How about taking a look at this circuit diagram:
https://www.diyaudio.com/community/...uild-awesome-prototype-material.256334/page-3
Just an example of a Totem Pole circuit:
Given a single B+ to Ground power supply:
The B+ current through a Totem Pole, from top to bottom is one current.
B+ has to supply that current.
If the top tube has 10 mA, then the bottom tube has 10mA (Series connection that is called a Totem Pole in this case).
That means the B+ has to supply 10mA of quiescent current.
Now, if the top device in the totem pole during maximum signal swing has to supply 10mA into the load, that means the peak current of the top device stands 10 mA quiescent + 10 mA signal output = 20 mA.
That means the B+ filter cap has to supply the extra 10 mA during the signal peak.
However, if the totem pole is Class A, when the signal polarity goes the other way, then the top device goes very close to cut off (close to 0 mA).
That means the Average current the B+ is supplying (20 mA + 0 mA) / 2 = 10mA.
The 10mA quiesent current that is in the bottom device comes from the top device (it does not come directly from B+, instead it comes from the top device).
The quiescent load on B+ is 10 mA, Not 20 mA.
Yes, I used 10 ma quiescent current load on B+, I did not use whatever quiescent current you will be using in your totem pole.
If the above does not explain it, you have Lots of study to do first.
If you want 20V Peak into 600 Ohms, that is 20V peak / 600 Ohms = 33.3 mA Peak
40Vpp into 600 Ohms is + 33.3mA, then - 33.3mA
It looks like the bottom device has to pull down 33.3 mA peak to ground.
Top device has to stand its quiescent current, then it has to pull up 33.3 mA peak.
Given a single B+ to Ground power supply:
The B+ current through a Totem Pole, from top to bottom is one current.
B+ has to supply that current.
If the top tube has 10 mA, then the bottom tube has 10mA (Series connection that is called a Totem Pole in this case).
That means the B+ has to supply 10mA of quiescent current.
Now, if the top device in the totem pole during maximum signal swing has to supply 10mA into the load, that means the peak current of the top device stands 10 mA quiescent + 10 mA signal output = 20 mA.
That means the B+ filter cap has to supply the extra 10 mA during the signal peak.
However, if the totem pole is Class A, when the signal polarity goes the other way, then the top device goes very close to cut off (close to 0 mA).
That means the Average current the B+ is supplying (20 mA + 0 mA) / 2 = 10mA.
The 10mA quiesent current that is in the bottom device comes from the top device (it does not come directly from B+, instead it comes from the top device).
The quiescent load on B+ is 10 mA, Not 20 mA.
Yes, I used 10 ma quiescent current load on B+, I did not use whatever quiescent current you will be using in your totem pole.
If the above does not explain it, you have Lots of study to do first.
If you want 20V Peak into 600 Ohms, that is 20V peak / 600 Ohms = 33.3 mA Peak
40Vpp into 600 Ohms is + 33.3mA, then - 33.3mA
It looks like the bottom device has to pull down 33.3 mA peak to ground.
Top device has to stand its quiescent current, then it has to pull up 33.3 mA peak.
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Quality content! Would you mind talking about how you landed on the 680R, 330R, and 82R values around the WCF?
Although it is now a few years ago when I designed my Hdph-amp, I remember that my design goal was an output power for "fairly loud" levels.
For my Sennheiser HD650, I used this as guide:
Also you can use the headphone power calculator by visiting this site:
https://gadgetmates.com/headphone-power-calculator
Then you can calculate the required currents for delivering those power levels to your headphone impedance using the specs of the headphone manufacturer.
And..... LTspice is your friend 🙂
When my pcb's arrived, I used ARTA and the sound card of my desktop computer to measure the THD of the chosen output level ("moderate" and "fairly loud").
In the end I came up with those resistor values to meet my requirements.
