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So there is this common emitter transistor and it says Ic = (Vcc-Vce)/ RL, what I'm not understanding is why Ve is totally ignored, I thought Ic = (Vcc-Vc)/RL. How come the Emitter resistor is totally left out of the calculation when it does affect the value of Ic.

The link to the whole page: http://www.electronics-tutorials.ws/transistor/tran_2.html

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  • \$\begingroup\$ For God's sake fix the question to show all of it. \$\endgroup\$ – WhatRoughBeast Dec 25 '17 at 1:00
  • \$\begingroup\$ There is no question in the picture, it's an explanation that I don't understand and I could not put all of the text in one pic, this is the link electronics-tutorials.ws/transistor/tran_2.html \$\endgroup\$ – BigRedMachine Dec 25 '17 at 1:15
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You are right. At DC, the collector current is the current thru Rl, which is (Vcc - Vc) / Rl.

Parts of what you snipped were cut off, but maybe the part you quote is talking about the AC characteristics. Note C2. It holds the emitter a at 0 V for the purpose of AC analysis, below some frequency.

Put another way, this circuit has higher gain at AC than DC. The equation for the AC component of the current thru Rl is different than that for DC.

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  • \$\begingroup\$ Hmm, I believe it's talking about DC because I still haven't got to the AC analysis and in the text it doesn't say anything about it, what could be other scenarios that Re is ignored? here is the link if you want to look at it: electronics-tutorials.ws/transistor/tran_2.html \$\endgroup\$ – BigRedMachine Dec 24 '17 at 23:37
  • \$\begingroup\$ That equation is talking about the upper figure in the page: electronics-tutorials.ws/transistor/tran43.gif \$\endgroup\$ – user16307 Dec 25 '17 at 1:37
  • \$\begingroup\$ Yes, Ic = (Vcc-Vc)/RL and there is a mistake in the tutorial. All the formulas are for a simpler version of the circuit where RE is omitted. The point of RE is to make the DC biasing less sensitive to variations of β, and RE is bypassed with capacitor to still preserve high AC gain. \$\endgroup\$ – Vesa Linja-aho Jun 27 '19 at 8:48
  • \$\begingroup\$ Yes - and that`s what you (and we all) should learn: Never blindly trust formulas and explanations to be found in internet contributions and - sorry to say - this applies also to some printed textbooks. \$\endgroup\$ – LvW Jun 27 '19 at 9:08
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When designing a common emitter amplifier, one of the first thing you do is assume a value for the collector current (usually somewhere between 0.5mA and 5mA). Then use your equation, RC = (Vcc-Vc)/Ic, to calculate the value of RC which, at the chosen collector current, puts the collector at mid-supply voltage.

Next values for RE and the base bias resistors are chosen to give an emitter current which is equal to the required collector current. (we assume that Ie = Ic).

So it is effectively the base bias resistors and the emitter resistor that set the collector current, and if the collector resistor has been sized correctly then this collector current drops the correct voltage across the collector resistor to bias the collector voltage to mid-supply for equal possible output swing in either direction.

So collector current,Ic = (Vcc-Vc)/Rc which is approx. equal to the voltage at the emitter divided by the emitter resistance, Ic = Ve/RE

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