Each output board get it's own connection to the power supply, each output board gets connected to the speaker terminals doubling the number of conductors for less resistance, perhaps more capacitance and inductance, who knows. I have not bench tested the amps, they are super stable, DC offset around 10 mv when thermal stability is reached. I hope this helps a few diy'ers. Cheers!
.350V or 0.7A pr device will give you a total bias of 2.8A on a 4 outputpair amp. not 5.6A.
Quiescent point 8 transistors conducting 0.7A. 5.6 A going from V+ to V-. That is what I understood from reading Master's F5 manual. If you only count one side of the push-pull configuration then the F5 has 1.3 A of bias and yes a F5T V3 will have 2.8 A bias.
Thanks for the photos.
By my calculations, if you wire it that way, it is unnecessary to use anything over 20 AWG.
For silver coated Belden 20 AWG, the Max Amps that may be carried is 6.5 Amps. As the boards are in parallel, the total amps that can be delivered to the speaker by the wiring is 13.
Finally, since P = I * V = I * I * R the max power supported by the wiring is:
169 watts into 1 ohms.
-
338 watts into 2 ohms.
-
776 watts into 4 ohms.
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1552 watts into 8 ohms.
Using 20 AWG and wiring the output boards in parallel does not pose an issue.
If the boards are wired in series, then the limit of 6.5 A is at the power supply side and the output side.
Now, the max power supported by the wiring is:
42.5 watts into 1 ohms.
-
85 watts into 2 ohms.
-
170 watts into 4 ohms.
-
340 watts into 8 ohms.
This may be a stupid question, but what determines the max power output after the amp exhausts the class A bias and transitions to class B operation? How is that calculated?
For example, I measure 320 mV across my source resistors, which means I'm running about .64 amps of bias current per output channel. This gives me about 104 W peak class A into 8 ohms and 73 W RMS. Will the amplifier just clip past this point, or will it transition to class B? If it transitions, how can I calculate how many extra watts are on tap?
For example, I measure 320 mV across my source resistors, which means I'm running about .64 amps of bias current per output channel. This gives me about 104 W peak class A into 8 ohms and 73 W RMS. Will the amplifier just clip past this point, or will it transition to class B? If it transitions, how can I calculate how many extra watts are on tap?
standar V2 = 2 outputpairs. 0.64A pr device = 1.28A bias. that is about 25W class A at 8ohm.
peak class A pover is 2 times bias. so 2.56A peak.
2.56/1.414=1.81Arms.
1.81x1.81x8=26.2W class A at 8ohm.
32V rails minus about 3V loss in the curcuit is 29V
29/1.414=20.5Vrms
20.5x20.5/8=52.3Wrms class A/B at 8ohm.
peak class A pover is 2 times bias. so 2.56A peak.
2.56/1.414=1.81Arms.
1.81x1.81x8=26.2W class A at 8ohm.
32V rails minus about 3V loss in the curcuit is 29V
29/1.414=20.5Vrms
20.5x20.5/8=52.3Wrms class A/B at 8ohm.
Exactly.
Of note is that class A/B simply means that the amplifier works in class A for part of the time. Remember the explanation of when the F5 leaves class A in the F5 manual? Look at the current flow this way:
At idle both N channel and P channel MOSfets are conducting the same amount of current.
When a speaker is attached and there is no signal ( the amp is idle ) the current *still* flows only through the transistors, and not through the load. Why?
Because the load is at 0 potential. It is less inviting for current to flow between the rails and zero than between the positive and negative rails. And biasing has balanced the amount of current that flows through each MOSfet at idle.
When a signal is applied, the current flow is unbalanced. Let's say the P Mosfet begins to close and the N MOSfet opens more.... there is more current flowing through the N MOSfet, and Less through the P.
The current that was once flowing through the P channel MOSfet needs to go somewhere.... it goes through the load. So you can think of the signal "steering" the current away from one MOSfet and through the load instead.
As the P channel restricts X amps, the N channel allows X amps. So, as half of the flow of the bias current in the P channel is subtracted, half is added to the N channel. When the P channel has blocked all current, the N channel has had it's current doubled.
At this point, the amplifier is operating in class B.... one transistor is conducting ALL of the current, and ALL of it is going through the load... the other transistor is pinched off.
As the signal polarity inverts, both MOSfets begin to conduct again, back into Class A operation, and the cycle repeats with roles reversed.
So Peak Class A current is twice the current flowing between the rails at idle-- which is twice the bias current.
Class B Peak is limited by the rail voltages.
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Of note is that class A/B simply means that the amplifier works in class A for part of the time. Remember the explanation of when the F5 leaves class A in the F5 manual? Look at the current flow this way:
At idle both N channel and P channel MOSfets are conducting the same amount of current.
When a speaker is attached and there is no signal ( the amp is idle ) the current *still* flows only through the transistors, and not through the load. Why?
Because the load is at 0 potential. It is less inviting for current to flow between the rails and zero than between the positive and negative rails. And biasing has balanced the amount of current that flows through each MOSfet at idle.
When a signal is applied, the current flow is unbalanced. Let's say the P Mosfet begins to close and the N MOSfet opens more.... there is more current flowing through the N MOSfet, and Less through the P.
The current that was once flowing through the P channel MOSfet needs to go somewhere.... it goes through the load. So you can think of the signal "steering" the current away from one MOSfet and through the load instead.
As the P channel restricts X amps, the N channel allows X amps. So, as half of the flow of the bias current in the P channel is subtracted, half is added to the N channel. When the P channel has blocked all current, the N channel has had it's current doubled.
At this point, the amplifier is operating in class B.... one transistor is conducting ALL of the current, and ALL of it is going through the load... the other transistor is pinched off.
As the signal polarity inverts, both MOSfets begin to conduct again, back into Class A operation, and the cycle repeats with roles reversed.
So Peak Class A current is twice the current flowing between the rails at idle-- which is twice the bias current.
Class B Peak is limited by the rail voltages.
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