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Interstage Transformer Nomenclature

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Hello,
I posted this question in other forums that I thought could help but no one responded so here I am. If you've seen this already excuse my multiple posts. But I really do need to know.
Could someone please explain the nomenclature used to describe an interstage transformer (IT). I've seen various designations like: 1:1, 1:2, 1+1:2+2, 5k:5K, etc... I understand even less about how one would choose an IT for a particular driver tube/power stage tube. I know I'm asking a lot, but please, if you have any insight in this matter, help me out. Regards,
David
 
I've seen all your posts at the other forums....as Sidewinder...and I did not reply because I thought someone who knows more than I do would answer.

So I'll try...
1:1 is that the turns ratio is the same on primary and secondary.

thus 2:1 means that there are 2 turns on the primary for every turn on the secondary...this in turn means that if you have a load of 5k at the output side...... the reflected load for the tube on the "primary" side will be 10k. etc. etc. etc.

1+1:2+2 means the same but there are 2 sections on both sides. This can be handy ...because if you connect the 1+1 side in parallel and the 2+2 side in series you would have 1:4 ratios...

I hope this is correct and if it is I hope it has helped... ;)




Cheers,
Bas
 
Just to add on, when impedances are quoted, they're assuming the secondary loaded with the specified resistance. If it's terminated with a different resistance, the impedance presented by the primary will change accordingly. So, if you have a 10K to 10K transformer and load it with 20K, the primary will present a 20K load.
 
When a manufacturer quotes an impedance, isn't that the optimum operating point for widest, flattest bandwidth and lowest LF distortion? For example, a Hammond 804 (we use often at work) can be strapped one-to-one as either 600:600 or 150:150. Terminating the latter's output with 10K will reflect a higher impedance to the primary but the low end still has to contend with the 804's low number of turns and DCR. I ask because after a couple of decades using line level trannies in a broadcast setting the simple turns ratio ratings used by most interstage manufacturers are still unclear to me.
 
rdf said:
When a manufacturer quotes an impedance, isn't that the optimum operating point for widest, flattest bandwidth and lowest LF distortion?

And/or power efficiency. Which isn't of concern here, but is in OPTs which are just beefy ISTs with a low-Z secondary. Unless you're driving a zero bias class B output and have hair-thin margins for drive capability. But when does that ever happen...

For example, a Hammond 804 (we use often at work) can be strapped one-to-one as either 600:600 or 150:150. Terminating the latter's output with 10K will reflect a higher impedance to the primary but the low end still has to contend with the 804's low number of turns and DCR. I ask because after a couple of decades using line level trannies in a broadcast setting the simple turns ratio ratings used by most interstage manufacturers are still unclear to me.

Correct; you also have parasitic (inter-winding and inter-layer) capacitance which becomes a greater concern to higher impedances. Any reactance really, inductive or capacitive, narrowing in the bottom and top end responses respectively.

In a perfect world, you could use 1 turn coupled to three turns for any frequency...sigh... ;)

Tim
 
1:1 IT All the same?

SY said:
For a 1:1, the primary impedance will look like whatever impedance is loading the secondary.

I apologize for all these basic questions, but I can't seem to get a firm intuitive understanding of ITs. Thanks for the patience. I kinda understand the concept of impedance reflection, but in an article by George Sanguinetti in Sound Practices Issue 15 he had used a Sowter 8423 IT with a 5k pri 1:1+1 between a 7788 driver and a vv52B that was choke load and parfed to a MQ2004opt. Later after he measured the plate resistance he and Brian Sowter decided that a 2k pri IT would be better. So the question is how are the 2 ITs different? Both are I assume are connected 1:1+1. Thanks
Regards,
David
 
The ratio of the reflected impedances is a function of the ratio of the number of turns. BUT.... this is something determined at midband. At the frequency extremes (for a given transformer), non-ideal elements like primary inductance and interwinding capacitance come into play, changing the frequency response and distortion characteristics of the transformer.

For example, I'm using some Jensen 1:1 input trannies that are optimized for 10Kohm loads. At 1kHz with a 10Kohm load on the secondary, the primary looks like a 10K resistor, the -3dB point at the high end is about 85kHz, and the rolloff is well-damped. Pass a 1kHz square wave through it and it looks pretty good.

Now load that same transformer's secondary with a 100Kohm resistor. At 1kHz, as expected, the primary reflects that load faithfully and looks like a 100Kohm load. But now the frequency response has a big bump at 50kHz and that same 1kHz square wave shows ringing on its leading edge.
 
Exactly, but for me that still leaves the question how to approach a transformer when the specifications are listed as simple turns ratios. Is normal practice in this case to contact the manufacturer for an opinion on the specific applicationin mind?
 
