# Back to basics - BJT transistor BC547B (Solved)

Thread Starter

#### PsySc0rpi0n

Joined Mar 4, 2014
1,545
Perhaps you are interested in the reason for this difference.
The reason is in your design which is a bad one. Let me explain:
Normally, the resistive voltage divider at the base is selected to be as low-resistive as possible because each good design should provide the required base voltage - independend on the pretty large tolerances of the current gain B (resp. beta).
(Rule of thumb: The base current should be max. 10% of the current through the divider chain).
The lower limit for both resistors is set by the allowed power consumption as well as the input resistance of the whole stage.

In your design, we see two resistors - each of 1 MegOhm.
As a result, the voltage at the base is strongly dependent on the base current which is different for the various transistors.
Example: I have simulated your circuit with the given values and a BC108 macro model.
The emitter current was 1.59mA.
For a better design with a low-resistive voltage divider and a suitable emitter resistor you will have an emitter current which will be much less dependent on the selected transistor type.
I think I understood the explanation in general, but there are a few details I want to explore. I'm aware of the bad design. I mean, not the design, but the values for the resistors, as the design is a common one, I guess, nothing special about it.

A few things I would like to discuss:
I don't understand when you say that the base current is different from transistor to transistor. I might be misinterpreting the purpose of the voltage divider there but, isn't it to take advantage of only have a single power supply so that we can have current at the base? I mean, if there were 2 power supplies, we could use one at the base and another at the collector but when we only have one, we use a voltage divider there so that we can use the single power supply to feed both terminals, right? So, if this is the purpose of the voltage divider, then we control the current going into the transistor base, therefore, not depending from one transistor to another.

Lastly, I wasn't expecting any difference at all, because if @Jony130 used simply math equations, then, results should be exactly the same.
I'm pretty sure no one will follow the steps I attached but, if no errors manipilating the variables, I should get the same values he found.

I'll just plug smaller values for R1 and R2 and see the difference in $$\displaystyle{I_{b}}$$.

#### LvW

Joined Jun 13, 2013
1,247
So, if this is the purpose of the voltage divider, then we control the current going into the transistor base, therefore, not depending from one transistor to another.
Perhaps there is a misunderstanding on your side:
It is the base-emitter voltage which sets and determines the emitter current - NOT the base current. This current may differ from one transistor to another - but when the base voltage is relatively independent on the (unknown) base current (and we have good stabilization provided by RE) , the DC operational point is independent on the selected transistor type.

Thread Starter

#### PsySc0rpi0n

Joined Mar 4, 2014
1,545
Perhaps there is a misunderstanding on your side:
It is the base-emitter voltage which sets and determines the emitter current - NOT the base current. This current may differ from one transistor to another - but when the base voltage is relatively independent on the (unknown) base current (and we have good stabilization provided by RE) , the DC operational point is independent on the selected transistor type.
I didn't that. I didn't say that Ib sets IE. Or at least, I didn't mean that.
What I meant was that R1 and R2 were using the same voltage supply as the Collector terminal to activate the transistor. And therefore, Ib would be set by those 2 resistors.

#### LvW

Joined Jun 13, 2013
1,247
I didn't that. I didn't say that Ib sets IE. Or at least, I didn't mean that.
What I meant was that R1 and R2 were using the same voltage supply as the Collector terminal to activate the transistor. And therefore, Ib would be set by those 2 resistors.
The supply voltage plays no role - it is the task of the two resistors to set the base VOLTAGE - as stiff as possible (which means: As independent as possible from the unknown base current). Remember: The currents Ie and Ic are determined by the voltage Vbe only - the current Ib is a kind of "byproduct".
This is the only reason to use two low-resistive parts for the divider chain (as low-resistive as allowed from the viewpoints of power consumption and input resistance).

Edit (add): The most serious problem is the temperature sensitivty of the current Ie (resp. Ic). This sensitivity is the reason for using a stabilizing resistor Re which produces a negative current-controlled voltage feedback: Rising current Ie increases the emitter potential Ve and, thus, reduces Vbe - if the voltage at the base (Vb) is a "stiff" voltage which is independent on the current Ib (which changes together with Ie).

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#### Jony130

Joined Feb 17, 2009
5,230
You made a mistake at the very beginning of your "work". And I did not notice it yesterday, because you made the same error in post #8.

$$I_B = \frac{I_E R_E}{\beta +1}$$

I hope that you see that

$$\frac{I_E R_E}{\beta +1}$$

Has a unit of voltage, not the current.

