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Old 8th January 2013, 05:40 PM   #21
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Quote:
Originally Posted by grommeteer View Post
Gingertube, something in your brain storming confused me.
To confirm what you did I re-engineered the single pair pp- circuit of the Philips data sheet. It runs of Ub=250V and puts out 7W.

10k : 8R - turns ratio 35.4
.
.

Peterfarrow, have you checked the turns ratio of yout OT yet?
The transformer is 2.5K:8 as there are four parallel pairs [10k/4 : 8] I will check the OT tonight on my signal generator... I spoke with Merlin Blencow, the well know valve design author, and we have been through the design in detail and he and I are both at a slight loss as to why the power is so low. I am thinking it might be the opt. so that is the next test...

Thankyou everyone for your contributions, please keep them coming!
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Old 8th January 2013, 06:04 PM   #22
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Quote:
I think the voltage over one tube is 200Vp, not 200Vpp.
Assuming that, I get 400Vpp over one tube, the voltage on the other tube is 400Vpp as well, but 180 deg. out of phase. So in total I get 800Vpp a-a.
I have been involved in several discussions related to output power, often prompted by the disbelief of some of the power measurements I have made on tubes. Recently I got well over 100 watts from a pair of 13GB5 / XL500.

Look at it this way. When the amp is driven to max voltage excursion, the plate voltage at one output tube is near zero, and the OPT CT is connected to B+ so the other plate must be near twice B+. Look at the OPT CT as the pivot point in an equal length lever, or See Saw. On the next half cycle the polarities reverse. This shows that the voltage excursion at each plate goes from near zero to twice the B+.

This can be verified with scope measurements. If your scope (or its probe) can't eat 600+ volts, make up a voltage divider with resistors....that can handle the voltage. I use five 100K 2 watt resistors in series since each is only rated for 300 volts. Attach one end of the resistor string to the plate, and the other to ground. Connect the scope across the single resistor at the ground end. Multiply all readings by 5.

Given the Philips example, each plate will swing from 50 volts to 450 volts, for 400 Vpp on each tube with the other tube being 180 degrees out of phase. This does agree with the claim of 7 watts into 10K assuming OPT losses.

It should be possible to get 40 to 50 watts into a 2500 ohm load from a 300 volt supply assuming that the output tubes can adequately drive this load impedance REGARDLESS of what tubes they are. The type of tube used will affect the absolute saturation voltage, which does affect the output power, and too small a tube (or combination of tubes) will lead to distortion as the tubes approach saturation. Adding additional tubes in parallel increases the drive capability and helps lower the distortion, but once the tubes are saturating properly will not increase the output power without lowering the load impedance.

Quote:
or have a 4-5K transformer which will be out of tune, but give me more swing and a few more watts...
A higher impedance OPT will give LESS output power given no other changes.

One thing that hasn't been discussed here is the cathode resistance. Do you have it bypassed with a capacitor. If not, try it. I have seen similar effects in pentode P-P amps with unbypassed cathodes.

It would seem that measuring the OPT is the next step. I use a simple method. Load the secondary with the proper resistance (8 ohms on the 8 ohm tap). Plug the full primary into the wall outlet. Measure the secondary voltage and calculate the ratio and primary impedance.

We have 120 volts at 60 Hz in the US, so this works with small OPT's. You seem to be in a 240 volt 50 Hz country, so if the OPT is 2500 ohms, you will see about 23 watts on the load resistor during this test (don't test small OPT's). Your OPT's are rated for 40 watts at 18 Hz so this is safe. You should see 13.55 volts across the load resistor.

Some Sowter OPT's have multiple windings. Are they interconnected properly? Do these OPT's have multiple secondary impedances (4,8,16 ohms). If so try all available taps and see if you get more power on a different tap.
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Old 8th January 2013, 06:12 PM   #23
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There you go!!!
The voltage ratio in a transformer equals the turns ratio.
The impedance ratio is the square of the turns ratio.
Have a second look at our combined brain stormings, please.
You will find, that the Philips (10k) OT has 35.4, your OT should have a turns ratio of 17.7.
So, you Need to use your OT with two and two in parallel and those pairs in series.
Your 4 in parallel would yield a ratio of 8.85.
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Old 8th January 2013, 08:21 PM   #24
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I spoke at great length with Merlin Blencowe today and Brian Sowter and the design is sound according to both so I am not going mad ( though some would disagree )

The cathode resistors are fully bypassed, the OPT does indeed have multiple taps, and I have arranged them as per Brian's notes to get 8 Ohms, as described above :

Here is the spec:
Its 2.5k Raa
And 120mA per side max
Drop throught style (half shroud)
40W at about 18Hz

43% UL taps

There are four windings on the secondary 1,2,3,4, Brian suggests 1,3,4 in series and 2 in parallel with 3 (as above!)

I will test the OPT now with my signal generator, don't fancy shoving it across the mains, even though I have a 2KW capable variac, if I do decide to do the mains test and you don't get another post from me after this one, well then the test did not go to plan

I'll report shortly...



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Old 8th January 2013, 09:36 PM   #25
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And there you have it.... the OPT is wrong.

17.2 volts peak to peak in from my signal generator give 2.36 volts peak to peak out.

that's a ratio of 7.3
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Old 8th January 2013, 09:55 PM   #26
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Just a "stupid" question before deciding the OT is no good - are you sure you tested to the anode connections and not the ultralinear taps, 2K5 Raa to 8 Ohms UL taps would yield about 7.3 turns (voltage ) ratio.
Cheers,
Ian
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Old 8th January 2013, 09:58 PM   #27
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Quote:
Originally Posted by gingertube View Post
Just a "stupid" question before deciding the OT is no good - are you sure you tested to the anode connections and not the ultralinear taps, 2K5 Raa to 8 Ohms UL taps would yield about 7.3 turns (voltage ) ratio.
Cheers,
Ian
Hi Ian, yes checked this, but I will buzz out the dc impedances to see if they are labelled wrong.

7.3 turns ratio gives a primary load of 426 ohms between anodes.... anyone tell me what the power would be with such a mis match...
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Old 8th January 2013, 10:05 PM   #28
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on my fluke 289 true rms rather than a scope measurement I have 7.7vrms in and 1.1vrms out

Looks pretty wrong to me..

I'll check the other unused opt


P.
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Old 8th January 2013, 10:09 PM   #29
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looks like the labelling on the opt is wrong,
centre tap to anode 22.66ohms
centre tap to ul tap = 47.88 ohms
same the other side, looks like the UL and anode labels are transposed...

P.
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Old 8th January 2013, 10:14 PM   #30
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Quote:
Originally Posted by gingertube View Post
Just a "stupid" question before deciding the OT is no good - are you sure you tested to the anode connections and not the ultralinear taps, 2K5 Raa to 8 Ohms UL taps would yield about 7.3 turns (voltage ) ratio.
Cheers,
Ian
Genius observation - I owe you a beer at least!

6.6v ac to 0.37 vac or 17.7:1

spot on now, connect across the "UL" labelled taps!!!!
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