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7th July 2018, 10:48 AM  #21 
diyAudio Member
Join Date: Dec 2008
Location: UK

No they are not the same! If the 10%Ra means 200R in this case it is quite different from voltage ratio which would result in 20R winding. 400R and 40R for cathodetocathode, respectively.
Once one has both can simulate both....simulation it's cheap while waiting for a response. I have sent them an email.... 
7th July 2018, 11:06 AM  #22 
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Join Date: Apr 2009

45, wether you specify a tertiary winding in impedance or voltage ratio, you will calculate the same inductance for that tertiary winding wrt primary inductance!
Your example of Tamura F2021. Primary impedance 5k aa; CFB winding 16 ohm. So the winding ratio of this is the root of (5000/16), right? Being 17.68 : 1. Let's say (I don't see it specified) that the F2021 primary inductance is 100 H. Then, the inductance of the CFB winding is (1/17.68)² x 100 = 0,32 H. Now the calculation with impedance. 5000 / 16 = 312,5. (1/312,5) x 100 = 0,32 H. Wow the same outcome! So, in terms of impedance the CFB is 16 ohm; in terms of winding ratio it is some 5,66 %. Just a matter of how the specification is done. When the 10% CFB of the Polish transformer is indeed 10% of 4k primary impedance, then the winding ratio of that CFB winding is 3,16 : 1, and the ratio in percentage 31,6%. In that case the inductance of the CFB winding is 58 H (0.316² x 580), each halve is 14,5 H then. Last edited by pieter t; 7th July 2018 at 11:36 AM. 
7th July 2018, 11:17 AM  #23 
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Join Date: Jun 2015
Location: Perth, Western Australia

I look forward to the response. I sent an email 3 days ago and haven't had a response yet.
Meanwhile I tried 28.92H for the CFB windings in my amp simulation and I could only get 15W with huge distortion out whereas I was earlier getting 80W with my modified generic transformer and 4dB of CFB. I'm rerunning my simulations without the CFB windings first to get a baseline. Then I'll redo it with 1.5H. 
7th July 2018, 12:35 PM  #24  
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Location: UK

Quote:
The reason why this manufacturer uses percentage of Ra might be that they make several transformers with different primary impedance but the same ratio for all of them. We will know soon.... 

7th July 2018, 01:01 PM  #25  
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Quote:
About confusion: not with me. I just try to explain, and take away some confusion here and there. The manufacturer can take away the last bit of confusion, it has nothing to do with how to calculate induction of windings. You seem to like confusion because you bring up an example which actually is perfectly clear in impedance ratio and therefore also in voltage ratio. No need to do that IMO. 

7th July 2018, 01:29 PM  #26  
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Join Date: Dec 2008
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Quote:
If no response drop me a pm. Quote:
You need to be careful because it makes a big difference 10% voltage ratio and 10% of Ra. Let's make the two examples again from the point of view of operative conditions for class A. 10% voltage ratio means each single cathode winding has got 20R (i.e. 1.45H). Cathode to cathode will be 40R (i.e. 5.8H). This means that the total primary impedance will be (SQRT(4000)+SQRT(40))^2=4840 ohm It's higher but not too different. The feedback factor will be SQRT(40/4840)=0.091=9.1% So actually it will a bit more than 40R (and so inductance) to get 10% CBf...but this is close enough to play with it. for 15W output the driver will have to swing bias + 9.1%of output voltage = bias + 17.4Vpeak. If the CFb is really defined as 10%Ra then it will be 200R for the single winding or 400R cathodetocathode. So the actual endtoend primary load will be (SQRT(4000)+SQRT(400))^2=6929 ohm Then the feedback factor beta will be SQRT(400/6929)=0.24=24% So the actual total load value is significantly higher and you need to tweak the operative conditions quite a lot respect to 4K (i.e. cathode winding not used) for this already. The much higher feedback factor means that the output stage has got rather lower gain. On one side this is good because it means much lower distortion and low Zout that very like will make ground feedback unnecessary but on the other end will be more demanding on the driver that has to swing a lot more volts to achieve the same output power you would get if 6929 were all at the plate. So if you have designed the output stage around 4K and the driver to deliver an undistorted swing just over bias value it WILL result in much higher distortion. With 6929R primary load output power will be likely lower anyway respect to 4K but easily class A and the driver will have to swing the bias voltage + 24% of the output voltage. For 15W output the driver has to swing the bias voltage + about 55Vpeak without distortion! At least, because of the relevant cathode feedback the input impedance of the power stage will be much higher. This means that the grid resistor of the power tube will be multiplied by (1+betaA), where A is the open loop gain (i.e. without CFb). P.S. If you accept an advice, instead of wasting time with simulations you could find and read all old articles by Crowhurst. Those will give you good understanding and the ability to make your amps with a bit of ink a paper rather quickly! At worst you will need to draw a few load lines to get practical numbers for calculations. But this is easy to do. I see the simulator al lot more useful for PSU calculations to get a good point to start from. Last edited by 45; 7th July 2018 at 01:54 PM. 

9th July 2018, 08:09 PM  #27 
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Join Date: Dec 2008
Location: UK

So it really is 10% of Ra. Here is the response from the manufacturer:
Hello. Thank You for contact us. There are two cathode windings  one for each half of the primary winding. TTGCFB4000PP has a 4k Raa (anode to andode)  2k between the anode and the center tap. Each CFB winding is calculated as a 10% of a half, so 10% of a 2k. Best Regards / Serdecznie pozdrawiam, Tomasz This means that each cathode winding is 200R and the output voltage will be 24% at the cathode and 76% at the anode when CFb is connected. Can't use constipated drivers... 
9th July 2018, 08:21 PM  #28 
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Join Date: Apr 2009

Are you convinced by this response?
Speaking about impedance: when primary aa impedance is 4k, what is the impedance then of the primary half wrt the secondary? Knowing the number of turns would give more certainty. 
9th July 2018, 08:42 PM  #29 
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Join Date: Dec 2008
Location: UK

The only thing one needs to do to get it right is to stick to their definition and calculate everything for 1/2 primary.
With CFb the total half primary impedance will be Za=[SQRT(2K)+SQRT(200R)]^2= 3465R. So the total primary load endtoend will be 6930R. If I start from Zaa=4K and Zcc=400R I get Zaa=[SQRT(4K)+SQRT(400R)]^2= 6930R!! No mistake. So going back to the initial question... If the simulator input is the total primary inductance of 580H then the total inductance of the cathode winding is 58H. If the simulator want to know the inductances of each winding both anode and cathode the inputs will be 1/4 of the values above. Last edited by 45; 9th July 2018 at 09:01 PM. 
9th July 2018, 09:08 PM  #30 
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Join Date: Apr 2009

The subject of this thread is to get the right values for the simulation.
When a half primary impedance spec is indicated to be 2k for a 4k aa impedance instead of 1k, I have my doubts. I calculate the half primary impedance to be 1732.5R (1k and 100R resp. in the equation). Full primary impedance wrt half primary impedance is 2² x half primary impedance; then you have the correct ratios wrt the secondary impedance. 
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