the LED to the collector, what's going to change in the calculations?
I explained that in post #2.

Finally, and I hope I don't cause anybody's fury by this question, but... Do the 470 resistor and the LED on one side AND the 10K + 2K on the other create 2 equivalent resistors connected in parallel?
No.

• gerases

#### gerases

Joined Oct 29, 2012
177
Do you know Ohm's law?
Yes. But applying it is not always easy. I'm learning.

Can you explain why not?

Also, is this wrong?

I believe your base current calculation is off. The equivalent voltage from the resistive divider at the base is 1.5V. The equivalent resistance at that point is 1.67KΩ. The base current is thus about (1.5 -.65) / 1.67K = 0.51mA.

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#### John P

Joined Oct 14, 2008
1,899
Sorry, I said the base current would be 0.1mA and really it's over 0.5mA. A result of careless estimation! With a base drive of 0.5mA, you might get the transistor pretty close to fully-on, but to keep the voltage drop as low as possible, a ratio of at least 20:1 and often 10:1 in base and collector current is usually recommended. That's because transistor gain isn't easily predictable, and generally the base current is pretty small under any conditions.

#### Ron H

Joined Apr 14, 2005
7,014
Sorry, I said the base current would be 0.1mA and really it's over 0.5mA. A result of careless estimation! With a base drive of 0.5mA, you might get the transistor pretty close to fully-on, but to keep the voltage drop as low as possible, a ratio of at least 20:1 and often 10:1 in base and collector current is usually recommended. That's because transistor gain isn't easily predictable, and generally the base current is pretty small under any conditions.
Yeah, if the transistor is saturated, a red LED would have about 2V across it. This leaves about 7V across the 470Ω resistor, so the collector current wilbe 7/470≈15mA. The base current should be ≈1.5mA to ensure saturation (per datasheet), but 0.5 mA will almost certainly saturate it.

#### gerases

Joined Oct 29, 2012
177
Having spend a few hours pondering about how common-emmiter and emitter-follower configuration work using iCircuit, I have the following question about the emitter-follower. Take a look at the attached picture. This is all purely theoretical. Arbitrary resistor values, arbitrary everything.

My question is how the heck do you analyze an emitter-follower circuit. If we remove the 10 Ohm resistor, we get a common-emitter circuit and there everything is more or less clear. I.e., the voltage divider of 1K resistors cuts the 9V voltage in 2 making it 4.5V. But because the BE junction can drop at most 0.7V or so, the rest goes to the top 1K resistor making its voltage drop 9-0.7. Then the current splits between the base and the bottom 1K resistor and we get some action in the CE junction. Please ignore the little details such as "but there are no current limiting resistor in the CE junction". That's not important for my question.

Now, let's put the 10 Ohm resistor back in the emitter path. And here's where I'm stuck. We seem to have to know what the voltage drop across the 10 Ohm resistor is first. Then we add 0.7 to it to get the base voltage, then if the voltage is less than 4.5, the difference goes to the top resistor, etc. So the analysis needs to be started from the collector emitter side, which is kind of backwards. Is there a way, taking this circuit as an example, to figure out the base voltage without analyzing the emitter resistor voltage first?

Put differently, if I just put together this circuit on paper, how would you figure out the base voltage? Say the transistor were a 2N2222?

#### Jony130

Joined Feb 17, 2009
5,241
Simply we use a circuit theory.
For example this one From the II Kirchhoff's law we can write

Vcc = I1*R1 + I2*R2 (1)

I1 = Ib + I2 (2)

I2*R2 = Vbe + Ie*Re (3)

And Ib = Ie/(β+1) (4)

And we can solve this for Ib.

$$Ib = \frac{R2Vcc - Vbe(R1+R2)}{(\beta + 1)Re(R1+R2) +(R1R2) } = 419.047\mu A$$

I assume β = 99 and Vbe = 0.6V and knowing Ib we can easy solve for Ie. Ve and Vb

But there is also a simpler way to solve this circuit by using thevenin's theorem.
We can replace the voltage divider (this gray rectangle) with his Thevenin's equivalent circuit

$$Vth = Vcc *\frac{R2}{R1+R2}$$

$$Rth =R1||R2 =\frac{R1*R2}{R1+R2}$$ And now we can solve for Ib

Vth - Ib*Rth - Vbe - Ie*Re = 0

And we also know that

Ie = Ib*(β +1)

so we end up with

Vth - Ib*Rth - Vbe - Ib*(β +1)*Re = 0

So Ib
$$Ib = \frac{(Vth - Vbe)}{ Rth + (\beta + 1)Re}$$

Ie = (β+1)*Ib

Ic = β*Ib

Ve = Ie* Re

Vb = Ve+Vbe

And that all we need.

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• #### WBahn

Joined Mar 31, 2012
26,398
Put differently, if I just put together this circuit on paper, how would you figure out the base voltage? Say the transistor were a 2N2222?
You generally start off making some assumptions, analyze the circuit under those assumptions, and then go back and check the validity of the assumptions.

In your example, I would assume that the base current is very small compared to the voltage divider current. Under that assumption, the base voltage is 4.5V (half of the supply voltage).

We then assume that the transistor is active with about 0.7V for Vbe, making the emitter voltage 3.8V.

The emitter current is thus 3.8V/10Ω = 380mA

Now to check the assumptions.

The Vce is 5.2V, so it is not saturated.

The base current is probably 3.8mA (β=100) or less (perhaps down to the 1mA) range.

The voltage divider current (assuming negligible base current) is 9V/2kΩ=4.5mA.

Uh oh. The assumption that the base current is much less than the voltage divider current isn't looking too good. So we need to look at things more closely.

If you can't lower the voltage divider resistances, then you have to take the base current into account. Let's assume a beta of 100.

If the base voltage is Vb, then KCL at the base becomes:

(Vcc-Vb)/1kΩ - (Vb)/1kΩ - (Vb-0.7V)/β10Ω

Solving for Vb, we have

Vcc/1kΩ +0.7V/(100*10Ω) = Vb( 1/1kΩ + 1/1kΩ + 1/(100*10Ω))

(Vcc+0.7V)/1kΩ = Vb*(3/1kΩ)

Vb = (Vcc+0.7V)/3 = 9.7V/3 = 3.23V

This would yield an emitter current of (3.23V-0.7V)/10Ω = 253mA

You can see how much impact a non-negligible beta can have.

BTW: Your sim results are using a beta of 1000, which is unrealistically high.

• gerases