What’s going on with the inductor in the Bobtavia?

[Rob Strand]

The original poster wrote:

I think this transformer is 1k:8ohms. Looks like the 386 is feeding the 8ohm side. So the signal should be stepped up on the 1K side. What happens here? Is the AC signal boosted by 1000/8 ?

I'm assuming it's a 1k ct to 8 ohm.   Pretty common back in 70's transistor radio times.

For example,
https://www.jaycar.com.au/1k-ohm-centre-tapped-8-ohm-output-transformer/p/MM2532

The part number 273-1380 is an old Radioshack/Tandy/Archer part number.

Go here, printed page 69,
https://www.tandyonline.com/doc/catalogues/Tandy-Catalogue-1974.pdf

1k ct to 8 ohm.

[Rob Strand]

Sorry I forgot to add AC boost by approx sqrt(1000/8) = 11.2 to outside terminals
then 11.2/2 = 5.6 times each side of the centre tap.

[Vivek]

Rob, is voltage ratio dependent on impedance ratio or turns ratio ?

[Rob Strand]

For the ideal case all the ratios are related by nature. The turns ratio is the only thing that is truely constant because you have physical turns.  So from that perspective all the other ratios depend on the turns ratio.

n = N2/N1 = V2/V1 = I1/I2 = sqrt(Z2/Z1)

For the non-ideal case things start to deviate.   The way you handle that is with the circuit model I posted before. 

Imagine a 10:1 transformer.   In the ideal case 10V in and 1V out.  Also if you put a 1 ohm load on the output it looks like 100 ohm on the input.

But now consider a simplified non-ideal case where the secondary has a resistance of 1 ohm, primary still ideal.  At no load we still see 10V in and 1V out.   However, when we place a 1 ohm load on the output we will see 200 ohm load on the input.  The true load on the "ideal transformer part" is the 1 ohm secondary resistance in series with the 1 ohm load.  From outside the transformer we have a 1 ohm load but as far as looking into the primary is concerned the "ideal transformer" has  2 ohm load and the impedance seen at the primary is (N2/N1)^2*(1+1) (N1/N2)^2*(1+1)=10^2 * 2 = 200 ohm.

For small transformers you do have high winding resistances so things do get messed up like this.   You might ask if a manufacturer quotes a 1k to 8 ohms on one of these small transformer what are they quoting? 

- If I put 8 ohms on the output do I see 1k at the input?  In other words the manufacture had reduced the transformer turns ratio to allow for the transformer resistance *so that* the input looks like 1k with 8 ohm present? 
If we did that putting a 1k on the input side and looking into the 8 ohm terminal would worse off than if the transformer ratio was reduce.   In other words you can only compensate the impedance with the turns one direction.

or, 

- Based on the impedance ratio the implied physical turns ratio is sqrt(1000/8) = 11.2.  So we would expect 1V in and 1/11.2 = 0.0893mV out.   Probably not.   However  0.0893mV plus some drops due to the primary (R1, Xl1) forming a divider with the core loss is more like it.   This definition is at least symmetrical.

To be honest, when I check some manufacturer's data it is not possible to work out what their transformers are doing down to this level.    In fact the data for many small transformers often don't give enough specs to split the difference, or the data contains errors!  In other cases the manufacture gives wide tolerances or don't give the test frequency, again making it difficult to check.  For larger transformers things tend to be closer to the theoretical results.

For 600 ohm telecommunications transformers it usually looks more like the second case.   However more often than not these are 1:1 or 1:1 ct, ie. simple ratios.

[Vivek]

Rob, I really appreciate your mathematical posts !!

Here is a thought experiment Question 1 :

We make a transformer as follows

Transformer 1:
Primary has 100 turns and DC resistance 10 ohms
Secondary has 1000 turns and DC resistance 100 ohms

Then we use wires of different thicknesses on the secondary and make

Transformer 2:
Primary has 100 turns and DC resistance 10 ohms
Secondary has 1000 turns and DC resistance 50 ohms

Transformer 3:
Primary has 100 turns and DC resistance 10 ohms
Secondary has 650 turns and DC resistance 100 ohms

Now we apply 1VAC to the primary. Place 200k ohms load on the secondary.

What voltage will be expect on the secondary ?



Question 2 :

How is following applicable

n = N2/N1 = V2/V1 = I1/I2 = sqrt(Z2/Z1)

when we can use different wire thicknesses on the windings and therefore delink turns with impedance ?