The correct equation should look like this:

$$\Large V_{CC} = (I_2 + \frac{I_E}{\beta +1})R_1 + I_2R_2$$

$$\Large I_2R_2 = V_{BE} + I_ER_E$$

Thread Starter

#### PsySc0rpi0n

Joined Mar 4, 2014
1,545
You made a mistake at the very beginning of your "work". And I did not notice it yesterday, because you made the same error in post #8.

$$I_B = \frac{I_E R_E}{\beta +1}$$

I hope that you see that

$$\frac{I_E R_E}{\beta +1}$$

Has a unit of voltage, not the current.

The correct equation should look like this:

$$\Large V_{CC} = (I_2 + \frac{I_E}{\beta +1})R_1 + I_2R_2$$

$$\Large I_2R_2 = V_{BE} + I_ER_E$$
Ohh ok. I see.
I'll redo it again.
Thanks

#### Audioguru again

Joined Oct 21, 2019
3,190
I don't understand the irony about buying hundreds of transistors and do what you said plus finding one to have 9μA.
A transistor has a wide range if beta. A BC547B has the "B" range of beta from 220 to 500 in your circuit where its collector current will be around 4mA when its 1M biasing resistors are properly calculated. A beta of 220 is medium and a beta of 500 is very sensitive.
Usually the base voltage divider resistors have a current that is 10 times what is needed so that the beta if the transistor does not affect how it is biased. Your very high value 1M for the base voltage divider resistors causes a huge affect with different betas.
Actually, you selected a B beta range of only 2 times then maybe your 1M resistors can have a current only 3 times more than is needed.

Datasheet I saw, said a gain between 20 and 200, iirc.
The datasheets of a BC547B from ON and from Philips show a beta from 200 to 450 when the collector current is 2mA and a graph shows the beta a little higher at the 4mA your circuit needs.

And of course I want every transistor to work. Why shouldn't they work? Aren't they meant to work? I don't understand, sir. I'm completely lost.
A transistor amplifier is designed with the base voltage divider resistors to have a current that is about 10 times more than is needed so that different betas do not affect the biasing. Your 1M resistors values are so high that a high beta transistor is saturated and a low beta transistor is almost cutoff. Then many transistors will not work in your circuit unless you buy and test the beta of many transistors to MAYBE find ONE that works.

#### MrChips

Joined Oct 2, 2009
23,528
The mistake you are making is that you are attempting to use linear equations to model a non-linear circuit.
Your initial assumption is that if you had the correct equations you can solve for the unknown currents and voltages.

Semiconductor devices such as diodes and transistors are highly non-linear devices. You require complete mathematical models in order to apply numerical analysis. Current gain beta is an approximation only. It varies with collector current, temperature, and from device to device.

Here are some issues that have already been pointed out to you.

1. R1 and R2 constitute a voltage divider, yes. The current through R1 and R2 by themselves would be 12V/2MΩ = 6μA.
With the rest of circuit connected, the voltage at the junction of R1 and R2 drops from 6V to below 2V. The most base current available from R1 alone is 12μA. Proper BJT base bias calls for available current that is at least 10 times the required base current. This is what we call a "stiff" base bias. As it stands, R2 has little effect on the circuit because of the much lower current it draws.

2. You don't know the current gain beta. As already stated, beta varies with collector current and with device. When AG says to buy hundreds of BC547B transistors he is not being flippant. What he is saying is that you would have to test your circuit with hundreds of devices in order to categorize the average as well as extreme behaviour of the circuit. You would have to select the transistor with the beta that you chose in your modelling.

3. Adding the load resistor and diode at the collector output upsets the apple cart.
Without this load, the maximum collector current is 12V / (1000Ω + 460Ω) = 8.2mA if the transistor is in full saturation.
With the load attached, most of the current through Rc is going towards this load.

Thread Starter

#### PsySc0rpi0n

Joined Mar 4, 2014
1,545
Ok, as I have a problem here at work and I can't do anything about it, I'll try to put here the equations and steps.

