Y.A.F5Tv3.B.T. Yet another F5Tv3 Build Thread

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Hi guys,
i've finally decided to collect all my "garbage" heavy stuff and make this amp.

I want to go the "hard way", so the rev.3 was choosen.

Dual mono-block, for sure.. I can machine on a proper CNC 3-axis my schizoid delusions, so in the end the amps will appear something like attached pic.

At the moment, i have on my desk 2 heatsinks : 300x400mm h.86, they weight 9.5Kg each, and by the specs, by natural convection, they give a nice 0.18 K/W figure.

I have also two big irons, 1400VA each, with only the primary winding wound with a 15% more of the nominal flux for that geometry. On top of that i have a screen winding and then, when i finally choose the rail voltages i'll bring them to a fellow's company that finalize the windings.

For the filtering bank i am not sure what path to follow... in my mind i have a CLC filtering, by simulation a 1mH coil will give a nice output (compared to a CRC)... any advices on that ? The capacitance will be donated by an array of 4x22mf each rail, for each side.

I need help in taking some decisions about the rail voltages / bias.
I have made some calcs, and they went that way :
Starting from the end i want an increase of 20 degC above room temp.
so, given 0.18k/W is 20 / 0.18 = 111W of thermal power.

The topology is a pushpull, so in best case i will have a power efficenciency of 40%. 111W * 0.4 = 44W of class-A power.

in making a 100W A/B class on 8ohm i will give -> sqrt(100*8) = 28V rms or
39v peak-t-peak.

44W / 39v = 1.12A bias. cut in half and i have 0.56A per bank of N and P.
0.56 / 4 elements = 0.14A each fet.

To that point, how many mistakes i have done ?? :D
If all the things are correct, i will have 27Vrms for the secondaries and 44w of full class-A and capability to arrive at 100w in AB.

Please, correct me if i am doing something stupid about.

I will follow with the developments.
Thanks !
 

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You should not use the 40% efficiency in your calculation.
Say you can dissipate around 120W each channel and you want 100Wpc class AB.
Use 44 volt rails. That gives a clean 40volt sine wave. Or 100 Watt in 8 Ohm.
120 Watt of heat, divided by 44Volts is around 2.8 Ampere of bias. Divided by 8 N and P MOSFETs is around 340mA bias per MOSFET.
Your amp will stay in class A the first 30 Watts in 8 Ohm, after that it will be class AB.
 
You should not use the 40% efficiency in your calculation.
Say you can dissipate around 120W each channel and you want 100Wpc class AB.
Use 44 volt rails. That gives a clean 40volt sine wave. Or 100 Watt in 8 Ohm.
120 Watt of heat, divided by 44Volts is around 2.8 Ampere of bias. Divided by 8 N and P MOSFETs is around 340mA bias per MOSFET.
Your amp will stay in class A the first 30 Watts in 8 Ohm, after that it will be class AB.

Can you elaborate a little more about how you derived 44 volt rails ?
if i do sqrt(100*8) = 28v rms = 40v p-t-p, you may have added 4 volt for losses ?
 
Correct, 4 volt for losses. You can't swing to the full 44 volt without added distortion.
So your transformer needs to be around 34Volt AC. 34 x 1.31 ( under load) = 44.5 Volts. Or use 32 volts trafo.

Oh, i see. I will make 0-34v 0-34v (under load) on the trafo.
I have a huge power reservoir, maybe i will try to increase the biasing as much as possible while keeping an eye on the final temperature of the heatsink.

I have made some thermal simulations on the heatsink, and despite the 0.18Rth with a properly distributed load on the surface i can keep the temperature a little bit (10%) less of the calculated one. So it's worth trying pushing bias a little bit.

for the mosfet i will match 4 by 4 a bunch of irfp240 and irfp9240 (btw, i have read anywhere that irfp240 is a 20a device, but in the datasheet from vishay i read that they are rated for 12a... what i am missing ?)

for the input diff. stage i will use the k170 / j74 combo (7ma Idss matched within 0.1ma).

What do you recommend for the cascode ? i have some trouble in getting the SC4793/SA1837. maybe there are some other viable substitution ?
 
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i've found SC4793 and SA1873 on tme.eu, maybe is useful to other EU fellows.

I am doing some calculation about che cascode.
Nelson suggest to be 1/3 of the rail voltage. In my case, will be around 15v (45v / 3)...

I think that the divider of interest is R25/R27 and R30/R26. I'm right ? If so, i don't see why they are different values...

If i fill the calculations :
TOP
Vcascode = [Vrail / (R25+R27)] * R27 -> Vcascode = 45v / (10k + 4.75k) * 4.75k = 12.8volt
BOTTOM
45v / (47k5 + 4k75) * 4k75 = 4.1volt.

They don't have to be equal for the top and the bottom half ?

... As usual, what i am missing ? (apart from a solid electronics background :rolleyes:)
 

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120 Watt of heat, divided by 44Volts is around 2.8 Ampere of bias. Divided by 8 N and P MOSFETs is around 340mA bias per MOSFET.
Your amp will stay in class A the first 30 Watts in 8 Ohm, after that it will be class AB.

Sorry to bother you, i am re-reading your line, and i want to be sure to have understood well.

you say 30w in 8 ohm because given the 2.8A bias, i need to divide by 1.414 to get Arms and then Irms*Irms*8 = 30ish watt... that's correct ?
 