For a 1:1, the primary impedance will look like whatever impedance is loading the secondary.

That's not always the case. An interstage transformer connected directly to the grid of the tube in the following stage sees no load (in Class A). The primary impedence (5K, for example) will indicate the load the driving tube sees terms of an impendence at a certain (usually low end) frequency based on the inductance of the primary winding. In that case you look at turns ratio and primary inductance to see if an interstage transformer is suitable for your purpose.

John
 
Yes, like rdf, I'm still confused. Furthermore upon reflection on the concept of secondary reflection, since an IT's secondary is conected to the grid of an output tube, doesn't it in theory has infinite impedance to dc, since it's not connected to anything? This is not so for a.c., but how is this impedance calculated?
Regards,
David
 
TerribleT said:
Yes, like rdf, I'm still confused. Furthermore upon reflection on the concept of secondary reflection, since an IT's secondary is conected to the grid of an output tube, doesn't it in theory has infinite impedance to dc, since it's not connected to anything? This is not so for a.c., but how is this impedance calculated?

Well, a transformer doesn't respond down to DC; it's limited at the low end by the primary inductance, which effectively shunts the primary. At midband, the inductance of a properly-speced represents a very high Z, enough so that it can be treated as an open circuit.
 
Here's a quick sketch of some equivalent circuits, very simplified (if you really want to understand this more rigorously and in more depth, the Radiotron Designer's Handbook is the place to go).

The top sketch shows the midband model, idealized. The source is a voltage source in series with its output impedance, Vin and Rs. The load is assumed to be a pure resistance, too, RL. The impedances (in this case, pure resistances) are transformed ny the square of the turns ratio.

The bottom sketch introduces the first level of non-ideality. The source and load impedances are still assumed to be pure resistances, but there are a couple of new twists. The first new twist is the finite inductance of the primary. In a practical transformer, the inductive impedance at midband and high frequencies is very large compared to N2RL so can be neglected. As we go down in frequency, the source starts to be loaded by the inductive impedance, which shunts more and more of the source current as the frequency goes down. By the time you reach DC, the voltage across the primary (idealized again!) is zero- the transformer doesn't transform DC.

The other new twist is the leakage inductance, represented by Lleakage in series with the load. At midband and low frequencies, its inductive impedance is negligible. As we go higher in frequency, we start to see its effect; its impedance gets larger with frequency and, acting as a voltage divider in series with the load, it starts dropping more and more voltage. At the extremes of high frequency, effectively all the output voltage is across it, so the load's share drops to zero.

Now, to complicate things more, there are various capacitances that need to be added to that model to show different forms of peaking and rolloffs (as shown in the simplified model, the HF rolloff is first order, whereas in a real transformer, it's third order). The goal of the transformer designer is, knowing the nature of the source and load impedances, balance off the winding geometry, core, number of turns, and wire resistances to get a flat response over the desired range. Deviate from the recommended source and load impedances and you'll see a non-flat response compared to running the transformer with optimized loads.

Does this make it any clearer?
 

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This is not so for a.c., but how is this impedance calculated?

The impedence X = (2pi)fL (where f is the frequency in question and L is the inductance in Henries). A good rule of thumb is that the impedence at 50 hz should be about two to three times the plate resistance of the driving tube or about 7 to 10 Henries of inductance per 1000 ohms of plate resistance.

For example, a 5687 driving a 2A3:

Rp = 2000 ohms

X = 6000 ohms at 50 hz

L = X / (2pi)(50) = 19 H

This is really a minimum requirement as setting X at 10-20 hz will give more headroom and cost more or you can save money by using tubes with lower plate resistance such as 6C45pi.

John
 
Going back to the Hammond 804's, measurements show driving transformers with too low an impedance can also cause problems. It's been a long while but I recall that too much source impedance rolled of the high end too quickly and too little leads to peakiness. That's what's bothersome about ambiguous turns ratio specs. Companies like Hammond, and Sy's example Jensen, are very careful to specify precise, optimum conditions for their product such as I/O impedance and operating levels for best performance. It wouldn't surprise me if the dislike for interstage transformers I sometimes read about on the forum comes from simple mis-application of a product based on lack of information.
 
User name

Hello,
TerribleT is Sidewinder. I had difficulty in registering using Sidewinder so I used TerribleT. Both are names of jazz tunes by my favorite trumpet player, Lee Morgan. Again, thanks for all the replies. I am beginning to understand. I need to further digest all the information given. But it seems, everytime I gain knowledge, it become evident I know less than what I thought I didn't know before. Here's another post: http://www.audioasylum.com/forums/magnequest/messages/1832.html[/URL]
 
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