CLC is a capacitor input filter.
LC is a choke input filter. Their performance is completely different.
That does not work.
4ohms driven from a 30-0-30Vac transformer will give about 100W to 120W.
That would normally require a 100VA to 200VA transformer.
Using a 300VA transformer to charge the smoothing capacitance will result in the capacitors recharging to very close to full voltage on every half cycle.
Very small capacitors which will have a lower ripple current rating will discharge deeply on every half cycle. They will overheat pretty quickly if you try driving the amplifier very hard.
You would need to use a transformer well below 100VA to reduce the ripple current on the smoothing capacitors to significantly reduce the heating effect.
Just double up the output stage SOA and you get a better amplifier.
ASKA for the CRC 0.15R resistor, any type, format you would recommended?
SMD, Wierwound, MOX and what wattage? In such an amp as this one?
Thanks
SB
SMD, Wierwound, MOX and what wattage? In such an amp as this one?
Thanks
SB
These 0.15R are the Vishay-Dale WSR2 series, 4827 smd package. Check them on Digikey; they are 2W and metal film.
I find them very useful for improving the power supply by quarantining the rectifier pulses into the first cap near the bridge.
Hugh
I find them very useful for improving the power supply by quarantining the rectifier pulses into the first cap near the bridge.
Hugh
I know I'm quite late to the party but I've just read most of the thread and I'm impressed by the great development work that has been done and by the really informative posts of Hugh, that helped me to better understand many technical issues in this hobby.
Thanks to all, you've helped me to improve my knowledge in audio circuits!
And now my request of further help: I was preparing myself to build the Inverted Circlophone but this VerySimple Quasi could be assembled with not much more effort since I've already got most of components and I could make use of the same power supply, thus letting me taste this little pearl.
Clearly I lack the boards and I'm asking if anyone of you, who has made some sample boards, has got some spare boards and could sell a stereo pair of them to me.
As a second option, I could etch them in boards with a single layer of copper; is available any file of definitive circuit suitable for the etching process? Could anyone of the designers supply it to me?
Many thanks in advance
Nicola
Thanks to all, you've helped me to improve my knowledge in audio circuits!
And now my request of further help: I was preparing myself to build the Inverted Circlophone but this VerySimple Quasi could be assembled with not much more effort since I've already got most of components and I could make use of the same power supply, thus letting me taste this little pearl.
Clearly I lack the boards and I'm asking if anyone of you, who has made some sample boards, has got some spare boards and could sell a stereo pair of them to me.
As a second option, I could etch them in boards with a single layer of copper; is available any file of definitive circuit suitable for the etching process? Could anyone of the designers supply it to me?
Many thanks in advance
Nicola
You commonly see a small 'c' before the LC, but the LC is the driver (push the 'c' too large and it reverts to capacitor input)
As Andrew pointed out you need a minimum standing (or bias) current to keep a choke input filter at 0.7 Vin for a fixed L. Otherwise it "converts" back to capacitor input (and Vout = 1.4 Vin).
The "magic" is a "Swinging" choke, whose L is not fixed. The L rises as the current drops, reducing the critical current required to keep it in "Choke input" mode.
The classic article is by "Cathode Ray" in Wireless World, December, 1957
A copy is visible here
Hi Nicola,
Dacz layout is latest and incorporates all changes that happened over time. So would be good to go, but i am not sure if it has been tested by Dacz/anyone. My layout is from earlier version. Also his schematic is very informative, so easier to understand optional components, pin-outs and set-up voltages... all the best for your build.
reg
prasi
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Hi Prasi,
thanks for your advice and support.
I've just downloaded the Sprint viewer and I can see I can print the copper layer for etching process.
I'll analyse it deeply before doing it but it would be helpful if Dacz (or anyone who has tested it) can confirm that layout is correct and works.
Still thanks
Nicola
thanks for your advice and support.
I've just downloaded the Sprint viewer and I can see I can print the copper layer for etching process.
I'll analyse it deeply before doing it but it would be helpful if Dacz (or anyone who has tested it) can confirm that layout is correct and works.
Still thanks
Nicola
there is errata in C11, it was shown as copper trace insted of silk. i'll post the latest when i come home
Dacz
Dacz
here is the latest layout and schematic
Attachments
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hi ranchu,hugh,
i have some leftover 2N5416 from previous leach amp build. can i use it in place of Q3 and Q5 vas buffer - phase inverter.
http://www.farnell.com/datasheets/296648.pdf
regards,
Dacz
i have some leftover 2N5416 from previous leach amp build. can i use it in place of Q3 and Q5 vas buffer - phase inverter.
http://www.farnell.com/datasheets/296648.pdf
regards,
Dacz
No DAZC,
This is slow; 15MHz and Cob of 15pF.
