# Transistor circuit Design

Discussion in 'General Electronics Chat' started by shreeram, Nov 29, 2012.

1. ### shreeram Thread Starter New Member

Nov 29, 2012
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How can i design the Base bias transistor circuit...
I found the answer that , use : Rc=(Vcc-Vce)/Ic
And Rb =(Vcc-vbe)/Ib
but my doubt is that how can we get Ic or Ib before the circuit design??

2. ### MrChips Moderator

Oct 2, 2009
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Show a circuit diagram.

3. ### Audioguru New Member

Dec 20, 2007
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A transistor usually has a voltage divider providing a certain voltage to the base. The current in the voltage divider is 10 times the typical base current of the transistor so the base voltage does not change much when the transistor has low hFE or high hFE.
The emitter has a resistor to ground so that a transistor with a low Vbe or a high Vbe makes little difference. The emitter resistor can be bypassed with a capacitor for high AC voltage gain.

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4. ### bretm Member

Feb 6, 2012
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If he's being asked to design a base-bias circuit, if it's anything like the textbook I'm looking at it would be like the voltage-divider circuit but with R2 removed. And since he said Rc=(Vcc-Vce)/Ic, it means Ve=0 which means short out R4 as well.

Not a good circuit, but intended for teaching the disadvantages of the circuit compared to voltage-divider biasing.

Vcc is known. Vce is known (you usually choose it to be Vcc/2 as your Q point). Vbe is known (you usually assume it's 0.7V). Ic is known (it's chosen as part of the design).

Is that enough to get you unstuck?

5. ### Audioguru New Member

Dec 20, 2007
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NOBODY biases a transistor with its emitter grounded and the base current is from a resistor to the positive supply.
It is a thermometer.
When the transistor is warmed (maybe by its own current) then its hFE increases.

It works only if the hFE is a certain amount. You cannot buy a transistor with a certain amount of hFE because they are all different, even if they have the same part number.

The circuit I showed has only a small effect by a different hFE or temperature change.

6. ### bretm Member

Feb 6, 2012
152
24
Nobody except textbooks teaching how bad this circuit is. I'm guessing this is just a homework problem.

7. ### shreeram Thread Starter New Member

Nov 29, 2012
29
1
Ok fyn..Whatever the circuit it may be (base bias circuit,or collector to base bias circuit or voltage bias circuit)..BUT I AM CONCERNED , How to design the circuit i.e to find resistor values to the circuit..
How can we get to know Ic,or Ib values before knowing the resistances..??
In solving problems they give Vcc ,Vce,(Ve),Ic,Ib,Hfe(or beta value) ....My doubt is How come they are knowing the Ic or Ib (currents)...
{no problem with Vcc. and (hfe which is given by manufacturer)}

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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First you go to shop and buy a BJT. For example you buy BC548B.
Next you look into the data sheet and check for Hfe_min = 200 and Hfe_typical = 250.
http://www.b-kainka.de/Daten/Transistor/BC548.pdf
Now you start design your circuit.
For example we have Vcc = 10V and we want Vce = 5V.
So we choose Ic = 10mA
Rc = (10V - 5V)/10mA = 500Ω so we use 470Ω or 510Ω resistor.
We calculate the base current
Ib = Ic/Hfe = 10mA/250 = 40μA
Therefore
Rb = (Vcc - Vbe)/40μA = (10V - 0.65V)/40μA = 233KΩ so we use 220KΩ or 330KΩ.
Next we build the circuit and we test it.
I build this circuit on the breadboard and use two different BC548B (form different manufacture ).
And the results of measurements are :

Vce = 5.7V and Vce = 5.1V for Rb = 330KΩ and Rc = 510Ω.

As you can see the result are not so bad as for such a simple and bad circuit.

We have a simpler situation in circuit when BJT need to work as switch.
The simply we select Ic/Ib = 20 ... 10 and we have the end of a story.

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9. ### shreeram Thread Starter New Member

Nov 29, 2012
29
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Thank you.. 80% of my doubt is cleared..
So you say that ...we randomly assume some Ic current ??
If you don't mind (can you say)is there any particular way to do so??

10. ### ramancini8 Member

Jul 18, 2012
447
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Normally, the transistor is driving a load, another stage, a speaker, a soleniod, etc., and the load determines the Ic, although sometimes indirectly.

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11. ### shreeram Thread Starter New Member

Nov 29, 2012
29
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Out of this topic !!!
how to post Signature in (allaboutciruit.com)
Can u say the steps??

12. ### Audioguru New Member

Dec 20, 2007
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The simple transistor circuit is a thermometer. Heat the transistor and watch it conduct more then cause distortion. Cool the transistor and watch it conduct less then cause distortion.

The simple circuit works when the hFE of the transistor is "typical" but the transistor is almost cutoff if the hFE is low and it is almost saturated when the hFE is high.

You cannot buy a transistor with a certain hFE, you get whatever they have. The hFE for a BC548B is selected to be a range in the middle of all BC548 transistors. A BC548A has a low range of hFE and a BC548C has a high range of hFE but they still have a range.

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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No, the load determines the Ic current that we choose in a particular application.

14. ### shreeram Thread Starter New Member

Nov 29, 2012
29
1
THANKS for replies...
Other answers are also welcomed...

15. ### #12 Expert

Nov 30, 2010
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It is true. Most circuits are designed from the finish to the beginning. The final load demands a particular IC from its driver, and you are designing a driver. The final load might be a speaker, another transistor stage, the input to an integrated chip, a length of shielded cable (at a particular frequency), or many other things. One form of thinking that helped me is to assume the circuit will work correctly, then calculate backwards to see what it needs.

Of course, you will try circuits that seem right, then find out it will get too hot or fail at the highest frequency you need. Then you change your design. Almost nobody gets it right the first time. Besides that, as a project gets larger, more parameters must be balanced from stage to stage and compromises must be made. Right now, you are doing one stage at a time. You are supposed to discover what happens when you focus on one aspect then calculate what will happen to other aspects because of what you chose at first. This is the experience that will eventually teach you how to guess very close to what you need the next time you need a circuit like this.

As Audioguru said, this circuit is useless in a production line design, but that is one of the things you are supposed to learn from this exercise.

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16. ### ramancini8 Member

Jul 18, 2012
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Japan designers made millions of circuits the way you described. Yes, they are beta dependent, but they selected the transistors for the correct beta. This was in the days when components cost more than labor. Yes, the circuits were temperature dependent, but they were not guaranteed to operate except at room temp.

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17. ### #12 Expert

Nov 30, 2010
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Good answer! You can make this work on a production line basis if you sort every transistor for gain and only promise a very narrow range of temperature. But that was then. In todays world, sorting transistors like that is no longer considered "production line" behavior.

ps, thanks for the education. I've seen single resistor bias in several devices over the last 40 years and wondered how the designer got away with that. Now I know.

18. ### Audioguru New Member

Dec 20, 2007
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I got one of the first "transistor" radios for my birthday. It was nice, small and had a leather case. It had 6 transistors. I betcha I still have it somewhere.

My friend HAD to have a better one so he bought one that had engraved on its cover "14 Transistors". Inside it had 8 transistors connected together in a circle. They were not part of the circuit.

19. ### MrChips Moderator

Oct 2, 2009
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Here is my 6 transistor radio. Still works!

Last edited: Dec 1, 2012

Nov 29, 2012
29
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thank you..