Starting from the 4 fixed equations already pointed by @Jony130, I want to see if I can get there by hand.

$$\displaystyle{\begin{matrix} V_{cc} = I_1 \cdot R_1 + I_2 \cdot R_2 \left ( 1 \right )\\ I_1 = I_b + I_2\left ( 2 \right )\\ I2\cdot R_2 = V_{bE} + I_E\cdot R_E \left ( 3 \right )\\ I_b = \frac{I_E}{\beta \cdot + 1 } \left ( 4 \right ) \end{matrix}}$$

So, to start, I replace $$I_b$$ from 4 and $$I_2$$ from equation (3) in equation (2), getting this new equation (5).
$$\displaystyle{I_1 = \frac{I_E}{\beta + 1} + \frac{V_{bE} + I_E\cdot R_E}{R_2}} \left ( 5 \right )$$

Now, I use equations (5) and (3) to replace $$I_1$$ and $$I_2$$ in equation (1):
$$\displaystyle{V_{cc}=\left ( \frac{I_E}{\beta + 1} + \frac{V_{bE} + I_E\cdot R_E}{R_2}\right )\cdot R_1 + \left ( V_{bE} + I_E\cdot R_E\right )}$$

Put everything on the right hand side over a common denominator
$$\displaystyle{V_{cc} =\frac{\left( I_E\cdot R_2 + \left ( \beta + 1\right )\cdot\left ( V_{bE} + I_E\cdot R_E\right )\right )\cdot R_1+ \left( \beta + 1\right)\cdot R_2\cdot \left( V_{bE} + I_E\cdot R_E\right )}{\left ( \beta + 1\right )\cdot R_2}}$$

Now I need to separate all terms to be able to isolate $$I_E$$. This is the tricky part.
I'll rpobably don't be able to finish this before leaving work, so I will finish it at home!

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Thread Starter

#### PsySc0rpi0n

Joined Mar 4, 2014
1,545
@MrChips I understand most of what you said.
But let's get some things clear.

I'm not looking to get results that can hold for all transistors. I'm not trying to match the theoretical results to the practical and real life results. I know this is a non-linear device, but we use some fixed and given parameters so that we can achieve approximate results.

So, this said, I don't need to consider for my theoretical case that beta changes with temp and with IC, or that other variable paramteres change in practical and real circuts because I will and want to assume these parameters won't change and that I start by assuming some other parameters as known parameters, such as Beta, Vcc, Vin, Load, etc.

I'm not here to model a real life transistor situation. In fact, the main goal was to find the right equations and check if I was doing the DC analysis the correct way.

When I get home, I'll try to finish the equations and plug in some values.

Edited;
Fixed as mentioned my @Jony130 here

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#### Jony130

Joined Feb 17, 2009
5,230
You made a mistake here:

$$\displaystyle{V_{cc}=\left ( \frac{I_E}{\beta + 1} + \frac{V_{BE} + I_E\cdot R_E}{R_2}\right )\cdot R_1 + \left ( V_{BE} + I_E\cdot R_E\right )\cdot R_2}$$

to be more precise here:

$$\displaystyle{\left ( V_{BE} + I_E\cdot R_E\right )\cdot R_2}$$

once again the units do not add up.

It shoud looks like this :

$$\displaystyle{V_{cc}=\left ( \frac{I_E}{\beta + 1} + \frac{V_{BE} + I_E\cdot R_E}{R_2}\right )\cdot R_1 + \left ( V_{BE} + I_E\cdot R_E\right )}$$

#### LvW

Joined Jun 13, 2013
1,247
In fact, the main goal was to find the right equations and check if I was doing the DC analysis the correct way.
...............
When I get home, I'll try to finish the equations and plug in some values.
Just one small comment: All the equations are not independent on the selected parts values. Your equations assume that the transistor is operated in its amplification mode (base-collector junction reverse biased, Vce>Vbe).

#### MrChips

Joined Oct 2, 2009
23,528
@MrChips
But let's get some things clear...

I'm not here to model a real life transistor situation. In fact, the main goal was to find the right equations and check if I was doing the DC analysis the correct way.
Your equations are incorrect.
You will not get the same results as the simulation. Your equations do not match the model used in the simulation.

Thread Starter

#### PsySc0rpi0n

Joined Mar 4, 2014
1,545
Ok, after fixing my post above, I'll continue this.

