Is there a Better 10K:10K Audio Transformer than...

[Paul Marossy]

Cool beans. Which transformer did you use?

Cool beans?! I haven't heard that expression for a long time. Anyhow, I went with the trusty old Mouser 42TM018. Just how the frequency response of these audio isolation transformers works is still kind of a mystery to me, looking at the quoted frequency response. In any case, I can only hear a very slight change in the bass response when going thru the transformer vs. it being bypassed. It would be an interesting experiment to try one of the Edcor transformers, though... 

[R.G.]

What can I say? I've seen about six generations of "clever" expressions, and sometimes I get them mixed up. It's a "cat's meow" kind of thing... 

Quick course in transformers.
1. The input to a transformer always looks to the driving circuit like a series inductor (the leakage inductance), a series resistor (the winding resistance of the primary plus the referred secondary resistance), a shunt capacitance, a shunt inductance, and the referred secondary load in parallel with the shunt capacitance and inductance.

2. The referred secondary load is what you want to drive. You also have to drive the other junk to do this because the other junk is in the way, hanging on in series and parallel. Transformers are designed in a way to make the other junk not matter very much.

3. The shunt inductance is the inductance of the primary winding all by itself. It's this that sets the low frequency response. That happens like this:

• Assuming the secondary load is the actual nominal loading - for a 10K:10K, that would be a 10K resistor, we would expect that 10K secondary load to appear on the primary through the transformer.
• ... and it does. But the primary inductance appears in parallel with it. As frequency goes down, the inductance has an impedance that gets lower and lower, and at some point, the inductor's impedance is the same as the 10K referred secondary load.
• Since the transfomer's nominally being used to match things, the driving circuit has some significant output impedance. If that is true, then at the frequency where the inductance and referred load match, the inductance is eating drive current equal to the referred load current - and thats by definition half the power.
• We have just discovered the low frequency rolloff point. The maker specifies the low freuqency response as the frequency where the primary inductance equals the rated primary impedance.
• As a side note, the high frequency response happens when the leakage inductance impedance equals that same rated primary impedance (at least to a first order approximation)

It's quite uncommon to do this, but I misused the circuitry in my splitter to get around the rated 300 Hz rolloff point. I noted that for what I was doing, I didnt have to match the primary and secondary impedances. 10K is too low an impedance for audio purposes anyway. So I used an opamp to buffer the input of the splitter, and I had this opamp output to drive the transformer. Opamps have a near-zero output impedance within the limits of their ability to pour out current. So what if the primary is eating half the drive current? Let it. There's lots more where that came from. Let the opamp just shove lots of current into the paralleled primary inductance/secondary load. So what if the inductor eats most of it? The referred secondary load gets enough to keep its voltage response up. And it works. It extends the low frequency response on that transformer from 300Hz down to just under 60Hz - enough for isolating the 82 Hz low guitar notes.

[Paul Marossy]

Thanks for taking the time to explain that RG, I appreciate it. I'm printing this out to chew on it some more. 

[Joe Kramer]

. . .  Quick course in transformers. . . .

Cool beans!     Thanks RG!

One question: Why are transformer specs given in ohms AND ratio?  I've seen 600:600 and 10K:10K used almost interchangably in isolation applications, ostensibly because they're both 1:1.  What's the rule of thumb in deciding which is the more important figure, ohms or ratio, or (God forbid) both?

Joe

[R.G.]

Why are transformer specs given in ohms AND ratio?  I've seen 600:600 and 10K:10K used almost interchangably in isolation applications, ostensibly because they're both 1:1.  What's the rule of thumb in deciding which is the more important figure, ohms or ratio, or (God forbid) both?

Good question.

Transformers don't have ohms. Transformers have ratios and limits. When you see ohms, there are unstated specifications about the transformer you are expected to know.

