[dconant]

How would I go about removing the spider from the cone?

[RodB]

The spider is glued on the cone and so is the voice coil. I think you can use a cotton swab dipped in acetone and carefully apply the acetone to the edge of the spider. Maybe someone else has a different suggestion.

[MrFixr55]

I can't remember, but there was a particular product for loosening glue on the more modern "spiders", used post-war, and also to loosen the glue that holds the cone to the speaker frame. Can't remember what it was but GC cements used to sell it. Probably the same stuff needed to remove the spider used on a 91.

[dconant]

I'll give the acetone a try and let you know.

[morzh]

I do not have much experience with speakers, but acetone sounds right. It does not affect paper but does dissolve about anything else that has polymers in it, like glues, plastics etc.

[dconant]

I have received my new output transformer (Hammond 125C). I thought I understood the wiring but now I'm not sure. Since my primary impedance is supposed to be 680 ohms I don't see anything in the table below how I would connect the secondary wires. For testing purposes I will be using a PM speaker measuring about 5.3 ohms and a resistor for replacement of the field coil. How do I hook up this transformer? I have included the specs for the K-6 speaker. Here is the link to the transformer. https://www.hammfg.com/electronics/trans.../125a-125e

[morzh]

Ok.

Don't confuse the tubes' expected load impedance and DC resistance of your primary.
I do not have the sch in front of me, so I will be pulling the numbers out of you know what, but here it goes:

Say, your expected pushpull load impedance is 10 kOhm (see tubes' datasheet).
Your voice coil, let's say, about 1.1 Ohm.
This means, you need 95 turns ratio (9,000 impedances ratio) to convert 1.1 to 10,000.

Your transformer will show the coil resistance on top and your load impedance in teh coil under it. Your ratio is 9,000.
9,000 times 1.1 is 10000. But you have 1.5, so look for the value app 1.5 x 9000 which would be 13,500. The closest is 16500. This will be achieved connecting your voice coil to Wires 3&4.

Now, if you know exact voice coil impedance, use that instead of 1.1 ohm.

[dconant]

I see where the 10k ohms comes from in the tube data sheet. Where does the 95 turns ratio come from?

If I use 5.3 times 9000 I get 47, 700. There is nothing anywhere near that. Can you give me some examples if my voice coil measures 5.3 ohms?

What happens when the mismatch is not close?

[morzh]

95 is roughly the square root 9000.

OK, so you have 5.3 Ohm.

Look at 4 Ohms as the closest.

Let's solve this:

the closest in 4 Ohms column is 8,800, when connected to 3 and 5.
Now, 5.3:4 is 1.325.
So your 8,800 become 8,800 x 1.325 = 11,660.

This is pretty close to 10K you need.

However, you could also use 4 and 6. This will give you 6,400 x 1.325 = 8,480.

Also, you need to add to these numbers this to your 680 Ohms of primary DC resistance, more or less (I am not getting into reactances and complex impedances)m which will give you 9,160 Ohms, which is also pretty close.

So, use either 4&6, or 3&5, either one is good enough, 4&6 being closer.

[RodB]

It sounds like you measured your voice coil at 5.3 ohms. That would be a DC resistance. Impedances are calculated because they involve resistance at a certain frequency, usually 1000 hz for audio circuits. Your voice coil would probably be 6 ohms impedance. You use the output tube load impedance and speaker impedance to get the turns ratio for the transformer then use the chart to decide what secondary taps to connect to the speaker.

[morzh]

Rod

Yep, and 6 ohm still make it 4&6

It is all pretty approximate.
Not only that, but if one wants to be real anal about these things, the transformer itself behaves in a pretty interesting fashions depending on the frequency range.


At low frequencies the impedance is pretty much that of the reflected resistive impedance of the speaker and the losses of the output winding, paralleled by the equivallent core losses and the primary inductance impedance, and then added to (serialized with) the losses (DCR) of the primary.


At midrange, the impedance is pretty much the same without the primary inductance factor.

And in highs, it is parallel of parasitic capacitance of the windings and the core losses, serialized with full stray inductance (prim and sec), and reflected resistive losses and the resistive load.

[dconant]

I don't really understand all of what was just said. I wish I understood how to determine the turns ratio. I used 4&6, subbed in new e caps and fired it up for the first time. It plays. There is a fair amount of work to do on this chassis which include replace all caps and out of tolerance resistors plus a lot of cleaning. The big canned ohm resistor is open in the 2700 section. I subbed in a 5 watt. Do you think that is enough or do I need a 10 watt. Schematic does not say.

I'm just happy the radio is playing on what looks to be all original globe tubes. Probably be a while before I dig into this radio.

Thanks all for the help so far.
Dan

[morzh]

If your resistor does not get superhot, 5W is enough.

here's about the turn ratio: I will try to make it simple. But you have to read it all.

You need to understand what a transformer (AC power or audio output type) is:
- The ideal transformer is an IMPEDANCE MATCHING DEVICE. That is, it takes an impedance, whatever type, and converts is into the same exact type of impedance, but of a different value. At its simplest, it will convert a resistance of X Ohms into a resistance of N-square * X Ohms.
Where N is the turn ratio.

Example: a transformer with the primary of 100 turns and the secondary of 1 turn (100 turns ratio) will convert the resistor of 1 Ohm connected to the secondary into 10,000 ohm at the secondary.
That is, we have just matched a 10 kOhm output impedance, as a Pushpull of 2x42 tubes would like to see, to a 1 Ohm load, which is a typical old speaker (Philco) resistance.
(Again, I simplified the impedance of the speaker, but this is just not to confuse you much).

Why is it? Transformer is the power transfer device. Which means, with the ideal transformer, the power you put into the primary, will be the same, as dispensed to the load from the secondary.
Example: same as above.
You do know by now (it is a common knowledge), that the voltage at the secondary is equal to the Voltage at the primary divided by the turn ratio.

So we have The ratio N = 100.
This means, that a 100V, put across the primary, will result in 1V output from the secondary.
Now, 1V into a 1 Ohm load will create 1W of power. Because 1V/1 Ohm=1A. this is at the secondary and the load.
But at the primary, given the power constant of 1W, you will have 100V and 10mA (0.01A). 100V * 0.01A = 1W.
Now let's look at the resistance: we have 100V/0.01A = 10,000 Ohms.
This is how we converted 1 Ohm into 10,000 Ohm. That is, the conversion factor is 10,000, which is the square of the 100, which is the Turn Ratio.

So, in order to match an impedance Z1 to an impedance Z2, we have to make a transformer with the turn ratio of (SQRT os Square Root)

N = SQRT (Z1/Z2).

[morzh]

Let's add here the power transformer. In principle, this is exact same type of the transformer as the Audio Out.

Only it matches the power of the AC source (Mains, 120V in the USA case) to whatever the load type is.
This is why you see a, say, 120V and 1A in the primary (at full load of a tube radio), and then the sum of the powers of A+ and B+ being equal to it.
You simply matching the load: low voltage/high current filaments' load and high voltage/low current plates, SG biases etc, that demand small currents but 300-400V voltages.

[dconant]

Thanks Morzh for taking the time, I will try to digest this. I really want to understand this.