# Back to basics - BJT transistor BC547B (Solved)

#### Audioguru again

Joined Oct 21, 2019
3,327
You are lucky to be using a BC547B that has its beta selected to be within 200 to 450 when its collector current is 2mA. A basic BC547 has a beta between 110 and 800. When you buy some, they can have a beta anywhere in between.

If you reduce the 1M values of the biasing resistors then the circuit will work nearly the same with a transistor having any beta within the range without customizing the resistor values to match the actual beta.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
You are lucky to be using a BC547B that has its beta selected to be within 200 to 450 when its collector current is 2mA. A basic BC547 has a beta between 110 and 800. When you buy some, they can have a beta anywhere in between.

If you reduce the 1M values of the biasing resistors then the circuit will work nearly the same with a transistor having any beta within the range without customizing the resistor values to match the actual beta.
On my breadboard I already did that. Those 2 resistors are now of 10kΩ. I understand what you mean. With smaller resistance at the base, the current gain will have reduced impact in the output current at the collector, right?

But this is not related to the question I asked from the book. We just deviated to somewhere else.
Anyway, I think my assumption is probably correct. They use 100mA because that's what the lamp is rated at!

#### Audioguru again

Joined Oct 21, 2019
3,327
The book stooopidly mentions a collector current of 940mA instead of showing the datasheet that says, "For a good saturation then make the base current 1/10th the collector current" because the beta of 100 is only when the transistor is not a saturated switch.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
The book stooopidly mentions a collector current of 940mA instead of showing the datasheet that says, "For a good saturation then make the base current 1/10th the collector current" because the beta of 100 is only when the transistor is not a saturated switch.
I don't think it's stupid. They show the mistake anyone can make when they are not familiar with the details. This is probably one of the best books in the subject out there, as far as I know! Maybe it's not suited to advanced people like most of you here, but I like to read it and I like the feeling it transmits to me with simple examples, real examples and simple language!

#### MrChips

Joined Oct 2, 2009
23,966
The book did not say to use beta = 100.

The book implies that if the beta is 100, then for a collector current of 100mA you need a base current of 1mA.
It then goes on to say, in order to be conservative, use a base current of 9.4mA, not 1mA, i.e. a beta of 10.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
From the same page you posted, here is what I read.

"Blind application on Rule 4 gives Ic = 940mA (for a typical beta of 100). This is wrong."

View attachment 235186
@MrChips , I'm sorry, but did you read the entire text I posted in that post? I typed there that it was wrong ! Did you see? I don't have any questions about understanding it is wrong, because, well, the books says it's wrong.
My question wasn't related to this. It was related to the reason why they use 100mA as an example to justify that the above reasoning was wrong!

I don't know, but sometimes I have the feeling we are not reading the same post/text. I ask a question about the 100mA current, and I get answers about something else! I mean, generally speaking. I'm not saying this about you, specifically, I say it in general as I get replies that are not directly related to what I asked.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
The book did not say to use beta = 100.

The book implies that if the beta is 100, then for a collector current of 100mA you need a base current of 1mA.
It then goes on to say, in order to be conservative, use a base current of 9.4mA, not 1mA, i.e. a beta of 10.
How come? They say there explicitly: "Blind application of rule 4 gives IC = 940mA (for a typical beta of 100)" Meaning that if beta is 100 and Ib = 9.4mA, IC will be 940mA...

#### MrChips

Joined Oct 2, 2009
23,966
You are looking at it backwards.

The design calls for a load of 100mA. That was not an arbitrary number. It states that beside the bulb in Figure 2.5
The book rightly explains that if the beta is 100 then you need a base current of 1mA.
But in order to make certain that the circuit always works, ignore the transistor's beta and assume the beta is 10.
Then you need to apply 10mA to the base.

What a prudent designer does is to derate the specifications given by the manufacturer where it matters.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
Wait....
I didn't say the book was saying/suggesting me to use beta = 100. I said the book was using beta = 100 for those values of Ib and IC.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
...
The book rightly explains that if the beta is 100 then you need a base current of 1mA.
...
I can't read that anywhere in that text.
What I can read is that when the switch closes, Vbe rises to 0.6V, making the voltage at the base 9.4V (10V - 06V), therefore, a current of 9.4mA will show up at the transistor base. And that if beta was 100, collector current would be 940mA. That's what the book says, literally. Then it says this reasoning is wrong and explains why, with the 100mA current and the 10V across the lamp.