Even a BD140 would be more suitable.
It is a pity because it is 300V, 1W and 1A. But a large die....
HD
This is slow; 15MHz and Cob of 15pF.
Even a BD140 would be more suitable.
It is a pity because it is 300V, 1W and 1A. But a large die....
HD
Hi. I'm not trying to hack this thread, but I think that this discussion could be profitable to the others members as well. In particular ASKA & AndrewT can comments on my assumption. Any comments are welcome, I'm still learning 😉
Hi. Often when I size an amp power transformer rating, I miss the target an end up buying a different model, an expensive error, and then I need to build another amp to re-use the other transfo 😉. Please could you check my calculations/assumption and tell me what do you think. Am I on the right track?
This is for a Class AB design, Volt on the Output Stage +/-75Vdc, target of around 250W max into 8 ohms, double if possible in 4 ohms...
From AndrewT Calculations:
Pmax = Vpk²/Rload/2
Vpk << Vsupply.
expect Vsupply to be about 3V to 6V below quiescent voltage.
Expect Vpk to be 2V to 5V below Vsupply
eg quiescent voltage to amplifier ±75Vdc
Vdrop on max power is 4V
Vloss through amplifier is 3V
Therefore Vsupply ~ 75 -4V ~ 71Vdc
Vpk ~ 71V-3V ~ 68V
Max P ~ 68²/8/2 ~ 289W into 8r0 dummy load.
A Plitron power transformer 110V primary, dual sec 50V,5A (500VA) will produce about 55V under 120V primary, in Qc, Canada (120/110 = 10% extra)
That gives 55 x 1.4142 = 77.7V - 1.4V (rectifier) = 76.3V supply, close enough to the target of 75V for the rail
But under 8 ohms, 289W is 6A... Hence a 500VA is too small a rating, a 625VA, with 6.25A current is better choice
Now under 4 ohms load:
A 625VA transfo has 6.5A output, to reach about 180% power into 4 ohms = 520W (180% max according to Cordnell...) we need more current... 520W under 4 ohms is 11.4A!
50V, 750VA transfo is 7.5A, seems to be a good compromise
50V, 1000VA transfo at 10A, seems the best choice, but it is quite large and heavy...
I'm sure that an amp is not working 100%, with a fix 1Kz tone all the time, hence it is peak current that we need, not continuous...
What do you think, is it very desirable to go that high in power, with risk to damage the final stage, or if a 750VA would be a more sensible choice, an a lighter one as well 😉
Thanks again...
SB
Hi. Often when I size an amp power transformer rating, I miss the target an end up buying a different model, an expensive error, and then I need to build another amp to re-use the other transfo 😉. Please could you check my calculations/assumption and tell me what do you think. Am I on the right track?
This is for a Class AB design, Volt on the Output Stage +/-75Vdc, target of around 250W max into 8 ohms, double if possible in 4 ohms...
From AndrewT Calculations:
Pmax = Vpk²/Rload/2
Vpk << Vsupply.
expect Vsupply to be about 3V to 6V below quiescent voltage.
Expect Vpk to be 2V to 5V below Vsupply
eg quiescent voltage to amplifier ±75Vdc
Vdrop on max power is 4V
Vloss through amplifier is 3V
Therefore Vsupply ~ 75 -4V ~ 71Vdc
Vpk ~ 71V-3V ~ 68V
Max P ~ 68²/8/2 ~ 289W into 8r0 dummy load.
A Plitron power transformer 110V primary, dual sec 50V,5A (500VA) will produce about 55V under 120V primary, in Qc, Canada (120/110 = 10% extra)
That gives 55 x 1.4142 = 77.7V - 1.4V (rectifier) = 76.3V supply, close enough to the target of 75V for the rail
But under 8 ohms, 289W is 6A... Hence a 500VA is too small a rating, a 625VA, with 6.25A current is better choice
Now under 4 ohms load:
A 625VA transfo has 6.5A output, to reach about 180% power into 4 ohms = 520W (180% max according to Cordnell...) we need more current... 520W under 4 ohms is 11.4A!