P = IV = I^2 R = V^2 / R
that applies for DC constant current/constant voltage

When you use AC and that signal is a sinewave, the formula becomes:
P = IacVac = Iac^2*R = Vac / R

You can use Ipk or Vpk for that same sine wave where the formula becomes:
P = IpkVpk/2 = Ipk^2*R/2 = Vpk^2/R/2

and if the signal is not a sinewave, then it becomes:
P = IrmsVrms = Irms^2*R = Vrms^2/R

pick the part of the formula for the information you have and only one unknown

eg. you know the Ipk = 2.8Apk and R = 8ohms.
Using P=Ipk^2*R/2 gives P= 2.8^2*8/2 = 31.36W into 8ohms. But check what voltage is required to push that current into your 8ohms load.
V= IR = 2.8Apk*8ohms = 22.4Vpk
You amplifier needs to achieve 22.4Vpk, when delivering 2.8Apk to get to that 31.36W into 8ohms.

Allow for losses/droops:
The amplifier will lose 3V to 7V delivering current into an 8ohms load and expect quite a bit more loss into lower impedance loads.
The supply rail voltage at the Amplifier Power Pins will droop from the quiescent voltage when no output power is being delivered to when maximum power is being delivered. Expect PSU & cable droop to be around 3V to 7V for 8ohms load and again a bit more for lower impedance loadings.

Thus to get 22.4Vpk your quiescent voltage would need to be around 6V to 14V higher, i.e. 28.4Vdc to 36.4Vdc to achieve that 31.36W into 8ohms from a ClassAB amplifier.

ClassA is different. The loss through the amplifier will be a bit lower.
The droop in supply voltage will be a lot lower. You may find that typical ClassA loss+droop might be in the range of 2V to 5V

But beware: source follower (mosFET) stage will have much higher losses than BJT, or common Source mosFET stages.
Source follower mosFET could need an extra 2V to 5V on the quiescent supplies more than for a BJT amplifier.
The F5 and variants is a common source output stage and the extra loss for mosFETs is not usually incurred.
 
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P=IV = I^2 R = V^2 / R
that applies for DC constant current/constant voltage
When you use AC and that is a sinewave the formula becomes
P= IacVac = Iac^2*R = Vac / R
and you can use Ipk or Vpk for that same sine wave where the formula becomes
P = IpkVpk/2 = Ipk^2*R/2 = Vpk^2/R/2

and if the signal is not a sinewave, then it becomes
P = IrmsVrms = Irms^2*R = Vrms^2/R

pick the part of the formula for the information you have and only one unknown

eg you know the Ipk = 2.8Apk and R = 8ohms
then using P=Ipk^2*R/2 gives P= 2.8^2*8/2 = 31.36W into 8ohms, But check what voltage is required to push that current into your 8ohms load.
V= IR = 2.8Apk*8ohms = 22.4Vpk
You amplifier needs to achieve 22.4Vpk when delivering2.8Apk to get to that 31.36W into 8ohms.
Then allow for losses/droops
The amplifier will lose 3V to 7V delivering current into an 8ohms load and expect wquite a bit more loss into lower impedance loads.
The supply rail voltage at the Amplifier Power Pins will droop from the quiescent voltage when no output power is being delivered to when maximum power is being delivered Expect PSU & cable droop to be around 3V to 7V for 8ohms load and again a bit more for lower impedance loadings.

Thus to get 22.4Vpk you quiescent voltage would need to be around 6V to 14V higher, i.e. 28.4Vdc to 36.4Vdc to achieve that 31.36W into 8ohms froma ClassAB amplifier.

ClassA is different. the loss through the amplifier will be a bit lower.
The droop in supply voltage will be a lot lower. You may find that typical ClassA lose+droop might be in the range of 2V to 5V

But beware: source follower (mosFET) stage will have much higher losses than BJT or common Source mosFET stages.
Source follower mosFET could need an extra 2V to 5V on the quiescent supplies more than for a BJT amplifier.
The F5 and variants is a common source output stage and the extra loss for mosFETs is not usually incurred.

illuminating ;) Thanks.
 
.............
about the cascode... i need to check for both dividers ?
or only the bottom one rules the voltage ?
Both cascodes operate for each of their supply rail and jFET.
But since both are exposed to the same voltages and the same currents, you will find that one prediction works for both upper and lower.

2sk170/j74 seem to work very well with 4Vds to 10Vds and this is the voltage the cascode needs to provide.
If you were using BF862 instead it has three main differences that will affect how it is implemented:
a.) Idss is generally higher = more heat
b.) Power rating is lower = less tolerant of heat
c.) Voltage rating is lower

The result is that you bias the BF862 to a lower current and you cascode to a lower voltage.
 
i've found SC4793 and SA1873 on tme.eu, maybe is useful to other EU fellows.

I am doing some calculation about che cascode.
Nelson suggest to be 1/3 of the rail voltage. In my case, will be around 15v (45v / 3)...

I think that the divider of interest is R25/R27 and R30/R26. I'm right ? If so, i don't see why they are different values...

If i fill the calculations :
TOP
Vcascode = [Vrail / (R25+R27)] * R27 -> Vcascode = 45v / (10k + 4.75k) * 4.75k = 12.8volt
BOTTOM
45v / (47k5 + 4k75) * 4k75 = 4.1volt.

They don't have to be equal for the top and the bottom half ?

... As usual, what i am missing ? (apart from a solid electronics background :rolleyes:)

That is an error on schematic.
They should both be 10k
 
That is an error on schematic.
They should both be 10k

i knew it ! :D

anyway... in my very particular case, with 45 volt rail, to get the sk170/sj74 in the best fitness condition i have to supply them 10volt (as per AndrewT reply)
if i am correct Vcascode = [Vrail / (R25+R27)] * R27.
Fixed R25 @ 10k -> R27 = 2k85. with a 45v rail i will get 10v on the input jfets.
 
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