# common emitter max current Ic

Discussion in 'General Electronics Chat' started by donut, Jun 11, 2012.

1. ### donut Thread Starter Member

May 23, 2012
51
0
I am taking the initiative to teach myself about common emitter amplifiers.

I find hyperphysics to have a practical example of a common emitter circuit.

link provided for common emitter circuit.
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npnce.html#c2

So far by reviewing hyperphysics common emitter circuit example I learned that.

1) transistor has to remain in active mode in order to be used as common emitter amplifier

2) Using the transistor load line chart (Ic vs.Vce) of a 2N2222 you can determine the operating point that will keep the transistor in active mode. Ic = 4.5ma and Vce = 9V. (see transistor load line attachment)

I soon get lost in the example when I am told tha Ic max = 9 ma. 9 ma is used to derive Rc and Re (see screen shot common emitter gain)

My question is how did they get 9 ma?

Is 9 ma max Ic the maximum amount of current that the collector can withstand before turning from reversed biased (Vc > Vb) to forward biased (Vb > Vc)?

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2. ### panic mode Senior Member

Oct 10, 2011
1,328
305
although correct, example is not straight forward or easy to digest.
the diagonal line show operating curve of the transistor. one such point is 4.5mA at 9V for Vce.
you can also pick any other point on that line and it will be valid too. for example if transistor is saturated so that Vce is small, we will get the 9mA current trough resistors.

if transistor shuts off (Ib=0) then Ic=0 as well and if there is no collector current, Vce becomes 18V

one way to look at this is from KVL perspective,

Vbb= V1+V2+V3

where
V1=Ie*Re
V2=Vce
V3=Ic*Rc

If Vce=9V, we know we have 4.5mA that means that sum of voltage drops on resistors is the remaining 9V (because Vbb-Vce=9V=V1+V3)

If we pick point where Vce=0
we get Vbb-0=18V=V1+V3

also Ie and Ic are nearly identical so we can write

18V=V1+V3=Ic(Rc+Re)
Ic=18V/(Rc+Re)=18V/2k=9mA

3. ### panic mode Senior Member

Oct 10, 2011
1,328
305
normally however, transistor voltage drop is never 0V, more like 0.2V so max current would be

(18V-0.2V)/2k=8.9mA

to get transostor into this condition, we need base current sufficiently high.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
All small signal amplifier need to work on active region.
But to design these amplifier circuit we don't need any "load line chart".

All we need to know, if we want to design a BJT amplifier is:
1 - Supply voltage.
2 - What voltage gain we want.
3 - Internal resistance of the signal source.

Additional from the BJT data sheet Hfe_min and sometimes Vbe and Hoe are also needed.

First we choose Rc resistor.
Normally we pick Rc = 0.1*RL

Next to make sure that BJT work in the linear region and to achieve a large voltage swing you need to set the bias point (voltage at collector) equal to 0.5Vcc (and this assumption automatically ensures that the BJT will work in the active region).

And this imagines show why we need Vce = 0.5*Vcc

Vce=4.5V

Vce=3V

Vce=1V

So to amplifier our input signal waveform without distortion we need to set dc bias point at 0.5Vcc.

Try to analysis this example also

Vcc = 12V; Vc = 6V (red plot ); Ve = 2V (green plot) ; Vinput = 1V (blue plot).

Now we choose Ic current based on Rc values we chose previous.

Rc < 0.1*RL

Ic = 0.5Vcc/Rc

For good thermal stability we choose Ve to be grater then Vbe.
So for good thermal stability we usually choose Re = (0.1 ... 0.4)*Vcc/Ic or Ve large then 1V (Ve>>Vbe).

Next we choose voltage divider resistors

R1 = (Vcc - Vb) / (11 * Ib)

R2 = Vb / (10 * Ib)

And Vb = Vbe + Ve

And Ib = Ic/Hfe_min

Ksin301, anhnha and adam555 like this.