Thanks Jony.100μF and 8Ω load resistane gives Fc ≈ 200Hz
And the phase shift at 1kHz will be around φ = acr cot(1kHz/200Hz) = 11°.
That phase angle will give as a time shift Δt, for the period of F = 1kHz---> T = 1ms
Δt = (11°/360°)*1ms = 30.5μs
And yor input signal slew rate is 500mV*1kHz *2*pi = 3.142mV/μs. Hence after 30.5μs, we have around 96mV.
And this is why yor output is starting at 90mV. And this has nothing to do with the dc-offset or distortions. This is just how a simulation shows the phase shift. So, not need to worry.
What do you mean by "correct"? Yor math is corrent.So C = 1 / 2 * π * f * RL = 1 / 2 * 3.14 * 1000 Hz * 8Ω = 19.90uF. is correct?
What do you mean by "correct"? Yor math is corrent.
Is this value is correct or not it will depend for example, the application the technical requirement or capacitor size or cost.
Also you shoud keep in mind that at this frequency (1000 Hz) the phase shift is equal to 45° degrees and RC circuit gain is 0.707.
Yes, because of the fact that the positive gain (NPN) is different than the negative gain (PNP).
No, not always. In your circuit, the negative gain is larger.Is the positive gain always greater than the negative gain?
Yes exactly Q1 was added to drive the output stage (via Ic1 current ) but you also want gain from this CE amplifer too.I don't really understand this
circuit configuration. I do understand that the CE stage provides both voltage amplification and bias current for the output stage but beyond that I am lost.
No, not always. In your circuit, the negative gain is larger.
Yes exactly Q1 was added to drive the output stage (via Ic1 current ) but you also want gain from this CE amplifer too.
Can you be more specific what you do not understand in this circuit?
The gain is different because the output transistors are not perfectly matched. Hence, the NPN voltage follower will have a slightly different gain than the PNP one.Why is my negative gain higher? What can I do to equalize the
positive and negative gain?
You are asking about what? The voltage at Q7 and Q8 emitters should be close to 1/2Vcc = +6V for 12V supply.Should the dc voltage between the Q7 and Q8 emitters not be 6 volts?
You are asking about what? The voltage at Q7 and Q8 emitters should be close to 1/2Vcc = +6V for 12V supply.
The Q10, Q7 and Q9,Q8 are nothing more than the voltage followers. And thank to this complementary connection Vin = Vout (no DC-offset)
http://www2.engr.arizona.edu/~brew/ece304spr07/Pdf/Class AB with VF.pdf
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by Jake Hertz
by Robert Keim
by Jake Hertz