Some of these are related to the limits. A transformer may be specified as 10K:10K and (to pluck a number out of the air) 100mW. That is, the transformer is being represented to you as delivering one tenth of a watt into a 10K load resistor attached to the secondary. You're expected to know or figure out the relevant parts of the following:

100mW into 10K is computed by ohm's law. Power = I * V, and we know that V = I *R, so we can pervert those equations into P = V^2/R (that being how I write "V squared divided by R). So we can also figure that V = SQRT(P*R) = SQRT(0.1*10,000) or 31.6V. In this case, the volts are RMS, so that 100mW transformer will transform a bunch of volts. There's probably some fudge factor there, too.

There's not much current in that. 31.6V into 10K ohms is only 3.16ma.

A transformer specified at 600:600 ohms 100mW is going to be able to do an output of V = 7.74V maximum and 12.9ma by the same calculations.

So far so good. But back to the limits. The transformer is not only limited in power delivery, it's limited in frequency as well. Most audio transformers are assumed to be voice/telephone grade unless you pay extra, and the implication of that is that they'll do 300Hz to 3kHz. So back at our 10K transformer, they will only have put enough windings on the primary to keep the core from saturating with 31.6V at 300Hz, and enough to keep the inductance over 10K at 300Hz, or a minimum of about 5.3H. They may have put more in there, but that's all you could reasonably count on in the absence of a datasheet. The leakage could be quite bad, but most small transformers have much better treble response than they have to.  The 600:600 transformer can have lower inductance because it has much lower volt-seconds impressed on its primary.

Note that the 100mW 10K to 10K and the 600 to 600 are probably the same sized core. All that changes is the number of turns and the wire gauge. The power handling is a property of the core and window area, not the impedance level (in a broad generalities way).

So when they tell you 10K:10K 100mW audio, they're really saying "31.6V to 31.6V unloaded, up to 3.16ma secondary current, at frequencies between 300Hz and 3kHz". Only they tell you the square of the turns ratio (that is, the impedance transformation ratio) times the expected load impedance they designed for.

Did that help?

[Joe Kramer]

Did that help?

I think so.

First off, it's generous of you to assume I'll understand the math, but I'm embarrassed to say my eyes literally glaze over at numbers and hieroglyphics.  That's my failing, and maybe someday I'll get my math act together. 

That said, here's what I do get (I think).  I understand that transformers don't have any inherent impedance or ohms, but that they only "reflect" the impedances from and to their respective driving or loading sides.  I know that in terms of ratios, small primaries into big secondaries step up voltage at the expense of current (gain versus drive), whereas big primaries into small secondaries step down voltage to the advantage of current (drive versus gain).  The "1:1" question seems to me where all the tricky stuff comes to the forefront (although I realize it's still applies elswhere).

IIUC, the 600:600 will have better current drive ability (at the expense of voltage). For maintaining some sort of fidelity when sending audio through cables and whatnot, that's more important than volts usually.

Still not quite sure why "The 600:600 transformer can have lower inductance because it has much lower volt-seconds impressed on its primary,"  but I'm certain it's already there in your two posts.  I just need to mull it over, get some kind of mental picture to understand it.  Thanks!

Joe

 

 

[R.G.]

Still not quite sure why "The 600:600 transformer can have lower inductance because it has much lower volt-seconds impressed on its primary,"  but I'm certain it's already there in your two posts.  I just need to mull it over, get some kind of mental picture to understand it.

Think of it like this:
Any given piece of ferromagnetic material can only be loaded up with so much magnetic flux before it saturates. That literally means all the little atomic electron-spin magnets inside are lined up the same way, so you can pump harder, but no more of them CAN align. So any further magnetic pumping is just wasted as heat.

In a transformer, the primary side pumps into the core, the secondary side sucks out of the core, and the difference between the two maintains the core at a given level of magnetic flux. That is, the transformer core eats some of the energy we put in as a cost of maintaining the magnetic coupling. It's a good deal, since the cost is less than a few percent.

If you put a voltage across an inductor, the current inside the transformer can't change instantly, but instead ramps up linearly over time. The inductance is one way to express how fast it can increase. Big inductors can only ramp up or down slowly, little inductors can be moved faster. The magnetic flux inside the core is proportional to the number of amperes through the winding and the turns of the coil, or ampere-turn product. Since the current can only increase proportionately to the voltage across the inductor and the time it spends there, then you can also rate the inductor's core ability by the volt time product. If you put one volt across it, then the time until saturation is ten times as long as if you put one volt across it.

