| tab30 |
Hello,
I don't like the huge Voltage dropp, in saturation 10V or more is lost over the amp.
At +-70 supplV. that means that you, if you can avoid this voltage dropp, get 600 W insted of 400 out.
THE IDEA
Puting some smal additional windings at my Ring-Core, generating 2 additional. Voltages of about DC 10-15 V and put it in series with the main Supp. and drive the driver-stage with it.
So the driver stage will work with 80 V and the output with 70V.
What do you think ?
Will this destroy parts in the amp ?
I don't sink so, even the Fets are protected by a zenerdiode for high gate V.
Anyone tried this before ? |
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| Elso Kwak |
Hi tab30, This technique is used the the GAS Son of Ampzilla and also in some Elektor 2000 amp. Don't remember the right name of the beast.
http://home.kimo.com.tw/skychutw/ampzilla/index.html
+/-51V on inputstage and VAS, +/-45V on outputdrivers and outputstage.:cool: |
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| Mattyo5 |
I'm using 1kva toroids with each channel of my sym amps. And 120kuf w/ each channel. I have not checked to see if there is voltage drop at 200 wpc....but then again, i have never gotten the things to 200 wpc either...too loud :) great design Anthony!
If you are looking to avoid voltage drop, try a bigger power supply before you go above the recommended voltage.
-Matthew K. Olson
this is one monoblock :)
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| tab30 |
@ Elso Kwak
I know the elektor Gigant 2000, I nearly have all the articles from this german magazin.
And you'r right they use the same trick.
But my problem is the realisation at the Holton Amp, if it is possible without getting problems.
@ Mattyo5
I'm still using Caps with 100.000 uF and and 700 VA Trans for one channel (only one is required - subwoofer ) .
The voltage downfall is at short strong power pulses in the region of 1-2 V.
The main problem are the Fet's they need an Gate voltage of 2-4 V to switch on in the case of high current this figure can reach 8V.
"Powerful" greetings Andi |
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| slowhands |
| quote: | Originally posted by tab30
Hello,
I don't like the huge Voltage dropp, in saturation 10V or more is lost over the amp.
At +-70 supplV. that means that you, if you can avoid this voltage dropp, get 600 W insted of 400 out.
THE IDEA
Puting some smal additional windings at my Ring-Core, generating 2 additional. Voltages of about DC 10-15 V and put it in series with the main Supp. and drive the driver-stage with it.
So the driver stage will work with 80 V and the output with 70V.
What do you think ?
Will this destroy parts in the amp ?
I don't sink so, even the Fets are protected by a zenerdiode for high gate V.
Anyone tried this before ? |
The FETs are laterals, so have high on resistance of 1-1.7 ohm. Their saturated drop is 10-12v at max current, but since you have 4 in parallel, you only drop 3-4 volts. Yes, a 10 V higher rail voltage for the driver stage would give you more output. But do you have the cooling for it? Be careful with that, you may need forced air cooling.
Most of the laterals do have built-in zeners on the gates, but it does not hurt to have them. The way this is designed is not quite correct, they need reverse blocking diodes. As they are, you can only get limited gate drive because the off side zener forward biases and conducts. Max gate voltage is clamped by the forward biased zener plus the bias voltage. |
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| Kilentra |
In most amplifiers built from this design, the output devices are IRF 240 type. Those have a gate source voltage of about 5.5 volts at the peak output current which would be somewhere under 5 amps per device. And then the on resistance is 0.18 ohms so that's an additional voltage drop of under half a volt. In other words the voltage swing from one of these amps could be 6db from the rail. So the RMS power into 4 ohms might be nearly 500 watts, if the rails stay at 70 volts under load. I have 4 ohm speakers and my power rails sometimes exceed 75V. :)
Were those calculations close enough? |
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| slowhands |
| quote: | Originally posted by Kilentra
In most amplifiers built from this design, the output devices are IRF 240 type. Those have a gate source voltage of about 5.5 volts at the peak output current which would be somewhere under 5 amps per device. And then the on resistance is 0.18 ohms so that's an additional voltage drop of under half a volt. In other words the voltage swing from one of these amps could be 6db from the rail. So the RMS power into 4 ohms might be nearly 500 watts, if the rails stay at 70 volts under load. I have 4 ohm speakers and my power rails sometimes exceed 75V. :)
Were those calculations close enough? |
I used the published schematic for the device which shows the Toshiba laterals. I understand the bias voltage can be set for verticals too in the latest revision.
Perhaps you mean IRFP240 and IRFP9240? (IRF240 is a metal TO-3 part). Yes, these have much lower on resistance, but of course the P-channel device on resistance is higher than 0.18 ohm, IR says 0.5 ohm.
http://ec.irf.com/v6/en/US/adirect/...ductID=IRFP9240
Also, perhaps you meant 6v, not 6db?
I'd run the numbers a little differently, just to show how increasing the front end voltage helps.
In this design, you can't get rail to rail drive, you lose perhaps 2.5V on each rail. Add to that the gate voltage required to draw 5 A, say 5.5V as you propose. That totals 8v, which is effectively a loss of output voltage; it is subtracted from the rail voltage to get the peak output voltage. That 8V loss is what you partially recover if you use a higher front end supply voltage.
However, source resistors and on resistances cause other fixed losses, which you can't recover by increasing drive voltages. Let's assume the source resistors are 0.5 ohm and the on resistance is 0.5 ohm, a total of 1 ohm. Then at our peak current of 5A, we have a 5V loss of output voltage.
Let's work out an example. Using a +-75v supply, assume we have the 5V fixed loss and assume we have boosted the front-end voltage 10v. Our output voltage is 70v p-p = 49v RMS. Power into a resistive 8 ohm load is 49*49/8, or roughly 300 watts. At 4 ohms, fixed losses roughly double because currents roughly double. So output voltage is say 65V p-p=45v RMS. Power is 45*45/4 or roughly 500 watts. Of course this is at clipping and power supplies will sag, so realistically we might measure a solid 250 watts at 8 ohms, or 400 watts at 4 ohms. I did a lot of rounding in this example to keep it simple.
Note the two output pairs could supply this power to 8 ohms resistive, but if the load is 4 ohm you need double that number of output pairs due to higher dissipation. |
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| Kilentra |
| Aha, I forgot the Source resistance. And yes I meant IRFP240/9240 and 6 volts loss. Thanks for clearing that up. |
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