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Step Up Transformers? Pros / Cons?

SIY

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The current Denon SUT's are a bit expensive - but look for the older Denon SUT's from the 60's ,70's and even 80's . I do wonder about using the DL-103r carts with a 15 ohm coil with SUT's designed to work with 40 ohm coil DL-103 carts ?

RC network on the secondary will take care of that, assuming there's nothing odd about the step-up ratio.
 

ClosDeLaRoche

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The current Denon SUT's are a bit expensive - but look for the older Denon SUT's from the 60's ,70's and even 80's . I do wonder about using the DL-103r carts with a 15 ohm coil with SUT's designed to work with 40 ohm coil DL-103 carts ?

Is the AU-300LC discontinued? If so there are tons of Japanese sellers on eBay selling them used for $200 to $250. This is a good price for a good SUT. It seems perfect for the 103 and 301 MK2, and should be good with the 103r as well.
 
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audiopile

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I can appreciate the 300LC should work fine with the 301-II and 103 - but the 15 ohm 103r - why would that work with the 300LC ? Not talkin about soldering matching resistors -just used stock ?
 

ClosDeLaRoche

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I can appreciate the 300LC should work fine with the 301-II and 103 - but the 15 ohm 103r - why would that work with the 300LC ? Not talkin about soldering matching resistors -just used stock ?
I think people get hung up about loading sometimes. I've heard of users loading a 103 at 1k ohms and loving it, and loading it at 100 ohms and loving it. 470 ohms for the 103r isn't ridiculous IMO. I'd expect it to sound damn good at about $550 for the pair.

I suppose I'll never know until I hear it but if someone told me they paired these two pieces of kit I wouldn't scoff at that
 
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watchnerd

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I think people get hung up about loading sometimes. I've heard of users loading a 103 at 1k ohms and loving it, and loading it at 100 ohms and loving it. 470 ohms for the 103r isn't ridiculous IMO. I'd expect it to sound damn good at about $550 for the pair.

I suppose I'll never know until I hear it but if someone told me they paired these two pieces of kit I wouldn't scoff at that


Should SUTs be used as a tone control?
 

LTig

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I had no records in 2011 when I bought my first TT.

Now I have about ~400.
I have to admit that I owned about 80 records when I bought the LP12 in 1994. In the following 5 years (or so) I bough some 1200 records ...
 

LTig

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That's a bit surprising.....at 33 ohms internal impedance, following the usual 10x rule of thumb, I would have expect loading north of 300 ohms to be the starting point.
The Denon 300LC has a winding ratio of 1:10 (means 1mv in / 10 mV out). The input impedance of the SUT is the input load of the phono preamp multiplied by the square of the winding ration which here is 1/100. A typical input impedance of 47 kOhm results in a (theoretical) load of 470 Ohm as seen by the cartridge, which would be fine for 33 Ohm source impedance.
 

Tom C

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I think level matching is important. The Rothwell site the OP referred to is a nice discussion of the basic considerations for matching SUT and cartridge. My first SUT was a Bob’s devices Sky 40, which I paired with an AudioTechnica AT1000 (handmade by a Japanese grandma in the Audio Technica basement). That cart has a rated output of 0.2 mV and internal impedance of 3 ohms. Bob Sattin likes to match impedance, so he recommended his Sky 40. My phono stage has an input impedance of 47000 ohms, so doing the math, a Sky 40 presents a load of about 29 ohms to the cartridge. That’s pretty darn close to the theoretical minimum of 30. So far, so good.
I was pretty happy with the result. My phono stage has separate MC and MM inputs. The MM is straight tube (12AX7), but the MC input adds a jFET gain stage between the cart and the first tube stage. The MC input actually sounds pretty good with the AT1000, but the volume is a little low because the cart only puts out 0.2 mV. With the Sky 40 into the MM input, it was a lot louder. But I think it was too much. The signal level presented to the MM input was 40x0.2= 8 mV. The input sensitivity of the MM input of the phono stage is rated at 4.5mV. So while the Sky 40 gave a good impedance match, I think it was overdriving the input stage.
I decided to try a K&K diy kit, because at the time Lundahl had just come out with a new series of transformers, one of which seemed a likely good match as it had a turns ratio of 1:24. That gave an output level of about 4.8 mV, which matched the input sensitivity of the phono stage much better. I’ve been using that setup since, and find it marvelous.
 
