Hi:
I am still trying to figure out some stuff with these Class AB amplifiers. Its driving me
nuts. I am so stressed that I have lost all my hair! (Seriously, I kid you not). Even
though I now resemble Kojak, minus the lolly pop, I am not giving up.
Attached is a sim.
According to my calculations:
Ibias = Vcc * 1.4 / 2* 180Ω = 51.67 mA
Isat = Vceq / RL = 1.25 A
IC_avg = Isat / π = 1.25 A / π = 1.25 A / 3.14 = 398 mA
Does IC_avg only apply when there is an input signal?
According to a textbook that I read:
Idc = Ibias + IC_avg = 51.67 mA + 398 mA = 449.67 mA
However, LT Spice says Idc = Ibias + Icq = 51.67 mA + 18.35 mA = 70.02 mA with
or without an input signal.
I want to know where the 18.35 mA comes from and how Spice came up with it?
Thanks,
David
I am still trying to figure out some stuff with these Class AB amplifiers. Its driving me
nuts. I am so stressed that I have lost all my hair! (Seriously, I kid you not). Even
though I now resemble Kojak, minus the lolly pop, I am not giving up.
Attached is a sim.
According to my calculations:
Ibias = Vcc * 1.4 / 2* 180Ω = 51.67 mA
Isat = Vceq / RL = 1.25 A
IC_avg = Isat / π = 1.25 A / π = 1.25 A / 3.14 = 398 mA
Does IC_avg only apply when there is an input signal?
According to a textbook that I read:
Idc = Ibias + IC_avg = 51.67 mA + 398 mA = 449.67 mA
However, LT Spice says Idc = Ibias + Icq = 51.67 mA + 18.35 mA = 70.02 mA with
or without an input signal.
I want to know where the 18.35 mA comes from and how Spice came up with it?
Thanks,
David
Last edited: