# Base Current on CC BJT Amplifier

Discussion in 'General Electronics Chat' started by ELECTRONERD, Jul 25, 2009.

1. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
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Hello Folks,

I have a standard CC amplifier with a voltage divider at the base. What I would like to know is how to determine the current of the base with both 10K resistors as R1 and R2? If I want a lower current couldn't I simply connect a resistor between the voltage divider and base? Thanks for all your help, I appreciate it!

2. ### Wendy Moderator

Mar 24, 2008
20,772
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Schematics are worth thousand words. They prevent misunderstandings.

Having said that, if I read you correctly, you need to figure the approximate input impedance of the transistor input, which is the ß X Emitter resistance. This will figure into the base voltage divider as a parallel resistor across R2.

Here is an illustration I used for my Virtual Ground article.

Last edited: Jul 25, 2009
3. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
1,146
16
Thanks Bill, sorry I didn't make out a schematic. So you're saying that the base impedance is β x Re? Being the case, if I want to lower the input current (thus increasing the impedance) could I place a resistor between the voltage divider and base as shown in the attachment? This seems like it would be fine, is there any disadvantages?

Thanks Bill, I hope I can be as intelligent as you some day You're setting a fine example for the rest of us!

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4. ### Wendy Moderator

Mar 24, 2008
20,772
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I ain't that smart, but I've got a lot of experience under my belt. This means I can recognize a mistake the next time I make it.

If you increase RL then the base impedance goes up. You could also use a Darlington to increase the gain (it would increase the Vbe drop too), but would have the same effect.

So if you have a ß of 100, and a emitter resistor of 1KΩ, then the base resistance is 100KΩ. From there it just becomes part of the math.

Putting a simple resistor in front of the base just creates another voltage divider. IMO not the best solution. Either increase the emitter resistance or the ß.

I used that drawing because it was the closest thing I had in my library. It isn't the best though.

5. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
1,146
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Okay Bill, I went ahead and designed my own circuit and it works pretty good! Now couldn't I build another CC circuit but to change the impedance? 1.5kΩ is a fairly high value, even with a gain of ten.

I use the "CircuitMaker" program and unfortunately, I'm not quite sure how to let wires come out to the output and input of the circuit, it won't let me do it unless I have it connecting to another component. Maybe you can tell me.

Thanks!

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6. ### Wendy Moderator

Mar 24, 2008
20,772
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Like I said, a Darlington would do it, you'll be suprised how high that impedance goes.

7. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
1,146
16
Hey Bill,

Okay, well I have the KSD985 from Fairchild:

(http://www.fairchildsemi.com/ds/KS/KSD985.pdf)

Will this dramatically improve the impedance? Would my suggestion help it even more (adding another CC circuit to simply change impedance)? I'll go ahead and try it. Let me know if this is a good transistor for this.

As you might have found out...I'm not that familiar with transistors, but I intend to design with them and increase my knowledge.

8. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
1,146
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Ok, I replaced the 2N3904 with the KSD985 Darlington transistor and I didn't notice any considerable difference. It sounded like it had the same volume level.

9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
A common collector amp is a unity gain amplifier, that it is has a gain very close to 1 (usually several 9's, as in 0.999). It amplifies current drive, it will take a 1KΩ load and make it look like a 200KΩ load.

One of my favorite uses is the simple CMOS circuit, where the CMOS chip has trouble providing 1ma on the output. The following circuit will provide the LED 20ma, while drawing less than a .5ma from the CMOS chip.

There are a lot of applications where you don't want to load a source much, but need to drive something else (with the same level) hard.

That are you trying to do?

10. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
1,146
16
Hello,

While experimenting on my CC amplifier I found that I could only have a maximum gain of 30 since my batteries could only supply about 22mA of current max. So once I completed the circuit I had good results, now I had a transformer nearby and decided to change the impedance. This is where I found something unexpected! At first I hooked the output of the amp to the larger coil of the transformer and the earphone to the smaller coil. Interestingly enough, it reduced the audio level! I quickly tried the inverse and it worked louder! That must conclude that the ceramic earphone is a high impedance earphone! How could this be? The earphone is sensitive enough to make crackling noises when you touch the end of the leads to another piece of metal, a high impedance earphone wouldn't make it sensitive would it?

11. ### mik3 Senior Member

Feb 4, 2008
4,846
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If you want to lower the input current you can lower the base voltage or increase the resistance of the load. By putting a resistor between the voltage divider and the base you will reduce the base current but the circuit will loose its properties as a stiff voltage follower. If this resistor has a high value the output voltage will depend on the transistors gain (taking into account that the voltage divider forms a stiff voltage source for this circuit).

12. ### hobbyist Distinguished Member

Aug 10, 2008
773
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At the top is a button with a plus symbol on it, that is the wire placement tool, you can use that like a bus wire as well,

single click and draw your straight wire,

then double click and the wire will stay on the screen without connecting it to anything.

13. ### ELECTRONERD Thread Starter Senior Member

May 26, 2009
1,146
16
Thanks Hobbyist! Appreciate the help.