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#### jigsaw55

##### New Member
hi
here, output voltage is VRL.
how can i find emitter current of Q5

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• hw 001.jpg
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#### audioguru

##### Well-Known Member
The amplifier circuit with negative feedback will try to keep the output DC output voltage at the same voltage as the input which is 0V.
Then the 10k load resistor has a current of 5V/10k= 0.5mA.
A tiny amount of current (only 2uA?) is the base current of Q2 which can be ignored.
Then the emitter current of Q5 is 2.5mA plus its small amount of base current (maybe 20uA) plus the 2uA.

#### jigsaw55

##### New Member
so thank u , one more question .... is there any more analyitcs way or analysis already begin with assuming Vo=0 DC.?

#### audioguru

##### Well-Known Member
Sorry, I didn't notice that R1 connects to -5V instead of to 0V. Then the negative feedback through R2 will make the output voltage as high as is possible which is about +1.4V. Then the emitter current of Q5 is about 2.75mA.

#### jigsaw55

##### New Member
sorry but i cant see the point. feedback wants to keep output equal to input ok .. but there are 2 inputs here..we set them 0 for dc then I m confused :S ..May u explain with math process, please?

#### audioguru

##### Well-Known Member
The input at the left side is connected to 0V through the 100k input resistor and signal generator.
The input at the right side has a 100k resistor to -5V (-50uA) so the output into the 1M negative feedback resistor R2 will try to go to 50uA x 1M= +50V.
The output cannot go to +50V so it goes as high as it can which is about +2.7V.

The math is simply Ohm's Law and addition.

#### jigsaw55

##### New Member
ok i see it cant be 50V , but where the upper limit 2.7 comes from i didnt find...do u assuume VBE=0.7 or something... i m sorry to busy u

#### audioguru

##### Well-Known Member
Here is how I calculated the voltages (without including very small bias currents):

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• biasing.PNG
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#### jigsaw55

##### New Member
ok ı missed saturation point .i got it .. so thank you

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