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watchnerd

watchnerd

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Lundahl LL1931 Moving-Coil Transformer Review & Test

https://www.avhub.com.au/product-re...31-moving-coil-transformer-review-test-490591

https://www.lundahltransformers.com/wooden-case/

ImageResizer.ashx


ImageResizer.ashx


ImageResizer.ashx


ImageResizer.ashx

I just ordered a 1931 SUT from K&K audio, partly to see if it lowers my noise floor, partly as a learning exercise / shitz & giggles.

My guess is that I'll have it in about a week.

Not sure how I'll measure it when it arrives, as I believe my volt meter will be problematic in the mV range.
 

maty

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Not sure how I'll measure it when it arrives, as I believe my volt meter will be problematic in the mV range.

Howto - Distortion Measurements with REW by xrk971

https://www.diyaudio.com/forums/software-tools/338511-howto-distortion-measurements-rew.html

index.php


index.php


How much impedance, aka resistor? 600 Ohms?

https://wallofsound.ca/audioreviews/analog/kk-audio-premium-mc-phono-step-up-transformer-review/

My curiosity got the better of me wondering if the K&K loaded to 50Ω would still beat the Cheapo. Yes, it did. With resistors selected to effect a 50Ω load, the properties that distanced its performance from the Cheapo were still in evidence. 50 ohms is likely the lowest loading suitable for the Ortofon Quintet Red. It is possible to raise the 50Ω loading of the Cheapo, but it requires a bit of surgery on the phono stage. The 47,000Ω resistor at the input of the amp could be changed out for one of a higher resistance.
 
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SIY

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JP

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Sigh. It's looking like I'm going to have to put together an article on care and feeding of MC stepups.

Please do.
 
OP
watchnerd

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I don't want to measure distortion, just output so I can the input level of the MM stage more precisely.

I thought I had read somewhere that standard multi-meters like Flukes weren't accurate in the mV range.
 

SIY

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If you have the basic specs for the transformer and cartridge, it's easy to calculate that. If you give me an example cartridge/ transformer, I can walk you through that later today.
 

Tom C

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It’s important to know that if you connect a multimeter to the SUT to measure resistance, the multimeter will send a DC current through the SUT. It has to in order to perform the measurement. That will magnetize the core of the transformer, which will alter the sound, causing distortion. The effect is transient. No permanent damage is done. It goes away awhile after you play music through the SUT. When people are choosing resistors to tune the secondary loading and analyze for ringing, an oscilloscope is used.
 
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watchnerd

watchnerd

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If you have the basic specs for the transformer and cartridge, it's easy to calculate that. If you give me an example cartridge/ transformer, I can walk you through that later today.

The cartridge is an AT33EV, .3mV output, 10 ohms resistance: https://eu.audio-technica.com/AT33EV


The transformer is a Lundahl 1931: https://www.lundahltransformers.com/wp-content/uploads/datasheets/1931.pdf

When built, the full SUT would be the "premium" one on this page:

http://www.kandkaudio.com/moving-coil-phono-step-up-kits/
 

SIY

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OK, there's a lot of steps involved in the calculation, but it's really pretty straightforward. My apologies for not doing figures here, but I'm not on my home computer. I'll normalize things to the cartridge output and then put units on at the end. So for simplicity, the cartridge output is 1. That transformer has two turns ratios, 1:8 and 1:16, the former achieved by putting the primaries in series, the latter by putting the primaries in parallel. Recall that for a turns ratio of N, the voltage transforms as N, the impedances transform as N^2.

As a rough guide, at standard velocity levels (3.54 cm/s, 1 kHz), the cartridge outputs 0.3mV, so for an ideal 1:8, the output would be 2.4mV. For an ideal 1:16, 4.8mV. The actual levels will be slightly below this (10% below is a rule of thumb), but neither are likely to cause overload or insufficient signal problems. So unless you're anal, stop here. Say 4.3mV or so for the 1:16 stepup.

