Choosing transistor quiescent current
I see in the guides to standard Bipolar Transistor circuits (emitter-follower, common emitter amplifier), they always say: Choose a queiscent current of 1mA.
My question is: Why 1mA? Where does this number come from? Can I make it less, and save power? Is there a reason not to?
It's quite arbitrary. It's a value that works well for typical small signal transistors and circuits. If power is a concern, then you can certainly operate at lower currents. Low power op amps have some of their internal transistors operating on microamps of current.
Just be aware that there's usually a tradeoff between current, and the frequency response, noise, and gain of a circuit. You seldom get something for nothing.
Carl
Curmudgeon Elektroniker
What is the minimum amount of current I can get good results with a standard 2N3904 or 2N4400? Is 0.250 mA too little?
Also - given a transistor's data sheet, is there anyway to find this out? (I couldn't find it listed, and, indeed, most of the graphs started at 1mA.)
The Vbe drop changes linearly -2.2 mV/°C, and in some cases it is is important to take this into account. In any case, the Vbe drop for silicon is 0.7 V, so (Ie) x (emitter resistor value) should be >0.7 V for temp. stability.
For a class A common emitter amplifier, I usually center bias the transistor so the collector voltage is 1/2 Vcc, that is with no input signal. The collector load resistor depends upon output impedance. So these two thing affect Ic and the quiescent current.
The great thing about electronics is unlimited ways to do the job. The only limit is one\'s imagination. I generally think my way is best.
Show me a different way. I have an open mind.
0.25mA should be fine depending upon the gain and frequency response you need. The Fairchild data sheet graphs for the 2N3904 from www.fairchildsemi.com/ds/2N/2N3904.pdf go to 0.1mA
Carl
Curmudgeon Elektroniker
For example, if the supply is 9V, the collector operating point is 4.5V and the collector current is only 0.25ma then the collector resistor is 4.5V/0.25ma= 18k. the output impedance is 18k and will be affected if the load is less than about 100k ohms.
If the current in the transistor is 10 times higher at 2.5mA then it will be able to drive a load as low as 10k ohms.
Uncle $crooge
The collector current affects the internal emitter resistance re', which in turn will affect the gain in a CE amplifier. 1mA is decent for a 3904, should give decent gain without sacrificing self heating and Vbe problems over temperature.
"Everything that is done in the world is done by hope." -Martin Luther
"There are two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle."-Albert Einstein
Right - but won't that require a Re one-tenth the size as well, which will make the input impedance ten times smaller (input impedance = hfe * Re)? So, you've gained the ability to drive more, but need to be driven by more as well.Originally Posted by audioguru
My concern in this circuit isn't saving power, but rather that the input signal is very small.
You can change the input and output impedances at the same time if you want.
Maybe you should use an opamp.
Uncle $crooge
i dont think i need to say anything but just to ask bugmenot a question i am forced to!
For the Problem:
Q point/operating point(DC Values) is set using Re and it also flows in collector(Rc). set by Ve/Re.
Ve=Vb-Vbe.
In a ce amp Av=Rc/Re.normally it is said to be Rc=10Re.Correct me about the gain thing if i am wrong.
Opamps are way better.
PS:Bugmenot! is the site bugmenot.com yours?
I just cant come up with a good one!
from
http://sound.westhost.com/no-opamps.htm
re (in Ω) = 26 / IE (in mA)
The LM4562 dual audio opamp has a distortion of only 0.00003% (gain at 1) with a load of 600 ohms or more.
Uncle $crooge
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