So, here I just use the commutative property and perform all the multiplications so that I have all parcels split a part.
$$\displaystyle{V_{cc}\cdot \left ( \beta + 1\right )\cdot R_2 = I_E\cdot R_2\cdot R_1 + \beta\cdot V_{bE}\cdot R_1 + \beta\cdot I_E\cdot R_E\cdot R_1 + V_{bE}\cdot R_1 + I_E\cdot R_E\cdot R_1 + \beta\cdot R_2\cdot V_{bE} + \beta\cdot R_2\cdot I_E\cdot R_E + R_2\cdot V_{bE} + R_2\cdot I_E\cdot R_E}$$

Here, I isolate $$I_E$$ and perform the remaining manipulations so that I can solve in order to $$I_E$$
$$\displaystyle{I_E \cdot \left ( R_2\cdot R_1 + \beta\cdot R_{E}\cdot R_1 + R_E\cdot R_1 + \beta\cdot R_2\cdot R_E + R_2\cdot R_E\right ) = V_{cc}\cdot \left ( \beta + 1\right )\cdot R_2 - \beta\cdot V_{bE}\cdot R_1 + V_{bE}\cdot R_1 - \beta\cdot R_2\cdot V_{bE} - R_2\cdot V_{bE}}$$

and finally I get $$I_E$$ and the number matches what @Jony130 got too. 1.8mA.
$$\displaystyle{I_E = \frac{V_{cc}\cdot \left ( \beta + 1\right )\cdot R_2 - \beta\cdot V_{bE}\cdot R_1 + V_{bE}\cdot R_1 - \beta\cdot R_2\cdot V_{bE} - R_2\cdot V_{bE}}{\left ( R_2\cdot R_1 + \beta\cdot R_{E}\cdot R_1 + R_E\cdot R_1 + \beta\cdot R_2\cdot R_E + R_2\cdot R_E\right )}}$$

Ok, I understand that this math worth nothing because transistor won't be working in the region the equations were supposed to work. But at least I got some practice going with reading the circuit and taking the equations.

Now that I got the same result as @Jony130 I will move forward.

#### Jony130

Joined Feb 17, 2009
5,230
Next time try to apply Thevenin's theorem, it will be much simpler. For your circuit we have Vth = Vcc * R2/(R1 + R2) = 6V and Rth = R1||R2 = 500kΩ thus:

$$\displaystyle{I_E = \frac{V_{TH} - V_{BE}}{\frac{R_{TH}}{\beta +1} + R_E}} = \frac{ 6V - 0.65V}{\frac{ 500kΩ}{201} + 460Ω} = 1.8mA$$

Thread Starter

#### PsySc0rpi0n

Joined Mar 4, 2014
1,545
Now, I am reading AoE about transistors and at 2.2.1 - Transistor switch, they say that when the switch is closed, the base rises to 0.6V and the drop across the base resistor is 9.4V, so the base current is 9.4mA. So, if $$\beta$$ was 100, we would have 940mA flowing through the LED. But this is wrong, and then they take 100mA as an example to continue the explanation.

My question is why do they pick 100mA and not any other value? Is it because it is the max current that the LED takes, according to the picture?

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#### Audioguru again

Joined Oct 21, 2019
3,190
The lamp (not an LED) draws 100mA when it is lit. It draws much more current when it is cold.
The transistor is a saturated switch with a very low collector-emitter voltage so the beta of 100 is not used since the beta is when the transistor is a linear amplifier with plenty of collector-emitter voltage. The transistor datasheet says it saturates well when its base current is 1/10th its collector current.

#### MrChips

Joined Oct 2, 2009
23,528
You are still obsessed into believing that beta has a fixed value. This is not the case.

Thread Starter

#### PsySc0rpi0n

Joined Mar 4, 2014
1,545
The lamp (not an LED) draws 100mA when it is lit. It draws much more current when it is cold.
The transistor is a saturated switch with a very low collector-emitter voltage so the beta of 100 is not used since the beta is when the transistor is a linear amplifier with plenty of collector-emitter voltage. The transistor datasheet says it saturates well when its base current is 1/10th its collector current.
As you can see, the book says 9.4mA of $$I_b$$ and 940mA of $$I_C$$, so they use $$\beta = 100$$. They even say it.

You are still obsessed into believing that beta has a fixed value. This is not the case.
I'm not. It's not me using the value of $$\beta = 100$$. It's the book. I was just saying what the book is saying too. The book mentions 940mA of $$I_C$$. That assumes $$\beta = 100$$. That's all.

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