That volt-time product also affects AC operation. On each half-cycle, the voltage goes in one direction for a while, then reverses and goes in the other for a while. I won't lead you through integrals, but take it on faith that there is a maximum voltage/lowest frequency sine wave that lets the positive (and then negative) half cycle just complete without saturating the core. If you raise the frequency, you can also raise the voltage; if you lower the voltage, you can also lower the frequency. You can see that the volt-time product is a limit.

But we had to produce a certain number of volts at the primary of the transformer to get the required volts out of the secondary, and we had to meet a certain low frequency spec when the transformer was designed.  That means that if we want to keep our highly paid and heavily perked transformer design jobs, we have to make designs that can accept the specified voltage at the specified low frequency point, and that is in fact another specification of the primary inductance. The primary inductance has to be big enough that it can hold up that many sine wave volts at that lower frequency. So we have not only an impedance level specification of the primary inductance, we have a saturation specification of the inductance. One will be more restrictive than the other, and the transformer will have to work with that one.

However, in general, the lower the primary voltage, the lower the primary inductance can be without saturation. So the lower voltage winding for 600 ohms can be lower inductance  than the higher voltage winding for the 10K, if the other criteria to be specified don't get in the way.

Did that help?

[Paul Marossy]

That being the case, a 10K:10K transformer is better for a guitar than a 600 ohm:600 ohm, right? Or does it matter once you put an FET/opamp in front of the primary? I know, I'm ignoring the output impedance...

[R.G.]

That being the case, a 10K:10K transformer is better for a guitar than a 600 ohm:600 ohm, right? Or does it matter once you put an FET/opamp in front of the primary? I know, I'm ignoring the output impedance...

Hmmm... good question. I think the fact that it's buffered makes the questions be decided by secondary things.

Jensen used a 10K:10K without a buffer on guitar. I can't see how they got away with that. True, the transformer is unloaded and the reflected impedance of the cable and amp input are high, maybe over 1M, but the primary inductance is still designed for a 10K load, so I think you'd get some bass cut. But the buffer fixes that.

You need to make sure that your primary voltage NEVER gets over the implied voltage level for the primary for good fidelity. The 600 ohm one is over 7V rms, which is plenty, even for "distortion" humbuckers. In fact, a driver working from 9V can't get there. 9V limits you to +/-4.5V peaks with no losses, and that's 3.18Vrms.With battery life and losses, probably you can't get more than 2Vrms into the primary, so the 600 ohm is OK for that.

Because the transformer looks undamped at very high frequencies, over audio, you really need to terminate it in its nominal impedance somewhere above 20KHz. That's what that 10K/0.01 network does. I'm guessing that there's not much audio content up there, but if you had ultrasonics being generated, or RF, the loading on the output gets significant and you'd drain your battery faster with a 600 ohm.

Other than 600:600 transformers being REALLY designed for 300Hz to 3KHz, there's not a lot to choose from if you put in an active buffer. Probably it works OK either way.

Unless there's just something I forgot. That happens.

[Paul Marossy]

Jensen used a 10K:10K without a buffer on guitar. I can't see how they got away with that. True, the transformer is unloaded and the reflected impedance of the cable and amp input are high, maybe over 1M, but the primary inductance is still designed for a 10K load, so I think you'd get some bass cut.

Yeah, I've seen those Jensen Transformer schematics before. They claim that their JT-11P-1 10K:10K transformer has a freq response from 20kHz to 100kHz if I interpret their specs correctly. That must be why they want nearly $100 for one. Anyhow, maybe that's how they get any with it?

EDIT: I see that on your Hum Free A/B/Y Splitter that you have a conjunctive filter(?) on the secondary of the transformer. What is the purpose of that? Should I have that on my little circuit?

[R.G.]

