THUNDERFUNK 750-A

chicago_mike · Post by **chicago_mike** » 22 Nov 2012, 08:04

These are the putput tranny pdf's

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 09:35

So I still need to test this transformer. But what Im not sure is how to test current draw.

Or

How to calculate needed current draw. At least for the power amplifier section.

I'll google, hope it comes up with info.

deltafred · Post by **deltafred** » 28 Nov 2012, 10:02

chicago_mike wrote: How to calculate needed current draw. At least for the power amplifier section.

I'll google, hope it comes up with info.

Power = amps x amps x ohms

This is for a steady RMS output, not peak, burst, or program power or any of the other statistical ways that advertising literature has of making something look better/bigger/more powerful than it actually is.

rearranging the formula

amps = square root of (power / ohms)

so say you want 200W into 4 ohms

amps = square root of 50 = 7 (well 7.07 to be precise)

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 13:25

Gracias!

So this transformer is pulling some serious current.

DrNomis · Post by **DrNomis** » 28 Nov 2012, 13:38

deltafred wrote:
chicago_mike wrote: How to calculate needed current draw. At least for the power amplifier section.

I'll google, hope it comes up with info.
Power = amps x amps x ohms

This is for a steady RMS output, not peak, burst, or program power or any of the other statistical ways that advertising literature has of making something look better/bigger/more powerful than it actually is.

rearranging the formula

amps = square root of (power / ohms)

so say you want 200W into 4 ohms

amps = square root of 50 = 7 (well 7.07 to be precise)

If my memory serves me right the RMS power is about .707 of the Peak power deltafred, but I could be wrong......

deltafred · Post by **deltafred** » 28 Nov 2012, 14:18

DrNomis wrote: If my memory serves me right the RMS power is about .707 of the Peak power deltafred, but I could be wrong......

Forget about peak power, once you get away from RMS you are into advertising speak (and car sound systems). (IMO, IME, YMMV etc. etc.)

If you are working with bass amps then RMS is THE only acceptable unit. (IMO ...... etc. )

Now voltage wise peak and especially peak to peak is important because the supply rails set your peak/peak voltage.

If you have a transformer with say 37v - 0 - 37 VAC and full wave rectify it you end up with about -52, 0, +52 VDC supply rails. (37 X 1.414) (forgetting about voltage drop in the rectifiers)

If you drive a sinewave so it nearly clips then you have about 104V peak to peak, divide this by 2.828 (2 X square root of 2, one of the numbers that should be embedded in an EE's brain) then you have about 37V RMS output (the transformer voltage

).

To find the power it will drive into a load you need to square the voltage and divide by the load impedance, so say 8 ohm load.

P = V x V / R
=37 x 37 / 8
= about 170 watts.

Then calculate the current required (as per my prev post)

Amps = square root of (power / ohms)
= sq root of 170/8
= 4.6 amps.

So for to drive 170W into 8 ohms you need an absolute minimum of a 37 - 0 - 37 volt 4.6 amp transformer.

(I haven't checked the maths so it may contain errors)

DrNomis · Post by **DrNomis** » 28 Nov 2012, 14:39

deltafred wrote:
DrNomis wrote: If my memory serves me right the RMS power is about .707 of the Peak power deltafred, but I could be wrong......
Forget about peak power, once you get away from RMS you are into advertising speak (and car sound systems). (IMO, IME, YMMV etc. etc.)

If you are working with bass amps then RMS is THE only acceptable unit. (IMO ...... etc. )

Now voltage wise peak and especially peak to peak is important because the supply rails set your peak/peak voltage.

If you have a transformer with say 37v - 0 - 37 VAC and full wave rectify it you end up with about -52, 0, +52 VDC supply rails. (37 X 1.414) (forgetting about voltage drop in the rectifiers)

If you drive a sinewave so it nearly clips then you have about 104V peak to peak, divide this by 2.828 (2 X square root of 2, one of the numbers that should be embedded in an EE's brain) then you have about 37V RMS output (the transformer voltage ).

To find the power it will drive into a load you need to square the voltage and divide by the load impedance, so say 8 ohm load.

P = V x V / R
=37 x 37 / 8
= about 170 watts.

Then calculate the current required (as per my prev post)

Amps = square root of (power / ohms)
= sq root of 170/8
= 4.6 amps.

So for to drive 170W into 8 ohms you need an absolute minimum of a 37 - 0 - 37 volt 4.6 amp transformer.

