The GAP in my slow coming knowledge...

This brings up another question from me (you knew it would :razz: ). I thought tube grids were, in general, pretty high impedance. Why do I see resistors across the secondary of many line inputs? For example, on an LA-2 you've got a 68k. If the transformer were 1:1, which I don't know but just as an example, wouldn't that make the input imedance 68k? Why bother with that and just run the input straight to the grid when the impedance would be R6 (68K) plus the grid impedance making a nice high impedance load? I'm missing something here :?

cheers - and great topic -
kent

> I thought tube grids were, in general, pretty high impedance. Why do I see resistors...?

A tube grid is a bit of metal in a vacuum. It has very low DC conductance (high resistance), but like any object in the universe, it has capacitance to everything else in the universe. Particularly to the Plate, which has an amplified reverse-phase version of the signal at the grid.

The input of a tube looks like 30pFd-300pFd to ground. So right there we may have 100KΩ of impedance at 20KHz.

The transformer also has capacitance, and it works out well if the unavoidable capacitance is similar to the grid capacitance.

So now we may have 50KΩ of total impedance at 20KHz. but 1,000KΩ of impedance at 1KHz.

Do we want an impedance which varies so much? Usually not. Usually better to not have huge variations. Adding a 68K resistor limits the impedance rise to maybe 67K mid-band, 27K at 20KHz.

Another, more subtle reason. The transformer has leakage reactance. On hi-Z windings, this inductance and the stray capacitance resonates, then falls-off badly. This sets the upper limit of usable response, and also the highest practcable impedance of a wide-band transformer. A common trick is to let it resonate, then add just-enough resistance to dampen the peak.

Aha, I think I get it! It basically relates back to NYD's rule:

Passive filters tend to be fairly critical as regards source and load impedances since these impedances usually form part of the filter itself in actual operation.

Click to expand...

You would end up with a filter which is all wrong for good audio response!

So how does one go about determing whether a circuit(involving transformer) requires termination or not?

For example, the API 312 2503 output transformer does not have termination, but the Neve L01166 does ( a 1.5K resistor in series with a .01uF cap in the case of the 1063 circuit)

How do it know? :grin:

ju

Testing with a scope and a sweepable signal generator will give you the answer more quickly and with greater certainty than any other method.

With a sine input, look for peaks/dips in the response with and without termination. Try a pulse or square waveform and look for tilt, overshoot, ringing or other such nasties.

Don't forget what I wrote earlier: sometimes the transformer is correctly terminated by the source (rather than the load) impedance.

ok I'm still confused. I designed a simple little line stage with a 1626 tube a while ago and the output ohms is 2k. If I ad another tube and make a WCF then the output ohms goes down to 200. How would I decide which output transformer to use for a balanced output on each, assumimg I want them to be able to drive 600 ohm inputs?

Kiira

I don't know. I just take em apart. Ask Paul.

http://www.comtran.com/an/an002.pdf

maybe helpful

To drive a 600-ohm load from a 200-ohm source, use 2400:600 (2:1 turns ratio).

[quote author="CJ"]http://www.comtran.com/an/an002.pdf

maybe helpful[/quote]

:? I didn't get all that reflected impedance stuff. foo.



[quote author="NewYorkDave"]To drive a 600-ohm load from a 200-ohm source, use 2400:600 (2:1 turns ratio).[/quote]

How would I figure that out though? Or figure out which trafo to use with the 2k output stage? I can't find that in any of the links, yanno like a formula that will just tell me what to use? Or is it just experimentation/ general rules that are used?

thanks,

Kiira

Reflected impedance just means that what ever impedance is on one side of the transformer, that is what you will see on the other side. For a 1:1 that is.
If the ratio is 2:1 you will see 1/2 the impedance on the other side of the transformer.

The actual henries of the transformer itself are just what is required to transfer this impedance from one side to the other without losing anything.

here, you can have this for free: