Y.A.F5Tv3.B.T. Yet another F5Tv3 Build Thread
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gionag
diyAudio Member

Join Date: Jun 2007
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 )
Attached Images
 cascode_res.PNG (83.2 KB, 288 views)

Last edited by gionag; 18th October 2017 at 10:28 PM.

gionag
diyAudio Member

Join Date: Jun 2007
Quote:
 Originally Posted by WalterW 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 ?

 19th October 2017, 09:48 AM #13 AndrewT   R.I.P.   Join Date: Jul 2004 Location: Scottish Borders 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. __________________ regards Andrew T. Last edited by AndrewT; 19th October 2017 at 09:55 AM.
gionag
diyAudio Member

Join Date: Jun 2007
Quote:
 Originally Posted by AndrewT 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.

 19th October 2017, 09:58 AM #15 AndrewT   R.I.P.   Join Date: Jul 2004 Location: Scottish Borders Note: I was still editing out the mistakes and typos when you read that. __________________ regards Andrew T.
gionag
diyAudio Member

Join Date: Jun 2007
Quote:
 Originally Posted by AndrewT Note: I was still editing out the mistakes and typos when you read that.
ah.. ok... i will review.
about the cascode... i need to check for both dividers ?
or only the bottom one rules the voltage ?

AndrewT
R.I.P.

Join Date: Jul 2004
Location: Scottish Borders
Quote:
 Originally Posted by gionag ............. 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.
__________________
regards Andrew T.

 19th October 2017, 05:38 PM #18 WalterW   diyAudio Member   Join Date: Oct 2012 Location: Utrecht Calculate the cascodes so you have around 10 to 12 Volts on the JFET's. They should be the same for both rails. The different values in the schematic is a typo. __________________ Pass F5, F5Tv3, ACA, Aleph J, DDDAC1794, DCB1, BA3-preamp, VFET CSX1, Pass F6, Pass M2-HPA, VFET-X balanced, Pass J2 'If you see me talking to myself, don't worry. I'm getting expert advice'
2 picoDumbs
diyAudio Member

Join Date: Sep 2013
Quote:
 Originally Posted by gionag 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 )
That is an error on schematic.
They should both be 10k
__________________
"If you leave this point unattached to some circuitry, an ideal constant current source will emit a small lightning bolt which will travel until it connects to something." Nelson Pass

gionag
diyAudio Member

Join Date: Jun 2007
Quote:
 Originally Posted by 2 picoDumbs That is an error on schematic. They should both be 10k
i knew it !

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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