Let's look at it a bit more exactly. First, the primary side. We'll use 1:16 as our example. The primaries in parallel are 1.8 ohm/2 = 0.9 ohm DCR. The secondaries in series have a DCR of 105x2 = 210 ohm. The secondary impedance is reflected back to the primary as (DCR of the secondary x 2 since they're in series + load resistance which is probably 47k)/N^2. Putting in numbers, the reflected secondary impedance is then (210+47000)/16^2 = 184 ohm. So that's the load the cartridge sees.

Now, the DCR of the cartridge forms a voltage divider with the primary impedance. Since the primary DCR is under an ohm, it's negligible compared to that 184 ohm, so we'll neglect it. The voltage delivered to the transformer for the assumed unity output then is effectively 184/(10 + 184) = 0.95. So we've lost a little bit.

On the secondary side, the reflected primary impedance is (10 + 0.9)N^2 = 2790 ohms. (The numbers represent the cartridge DCR plus the primary DCR). This, in series with the secondary DCR of 210 ohm, form a voltage divider with the preamp input of 47k. So... we have 0.95 (calculated above) times 16 (the turns ratio) in voltage driving that divider. Effectively the preamp sees (0.95)(16)(47000)/(47000+2790) = 14.35 for that input of 1.

So... for your cartridge output of 0.3mV, the voltage the preamp would actually see is (14.35)(0.3mV) = 4.305mV, which is what the simple rule-of-thumb predicted.
 
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watchnerd

watchnerd

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OK, there's a lot of steps involved in the calculation, but it's really pretty straightforward. My apologies for not doing figures here, but I'm not on my home computer. I'll normalize things to the cartridge output and then put units on at the end. So for simplicity, the cartridge output is 1. That transformer has two turns ratios, 1:8 and 1:16, the former achieved by putting the primaries in series, the latter by putting the primaries in parallel. Recall that for a turns ratio of N, the voltage transforms as N, the impedances transform as N^2.

As a rough guide, at standard velocity levels (3.54 cm/s, 1 kHz), the cartridge outputs 0.3mV, so for an ideal 1:8, the output would be 2.4mV. For an ideal 1:16, 4.8mV. The actual levels will be slightly below this (10% below is a rule of thumb), but neither are likely to cause overload or insufficient signal problems. So unless you're anal, stop here. Say 4.3mV or so for the 1:16 stepup.

Let's look at it a bit more exactly. First, the primary side. We'll use 1:16 as our example. The primaries in parallel are 1.8 ohm/2 = 0.9 ohm DCR. The secondaries in series have a DCR of 105x2 = 210 ohm. The secondary impedance is reflected back to the primary as (DCR of the secondary x 2 since they're in series + load resistance which is probably 47k)/N^2. Putting in numbers, the reflected secondary impedance is then (210+47000)/16^2 = 184 ohm. So that's the load the cartridge sees.

Now, the DCR of the cartridge forms a voltage divider with the primary impedance. Since the primary DCR is under an ohm, it's negligible compared to that 184 ohm, so we'll neglect it. The voltage delivered to the transformer for the assumed unity output then is effectively 184/(10 + 184) = 0.95. So we've lost a little bit.

On the secondary side, the reflected primary impedance is (10 + 0.9)N^2 = 2790 ohms. (The numbers represent the cartridge DCR plus the primary DCR). This, in series with the secondary DCR of 210 ohm, form a voltage divider with the preamp input of 47k. So... we have 0.95 (calculated above) times 16 (the turns ratio) in voltage driving that divider. Effectively the preamp sees (0.95)(16)(47000)/(47000+2790) = 14.35 for that input of 1.

So... for your cartridge output of 0.3mV, the voltage the preamp would actually see is (14.35)(0.3mV) = 4.305mV, which is what the simple rule-of-thumb predicted.

Thank you so much for putting the work into modeling this and writing it up. It's incredibly informative.

Given all of the above, and knowing that my MM input is configurable for inputs from 1mV to 15mV, is there any benefit to using the lower 1:8 ratio instead of the 1:16 ratio?

Also, do I need to worry about capacitive loading, like with a "true" MM cart, or does that not apply to stepped up MCs?
 
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