[antonis]

Winding DC resistance is irrelevant to turns ratio hence voltage transformation and only counts for achievable power adaption..

[Rob Strand]

If I wind a transformer with N1 turn primary and an N2 turn secondary I don't think you have a problem accepting that V2/V1 = N2/N1.   So applying V1 to the primary we get V2 = V1 * (N2/N1) at the secondary.

For the ideal transformer you can also work out that the current ratio is the inverse, I1/I2 = N2 / N1.

Where you can misinterpret things is with the impedance *spec* of the transformer.    For the ideal transformer that has nothing to do with the DC resistances of the windings.     Despite all the derivations and calculations - that's the main message of the post!

Here's how it works,  (no *winding* DC resistance mentioned at all)

If I put a load on the secondary RL2 the secondary voltage and current must follow ohms law,

         RL2  = V2 / I2

If we write V2/V1 = I1/I2 = N2 / N1 in terms of the V2 and I2 we get,

             V2  = (N2/N1) * V1 and I2 = (N1/N2) * I1

We then can write,

       RL2 = V2 / I2 =  [(N2/N1) * V1] / [(N1/N2) * I1]  = (N2/N1)^2 (V1/I1)
=>   RL2  = (N2/N1)^2 (V1/I1)

If we apply V1 to the primary with RL2 present a current I1 flows into the primary.  So that means the impedance looking into the primary when RL2 is present is,

     RL1   = V1 / I1    =  RL2 *(N1/N2)^2

which we can write as,

    RL1 / RL2  = (N1/N2)^2

or

    N1 / N2 = sqrt(RL1/RL2)

That's where N1 / N2 = sqrt(Z1/Z2) comes from.    It says how a load at the secondary appears at the primary.

So when you specify a 10k ohm to 100ohm it is really saying the turns ratio is N2/N1 = sqrt(10000/100) = 10.
If I specify a 100 ohm to 1 ohm transformer I also get N2/N1 = sqrt(100/1) = 10.

In theory there is no difference between these two transformers.   Both transformers could be wound with with a 10 turn primary and a 1 turn secondary.   

In practice it's a different matter.   So here's where it gets complicated.   

In order to make a practical transformer we need to use real windings.   We also have to chose the number of turns.  We have to choose the size of the transformer.  The number of turns and the size determines the DC resistance.  If we fix the core size we can have a primary 10 turns a secondary 1 turns with thick windings and a low winding DC resistance, or, we can have a 10000 turn primary and a 1000 turn secondary of thin wire with high winding DC resistance.

The DC resistance is what determines specifying the transformer as 10k ohm to 100 ohm vs 100 to 1 ohm.   If we are going to use a 100 ohm load we want the transformer to look ideal as possible so we might aim for DC resistances less than 1/20th to 1/10th the intended impedance, say to 5 to 10 ohm for the 100 ohm winding and 500 ohm to 1k ohm for the 10k winding.   Normally we wouldn't want a 10 ohm DC resistance winding feeding a 1 ohm load because you will get a big voltage drop at the output.

Beyond that we get into fine details of transformer design (and there's a heap of details I've left out).

Here is a thought experiment Question 1 :

The simple and approximate answer ignores the DC resistance.  We can even justify it because the DC resistances are so much lower than the load resistance.

V2 = V1 (N2/N1)

        N1    N2    V1     V2
T1    100   1000 1       10
T2    100   1000 1       10
T3    100   650   1       6.5

If you wanted to account for the DC resistances the first step is to move the primary DC resistance to the secondary side of the transformer.   This is called reflecting the impedance,  and we use the formula.

            R1'     =   R1  * (N2/N1)^2                   ;effective primary resistance reflected to secondary side

Then that appears in series with the DC resistance of the secondary given a total effective winding impedance of,

           RT'      = R1'  + R2

It's like a resistance RT'  has been placed in series with the secondary winding.  (I don't know if it helps to look at the transformer equivalent circuit I gave earlier.)

This resistance RT' forms a voltage divider with the load resistance RL at the secondary,

      V2_loaded  = V2_unloaded  *    RL  / (RL  + RT')

Where V2_unloaded is V2 from the table calculated from the turns ratio (the ideal transformer calculations).

The thing to note here is if RL is a low value, say in the order of RT', the loaded output voltage will drop compared to the unloaded (and expected) voltage.   This is exactly the issue when the winding resistances are too high for the intended load impedance.

