Choosing transistor quiescent current

[bugmenot]

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?

 

[crutschow]

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.

 

[bugmenot]

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.)

 

[Willbe]

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.

 

[k7elp60]

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.

 

[crutschow]

0.25mA should be fine depending upon the gain and frequency response you need. The Fairchild data sheet graphs for the 2N3904 from https://www.electro-tech-online.com/custompdfs/2008/09/2N3904.pdf go to 0.1mA

 

[audioguru]

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.

 

[Analog]

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.

 

[bugmenot]

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.

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.

My concern in this circuit isn't saving power, but rather that the input signal is very small.

 

[audioguru]

You can change the input and output impedances at the same time if you want.
Maybe you should use an opamp.

 

[silvarblade]

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?

 

[Willbe]

from
**broken link removed**
re (in Ω) = 26 / IE (in mA)

 

[audioguru]

The LM4562 dual audio opamp has a distortion of only 0.00003% (gain at 1) with a load of 600 ohms or more.

 

[vuong]

Please help me ,

audioguru said: ↑
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.

how to i calculate the load , in this example how to i find value 100k ohms depend on supplr power : 9v, orerating point 4.5v . collctor current 0.25ma ?

 

[KeepItSimpleStupid]

If your looking for a magic formula to give to 100K, look elsewhere.

You can however judge the effects of a parallel resistor and 18K i.e 18K vs 18L||100K

You know that 18K||18K is definately too low because it results in 9K. At some point 18K||Rx becomes significant". What "significant" means is anybodies guess. 100% error or 0.0000001% error.

 

[vuong]

sorry, i'a little english , can you teach me how to i calculate THE LOAD when i have output impedance (example Rc = 18 K )

 

[audioguru]

How do i calculate the load , in this example how to i find value 100k ohms depend on supplr power : 9v, orerating point 4.5v . collctor current 0.25ma ?

With a 9V supply and a 100k ohms load then the collector resistor value should be 10k at most then the output level and gain will not be reduced much. The idle current should be 0.4mA for a symmetrical output swing.
For an idle current of 0.25mA then the voltage drop across the collector resistor is 10k x 0.25mA= 2.5V which will result in a reduced non-symmetrical output swing.
If the collector resistor is 16k then the voltage across it is 16k x 0.25mA= 4V and if the voltage across the emitter resistor is 1V then the output will have the maximum symmetrical swing, the maximum output level will still be pretty high and the gain will not be reduced much.

 

[vuong]

With a 9V supply and a 100k ohms load then the collector resistor value should be 10k at most then the output level and gain will not be reduced much. The idle current should be 0.4mA for a symmetrical output swing.
For an idle current of 0.25mA then the voltage drop across the collector resistor is 10k x 0.25mA= 2.5V which will result in a reduced non-symmetrical output swing.
If the collector resistor is 16k then the voltage across it is 16k x 0.25mA= 4V and if the voltage across the emitter resistor is 1V then the output will have the maximum symmetrical swing, the maximum output level will still be pretty high and the gain will not be reduced much.

thanks to you , but what role of "100k ohms load" in this calute ?
is 'the load" parallel with Rc and we have Z= (the load)*Rc/[(theload)+Rc], the load is large so Z ~ Rc?

and is "the load" is another transistor (or same it) or "the load" is earphone , headphone,speaker or somethingelse ?

 

[vuong]

And example i have preamp

input is micro phone . how to i choose Ie for TR-1 , and TR-2 ?

 

[audioguru]

The load is whatever this transistor circuit is driving.
A speaker is not 100k ohms but another circuit's input might be.