(I haven't checked the maths so it may contain errors)

I think this is all good stuff to keep revising and remembering deltafred, thanks for that cause I feel I've learnt something here......

deltafred · Post by **deltafred** » 28 Nov 2012, 15:26

It is all based around

1. Power = Amps X Volts
2. Ohms law - volts = amps X ohms, or rearranged to ohms = volts/amps, or amps = volts/ohms.
3. The RMS voltage of a sinewave is the peak voltage divided by 1.414

To calculate the current required where voltage and impedance are known - all values are RMS.
power = amps x volts (1)
but from (2) volts = amps x ohms so substitute this into the above and you get
power = amps x amps x ohms
rearrange it to separate amps out
amps = square root of power/ohms

To calculate the voltage to drive a known power into a specific impedance - all values are RMS.
power = amps x volts (1)
but from (2) amps = volts/ohms
so power = volts x volts/ohms
rearrange to separate volts out
volts = square root of (power x ohms)

The transformer voltage should be at least + and - the voltage calculated above as there will be voltage dropped across the rectifier diodes and ripple on the smoothing capacitors dependant upon capacitance and current drawn. This involves another load of maths if you want to calculate it.

The capacitor working voltage should be in excess of transformer voltage x 1.414 (from 3)

The rectifier reverse breakdown voltage should be in excess of transformer voltage x 2.828

I hope this makes sense (and that I haven't made too many mistakes).

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 15:34

After rectification, on the power supply, we see + and - 92 volts.
The amp puts out 750watts into a 2ohm load rms.

So 92x92/2?

And 750/2 sqrd?

I know I missed something.

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 17:39

So I am putting a shopping list together for this thing..

The one item I cannot find is the inductor on the power amp module.

Its 4uH. but I don't know who makes it.

deltafred · Post by **deltafred** » 28 Nov 2012, 19:03

Sorry for the delay, I've had other stuff to attend to.

So 750W, 2 ohms

volts = square root of (power x ohms)
= sq root 750 x 2
=sq root 1500
= 39 volts so a 40 - 0 - 40 (or a bit more) transformer would provide the theoretical minimum.

Looking at https://www.freestompboxes.org/download/ ... &mode=view
it states that the supplies are +/- 78 volts which is probably no load voltage. A transformer with 55 - 0 - 55 volts will give that

It will account for ripple in the smoothing capacitors (value unknown) and transformer regulation, normally taken as 5% at full load.

amps = square root of power/ohms
= sq root 750 / 2
= sq root 375
= 19 amps, some serious current there!

So a 55v - 0 - 55v @ 19A to give 750w continuous. I don't know the spec on the Thunderfunk (whether it continuous or not).

to double check
power = amps x volts
= 19 x 39
= 741 watts

which is reasonable as I have been rounding to whole numbers.

The current still stands at 19A, some serious current there!

To build a 750W power amp is no mean feat and could go horribly wrong unless you do an exact copy of the Thunderfunk (both electrically, and thermally/mechanically), just saying.

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 20:00

My thoughts too.

My concern is to get the inductor now.

I can get everything else. The heatsink Dave uses is from way back in the AMP 420 days. So I could always get in touch with Russ Allee about heat sink suppliers.

The enclosure is all aluminum. And the fan can run continuous.

19 amps....oof!!!

deltafred · Post by **deltafred** » 28 Nov 2012, 20:12

chicago_mike wrote: My concern is to get the inductor now.

19 amps....oof!!!

I don't see an inductor, probably not looking at the right schematic, do you have a link.

Yea 19A is heavy current stuff.

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 20:25

Its on the amplifier board.

The coil. A 4uH one. has a resistor in parallel with it. I think it should be on the PDF.

bajaman · Post by **bajaman** » 28 Nov 2012, 22:00

that's just a simple output inductor - about 20 turns around a screwdriver of 1.5mm diameter enameled copper wire should do the trick

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 22:13

deltafred · Post by **deltafred** » 28 Nov 2012, 22:54

Ok, I've read the whole damned thread and don't see a schematic anywhere for the output stage with an inductor on it, just a photo of an anti-parasitic one as Baja says, here https://www.freestompboxes.org/download/ ... &mode=view. I may of course have just failed to see it, it wouldn't be the first time.

Also you are talking about +/- 92 volts, where the f... does that come from. The schematic I was looking at is +/- 78v. (The one I linked before.)

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 23:03

The 92volts is on the actual PCB.

The schematic is for an older thunderfunk. Notice on the piccies..there are 6 sankens instead of 4.

I was indeed referring to the parasitic inductor. I figure I'll just make my own.

The schematic I got from Dave doesn't show all the updates he has done.

I think I got everything

chicago_mike · Post by **chicago_mike** » 28 Nov 2012, 23:06

Hmmmmm.

Okay the PCB says 92volts + -.

But the schema says what 73...

Could it be still 73volts even if the power supply PCB says 92?

hmmmm

deltafred · Post by **deltafred** » 28 Nov 2012, 23:49

Ah that makes more sense now, I remember you mentioning 6 op trannies and the schematic I was looking at had 4.

The choke isn't critical, as Baja says wind a few turns of insulated copper wire round a drill shank or screwdriver, that is all the manufacturer does. Just estimate the wire gauge from the photo (or use what you have) and put the same number of turns as in the photo.