Question 2 :

How is following applicable

n = N2/N1 = V2/V1 = I1/I2 = sqrt(Z2/Z1)

when we can use different wire thicknesses on the windings and therefore delink turns with impedance ?

So that was covered in my opening comments about where the DC resistance comes into the picture.

[pinkjimiphoton]

i still have a rat shack transformer waiting for me to build another one.

its an octave up. the diodes are used  like that so that on either side of the transformer so that it cancels out most of the fundamental.
if ya lift the ground, you get a nice wooly fuzztone

yes, the output jack as shown is wrong

the 386 has plenty of power to drive the diodes its still loud as hell

ge will weaken the effect and output volume significantly

this was like, my 4th or 5th build, i made a bunch of them, its a great sounding circuit that was designed to be built with readily available radio shack parts

i can measure the transformer if ya needs it

[iainpunk]

holy sheeeeet!!! this might be the final solution to my "Jawari wiring'' s gating problem.
i have had one of my guitars wired this way:



and it gates the output quite a bit, which is why i barely use the switch to activate it.
if i can find a way to fit single small battery inside the already cramped body cavity, and also how to wire it up, ill be so happy if it works out

youll be hearing from me

cheers

[soggybag]

I was just tinkering with my build. I was getting weak output unless the input signal is strong. With the gain all the way up the output is weak but the volume jumps when you hit a note on the low string or hit a chord. This seems like how it should sound all the time.

If I put a fuzz box or boost in front of the bob and crank it up the bob sounds like what I would expect.

Im guessing it's something about getting the output from the transformer up to and over the forward voltage of the series diodes.

I added a jumper across pins 1 and 8 of the 386 to put it in blow out mode and now it sounds like a super octave fuzz all the time. This mod makes the gain control mostly irrelevant.

[iainpunk]

i think i found a way.

im relying on the volume pot to limit the current, and its only a first drawing of the idea, maybe ill add a huge value resistor in series with the battery to make the current negligible. im not afraid of DC pop when switching, its an experiment anyway.




cheers

[soggybag]

Question: when AC is across one coil of the transformer the other coil mirrors the signal and might step the voltage up or down. Does that mean the signal is inverted at the bottom of the coil?

[antonis]

I'm not sure what you ask, Mitchell..

When there is no marking (either Dot convention or H - X labeling) we can't say anything about phase shift..
(the above actually state if both windings turns are wound in the same direction..)

In case of marking matched, there is no phase shift (0^o)..
In case of marking mismatched, there is 180^o phase shift..

[PRR]

... ... ... Does that mean the signal is inverted at the bottom of the coil?

It is a center-tap winding for a push-pull output. The two ends are out of phase, of course.

[soggybag]

Thanks, I had an idea that was happening. It was more intuition than a strong understanding.

What happens if the center tap is not connected to ground?

[anotherjim]

I might be wrong, but I think the phase "dots" will be diagonally opposite. At least it is on the small audio transformers with pin connections that  I have (from the pinout diagrams - not marked on the traffys themselves). Of course, this isn't so simple if it's wired and you can't see where the wires go to the windings. You could easily check with a 'scope but for a full-wave rectifier frequency doubler, the true phase from input to output probably won't matter.

Isn't it that the coils are laid in the same direction but magnetic coupling makes a 180deg phase shift from primary to secondary?
Then I found this...

... which I would assume puts the phase dots on blue and red, not diagonally opposite, but it's only schematic, not the physical layout.

[pinkjimiphoton]

Thanks, I had an idea that was happening. It was more intuition than a strong understanding.

What happens if the center tap is not connected to ground?

you get a standard fuzz sound

[soggybag]

Is my thinking correct here:

The center tap could be connected to ground and the octave would still be working because the two ouputs on either side would produce inverted signals from the other?

In Bobs case the center tap is connected to 9v through a 100k resistor. This produces the same effect but offset by the voltage at the center tap?

The voltage at the center tap is used to bring the output up to the forward voltage of the diodes in order to get them to conduct sooner?

I measured 5v at the center tap. The voltage dropped about 0.5v across each of the diodes. I measured 4V where the last two diodes meet before the output.

Was the goal here to get the resting voltage at the output to be at the middle of the signal swing?

[Vivek]

Try placing a cap across the resistor used to connect centre tap to ground

[soggybag]

Try placing a cap across the resistor used to connect centre tap to ground

I will give this a try later today.