#### MrChips

Joined Oct 2, 2009
23,966
You are misinterpreting the book.

Ic = β x IB

or conversely
IB = Ic / β

The book is saying, "what if β = 100", what is the value if IB if Ic= 100mA and vice-versa?

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
You are misinterpreting the book.

Ic = β x IB

or conversely
IB = Ic / β

The book is saying, "what if β = 100", what is the value if IB if Ic= 100mA and vice-versa?
Ok, anyway, let's move on.
I see what you mean and I understand, but the words in the book doesn't seem to express that idea.
I'm reading now in the next page, about the transistors as switches.

#### MrChips

Joined Oct 2, 2009
23,966
I cannot make apologies for the book.
You are taking equations too literally.

Ic = β x IB

The book says that if IB = 9.4mA and β = 100 then Ic ought to be 940mA. But that cannot happen because the load is 100Ω.
The most you can get from a 10V supply through a 100Ω load is 100mA.

Hence you cannot take this equation for granted:

Ic = β x IB

Real β of the transistor has not changed. You just can't get current gain of 100 in this scenario because the circuit is already in saturation mode.

#### hrs

Joined Jun 13, 2014
334
Most if not all BC transistors saturate at a beta of 20 not 10.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
I cannot make apologies for the book.
You are taking equations too literally.

Ic = β x IB

The book says that if IB = 9.4mA and β = 100 then Ic ought to be 940mA. But that cannot happen because the load is 100Ω.
The most you can get from a 10V supply through a 100Ω load is 100mA.

Hence you cannot take this equation for granted:

Ic = β x IB

Real β of the transistor has not changed. You just can't get current gain of 100 in this scenario because the circuit is already in saturation mode.
Ahh, that way I agree. That is in fact what is written on the book and what I understood.

Now, I was trying to follow what the book suggests as for a switch circuit in the next page. I'm trying to apply the same idea with my BC547B.

For my case, I'll assume the following:
Vcc = 12V
VLed = 1.94V

I'll target for a 5mA current through the led.
So, RC = (12V - 1.94V )/ 5mA ≈ 2000Ω

And for the example in the book, they assume β > 25. I'm not sure what to consider for the BC547B.

#### MrChips

Joined Oct 2, 2009
23,966
And for the example in the book, they assume β > 25. I'm not sure what to consider for the BC547B.
Again, you are misinterpreting the value for β.
It is not to assume that β = 25 or any value.

What they are saying is, if we calculate using IB = IC / 25, then this circuit will function as required for any transistor whose β > 25.

Semantics.

• Delta Prime

#### Audioguru again

Joined Oct 21, 2019
3,327
Most if not all BC transistors saturate at a beta of 20 not 10.
That is because their saturation Vce voltage loss is higher when the base current is less than 2Nxxxx American transistors.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
Again, you are misinterpreting the value for β.
It is not to assume that β = 25 or any value.

What they are saying is, if we calculate using IB = IC / 25, then this circuit will function as required for any transistor whose β > 25.

Semantics.
I understand that but still I have to go with some β value, right? Otherwise, how you find the resistor value for Rb???

I tried to go with a β of 100 and got Rb = 53kΩ. To make things easier I used a 47kΩ + 2.2kΩ + 2.2kΩ = 51.4kΩ (real value is 50.78kΩ).
I put these values on the breadboard, and I get:
IC ≈ 5.02mA (I think it can't get any better)
Ib = 55μA (expected 50μA)
Vbe = 0.689V ( I used 0.65V in calcs)

So, I think a gain of 100 was kinda ok to use!

#### Audioguru again

Joined Oct 21, 2019
3,327
You so not show a schematic for the circuit (post #59) on your breadboard with a BC547B transistor with a 5.02mA collector current and a bunch of series resistors at the base with a total of 50.78k. The voltage feeding the resistors is unknown and how can the beta of the BC547B be only 100?