50V, 750VA transfo is 7.5A, seems to be a good compromise
50V, 1000VA transfo at 10A, seems the best choice, but it is quite large and heavy...
I'm sure that an amp is not working 100%, with a fix 1Kz tone all the time, hence it is peak current that we need, not continuous...
What do you think, is it very desirable to go that high in power, with risk to damage the final stage, or if a 750VA would be a more sensible choice, an a lighter one as well 😉
Thanks again...
SB
Salut SB,
Do you really need 250W @ 8 ohm / 500W @ 4 ohm ?
Have you heard how loud a 50-100W really is with normal efficiency spkrs ?
Maybe I'm issing something here but I've never felt the need to have a very powerful amp.
BR,
Eric
Do you really need 250W @ 8 ohm / 500W @ 4 ohm ?
Have you heard how loud a 50-100W really is with normal efficiency spkrs ?
Maybe I'm issing something here but I've never felt the need to have a very powerful amp.
BR,
Eric
This is not directly related to this amp, but to the typical calculation we have to do when trying to size a toroidal transformer...
SB
SB
If not related to this amp then maybe you could post in the Power Suppies section of this forum.
BR,
Eric
BR,
Eric
Algar,
Figures look good but with some provisos.
In a general sense, for power output you figure it from the peak to peak voltage at the highest output voltage.
For 75V rails, and using bipolars, you should use at least four pairs of outputs. This has an advantage as the drive to the individual output devices and their emitter resistors is rather less than single very large devices, giving you more power but at a cost of mechanical and thermal complexity.
For mosfet amps, you can use some of the very large mosfets available from IXYS, Vishay SGS and Fairchild. Assuming you use 1Kw mosfets in Class AB (and be aware you should water cool them!), you could expect to lose quite a bit of voltage from the rails because the Vgs of these devices is around 5V and the high current output, particularly with the 4R loads, will drop considerable voltage across the source resistors. Choose 0.33R resistors, and assuming you are running 15A through them. This is a lot of drop from gate down to output; 5V plus (15 x 0.33) a total of 10V. You must also realise that the rail voltage will sag with heavy current; maybe up to 5V as well. So, with 75V at idle you must drop 5V from the gate/source drive, then another 5V from the source resistors, and add the 5V sag of the rail. This gives you a total voltage of 60Vpeak, or 120Vpp waveform output.
Now, for 8R loads, square this figure and divide by 64. This gives us 225W, maybe a litte more if you have a stiff supply, and theoretically 450W into 4R. However, with 4R I assume not the double power, but 1.8, so it would be 405W.
This is a lot of audio power, and normally used in sound reinforcement situations, like large auditorium, or maybe Texas level listening rooms with 50 people!
I cannot stress how difficult it is to design very large amplifiers which sound good. It is a huge problem to hold high resolution with such amplifiers.
Hugh
Figures look good but with some provisos.
In a general sense, for power output you figure it from the peak to peak voltage at the highest output voltage.
For 75V rails, and using bipolars, you should use at least four pairs of outputs. This has an advantage as the drive to the individual output devices and their emitter resistors is rather less than single very large devices, giving you more power but at a cost of mechanical and thermal complexity.
For mosfet amps, you can use some of the very large mosfets available from IXYS, Vishay SGS and Fairchild. Assuming you use 1Kw mosfets in Class AB (and be aware you should water cool them!), you could expect to lose quite a bit of voltage from the rails because the Vgs of these devices is around 5V and the high current output, particularly with the 4R loads, will drop considerable voltage across the source resistors. Choose 0.33R resistors, and assuming you are running 15A through them. This is a lot of drop from gate down to output; 5V plus (15 x 0.33) a total of 10V. You must also realise that the rail voltage will sag with heavy current; maybe up to 5V as well. So, with 75V at idle you must drop 5V from the gate/source drive, then another 5V from the source resistors, and add the 5V sag of the rail. This gives you a total voltage of 60Vpeak, or 120Vpp waveform output.
Now, for 8R loads, square this figure and divide by 64. This gives us 225W, maybe a litte more if you have a stiff supply, and theoretically 450W into 4R. However, with 4R I assume not the double power, but 1.8, so it would be 405W.
This is a lot of audio power, and normally used in sound reinforcement situations, like large auditorium, or maybe Texas level listening rooms with 50 people!
I cannot stress how difficult it is to design very large amplifiers which sound good. It is a huge problem to hold high resolution with such amplifiers.
Hugh
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