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These are my notes from
https://www.youtube.com/watch?v=WZD9RZoMhVE
I am not an expert, use at your own risk.
Vc > Ve, Vbe > 0.6V
Ic = beta * Ib where beta is about 100
Ie = Ic + Ib
beta depends, but we are ok if we underestimate, so 100 should be fine.
Vout is always Vin - 0.6
We control Vout independent of Vcc
We can say that the input impendence is much larger than the output impendence
For any circuit Rin >> Rout or the signal will be distorted at least Rin = 10 * Rout.
Zout = Zin / ( beta + 1)
Therefore this way a weak input signal (e.g. a digital one) can drive a large output e.g. a relay.
For amplification we have to add DC to the input or we would cut the lower halves of the sine waves, because the transistor will not be open iv Vb < 0.6 V
The large resistors on the base make a voltage divider. The added DC depends on their ratio.
The capacitor is for filtering the frequence (for e.g. an analog input like sound). We do not care about already added DC offset.
The output diode is 2V 20mA, I want to allow max 1.5V and 10mA.
Vout = Vin / 3.33
Iout = Vout / Rout
0.01 = 1.5 / Rout
Rout = 150 ohm
But I want to keep parallel resistors to control the flow 1/R = 1/Ra + 1/Rb
1/150 = 1/300 + 1/300
The V at base is Ve + 0.6 = 1.5 + 0.6 = 2.1, but this is the value in the lowest level, so in fact I have to calculate 3.1 V
The impendence on base is about 100 times larger 150 ohm * 100 = 15k
the impendance of the DC source should be at least 10 time less than the base impendence. therefor 1.5k.
For the divider we need 3.1V at base
The ratio is 3.1/1.9 = 1.63
R1 || R2 = 1.5K
R2 = 1.5k * 2.63/1.63 = 2420
R1 = 2420/1.63 = 1484
In reality I should go for more
1/R = 1/R1 + 1/R2
R1=2580
R2=3580
The kapacitor
fin = 1/(2piRinC1)
the divider an the base 1.5k Ohm || 15k Ohm =1363
1kHz cut of capacitor
1/(2pi * 1000 * 1363)
1.17e-7 so 117 nF should be -3 dB -> which should divide the signal by 2 - looks correct.
0.000000117
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