It's not a conjunctive filter - at least it's not intended to be. It's a frequency selective load. As I mentioned, it's to terminate the transformer secondary (and hence its primary also) in a design-matching 10K resistance at ultrasonic frequencies. If that wasn't done, there would be a big hump in the response somewhere. Often it's ultrasonic, as it was for the Mouser transformers, but it can get down into the audio as well. You really don't want a lot of ringing and underdamped response even if you can't hear it.

[Joe Kramer]

Did that help?

Again, I think so.   Less windings = less inductance = less low frequency roll-off at a given voltage (sort of).   Anyhow, I'm certain what you're saying is on the money, but I don't want to risk exhausting your patience, because it looks likes what I really need to do is bone-up on the whole subject of inductance, reactance, and how all that comes into play especially regarding frequencies at different voltage/current levels.   Paul's practical question of "what's better," and a ballpark answer from a more knowledgeable source (you) works for me at this point though.  Many thanks!

Joe
 

[Paul Marossy]

It's not a conjunctive filter - at least it's not intended to be. It's a frequency selective load. As I mentioned, it's to terminate the transformer secondary (and hence its primary also) in a design-matching 10K resistance at ultrasonic frequencies. If that wasn't done, there would be a big hump in the response somewhere.

Ah, I see. I remember that you basically said that earlier, but it didn't sink in until now. 

Often it's ultrasonic, as it was for the Mouser transformers, but it can get down into the audio as well. You really don't want a lot of ringing and underdamped response even if you can't hear it.

Yeah, I guess ultrasonic oscillations in any kind of circuit isn't good. And certainly not in the audio band. 

it looks likes what I really need to do is bone-up on the whole subject of inductance, reactance, and how all that comes into play especially regarding frequencies at different voltage/current levels.

That's basically the whole basis on how the wah circuit works - inductance, reactance, capacitance, apparent capacitance, capacitive reactance, etc. - they're all in there. 

[Smooth]

Hello, I am building a three FX loop box in a 1590DD, with an output splitter after the third loop, having two outputs for a wet/dry rig. One output will be isolated, with a phase switch.

I am trying to decide between a 600/600, or a 10K/10K isolation transformer.

A 600/600 transformer will drive with a lower output impedance than a 10K/10K. However, there is the really low 600R input impedance to consider, and it will require an IC buffer right in front of the split. However, I think an IC buffer would be best, even with a 10K/10K, because most of my pedals in the loops do not have an output buffer.

I am extremely low vision, so I can't make my own PCB. I must use a transformer with termination my helper can apply to my project, and I must use buffers I can source, like the GGG IC buffer, that comes with a PCB and a schematic/wiring diagram.

I will be running this looper at 18V from an isolated power source, because a couple of my overdrives can be ran at 9-18V, and I also have a booster that runs on 9-18V. I assume a transformer rated at 10Vrms is enough headroom for 18V pedals?

Is the 600/600 the better option just because it has a lower output impedance? If the output impedance of the IC buffer is extremely low, say 75 ohms, then it should be able to drive the 600R input? However, 750R would be 10 times larger, than the 75 ohm output impedance of the buffer. Plus, if the main output jack were ever to be connected into a lower than ideal impedance the output of the buffer would be farther loaded down.

On second thought, should I just use the 10K/10K, and use the isolated output into my wet effects? This way, the 10K output impedance can go to my delay, with a short cable. I can use the main output to my dry amp. I think all my wet effects have a high input impedance, and a low output impedance. The last one can even run in stereo, when I want that option. Basically, what I'm saying, is the wet effects would be buffering between the secondary, and the amp(s) they will be driving.

I will be using the Edcor WSM series 0.5W transformer with spade connectors, rated at 10Vrms. I think it'll fit in my 1590DD, even with three stomp switches, two buffers, 9 jacks, DC jack, and switches.
https://edcorusa.com/products/wsm-series-balanced-or-unbalanced-line-matching-transformers?variant=41117605167291

or the WSM-LD, with 9 inch lead wires.
https://edcorusa.com/products/wsm-ld-series-1-2w-balanced-or-unbalanced-line-matching-transformers?variant=43372828197113