Results don't match the theory

[ej_whyte]

I've already talked a bit about this in another thread, but i thought i would make a new one as the title of the old one was unrelated. The original thread is here: http://www.groupdiy.com/index.php?topic=44241.0.

I recently made my 1st PCB, and it all works fine in terms audio passing, no noise or any unwanted nastiness, however the levels I've measured on the scope don't match what I expected from the theory, and I can't figure out why. Speedskater has already given me some guidance on possible grounding issues but this doesn't explain the mismatched levels (I don't think?).

The circuit is basically a 4 channel summing bus, shunted with a 680Ω resistor to bring the level and impedance down, before running in to a Cinemag 1:5 with a 150KΩ load resistor. This circuit is repeated 4 times on the PCB, exactly the same except for the addition of a molex header for running to a bypass switch.

With a 50Ω source, from the schematic I got that there would be 12.04dB loss from summing 4 channels, and then a further ~18.5dB loss due to the 680Ω shunt, and then 13.97dB gain from the transformer (1:5) :

25Ω source + 10kΩ summing (each leg) x4 in parralel = 2506.25Ω
2506.25Ω x 2 = 5012.5Ω (2 legs)

Potential divider loss from 680Ω shunt = 20log(680/5012.5+680) = -18.45dB

So the total loss should be -12.04 + -18.45 = -30.49dB

Add in the transformer gain and the overall level should be -30.49 + 13.97 = -16.52dB

Yep?

However, here are the scope readings I got, obviously not extremely accurate as its an analog scope and human reading error. The top line is voltage, all taken differentially until it is unbalanced at the transformer. The next line is loss from the previous stage, and the bottom line is total loss from input. w/o 680Ω is where I lifted the shunt resistor so I could measure just the summing loss.

Input  w/o 680R  Post 680R  Output
1.6v        0.55v          0.16v          0.85v
Stage  -9.275dB    -10.724dB      14.505dB
Overall  -9.275dB      -20dB          -5.494dB

So where the reading is -9.275dB it should be -12.04 from summing loss. And where it is -10.724dB it should be -18.45 from the 680Ω shunt. The transformer seems ok, only 0.5dB out from theory.

I did some calcs to try and identify the problem, and found that the summing loss from 3 inputs is -9.54dB, close to my reading so i thought maybe an input isn't connected properly, but nope, they all check out ok. Further to this, if there was only 3 inputs connected, then the loss from the 680Ω shunt would have increased, but the readings actually show it as decreasing.

Thats about as far as I got, any help appreciated.

Thanks

 

[zebra50]

Morning!

With a 50Ω source...

Probably a daft question - did you use 50 ohm source resistors on the other channels? Your analysis treats them all as though they do.

 

[abbey road d enfer]

Yes, exactly what zebra said:
Your assumption of 12dB attenuation for 4 inputs is true only if the unused inputs are not left open.

 

[ej_whyte]

Yes, I was testing only one of the 4 independent summing busses, but all 4 inputs were fed from the same output of the dScope using a special cable I made - 1 xlr female to 4 TRS. Does it matter that they are fed from the same source or do i need separate sources?

Thats what I said about thinking maybe one of the jacks had a bad connection and I was only actually connecting 3 sources, but i tested all the jacks on the cable and all the sockets on the board:

I did some calcs to try and identify the problem, and found that the summing loss from 3 inputs is -9.54dB, close to my reading so i thought maybe an input isn't connected properly, but nope, they all check out ok. Further to this, if there was only 3 inputs connected, then the loss from the 680Ω shunt would have increased, but the readings actually show it as decreasing.

Cheers

 

[ruffrecords]

The usual way to test this is to short all the unused inputs and just drive one. Try that and see what you get.

Cheers

Ian

 

[ej_whyte]

Do you mean short the tip to the ring?

Cheers

 

[TomWaterman]

Elliot! Boards look good mate!

Try connecting a 50ohm resistor across tip to ring in 3 empty jack bodies and plug them into channels 2-4 and connect the 50ohm output from the dscope just to channel 1...

Just gor a d-scope at work too by the way... hope everything is good with you down there!

-T

 

[Jean Clochet]

That would be the correct way to do it yes but 50R is so small compared to the 20K build outs that you could just short the +tve & -tve terminals together for the unused inputs if you are impatient

As a quick calc., I get about -16.5dB for one channel through the whole network & transformer.  Same as you did with your maths.

Another way to look at the losses in your calculations btw is to take your shunt of 680 and parallel the three unused inputs with it.  So you have 20K050 X 3 all in parallel with the 680R.  This gives you a new value shunt of about 617R.
As a straight voltage divider, you then have a 20K05 on the top and a 617R on the bottom.  which is - 30.23dB, about what you got with your method.  Add the 13.97dB (ideal) from the transformer and it's -16.5dB.  Again, what you got with your maths. 

Good luck with finding the issue, the board looks nice btw.

 

[ej_whyte]

I have about 40 spare jacks around so ill make up some with 50Ω between tip & ring and see what i get with that.

Cheers Tom, yeh the boards turned out good, except for messing up the transformer pinouts, buts thats ok cos it still all works The placement is going good, learning a lot and i've only blown one thing up so far (to be fair it wasn't my fault either lol). Gonna get this board all tested out and check its as it should be then breadboard the 2520 gain stage i think.

 

[TomWaterman]

Good times, I might come for a visit sometime this year!

Good point Jean RE buildout values... either way would get you moving forward Elliot!

-T

 

[ruffrecords]

Do you mean short the tip to the ring?

Cheers

No, I mean short both the tip and ring to the sleeve. Your input sockets should be arranged to do this when nothing is connected to them.

Cheers

Ian

 

[ej_whyte]

Thanks everyone, made up some 50Ω jacks and it comes out pretty close to the theory

31.53dB total loss, just about 1dB more than the 30.49dB theory. Transformer is running at 13.82dB.

When i knew it was working i did a load of tests on it, partly just to learn how to use the dScope really but also to keep records for when i have to write all this up. THD+N, IMD, Freq response, Square wave response. I was running the output through a simple line driver from the THAT 1646 application notes, to balance it up for listening purposes mainly. The line driver includes a a trim pot so i can remove the 6dB gain from balancing, just so i know that whatever level i get on the scope is the actual output level of the circuit under test. The tests included the line driver, but i don't think it has affected the results much, for example i did a freq response plot for the driver and it never varied more than ~0.05dB.

Freq Response @ 0dBu

Square Wave @ 1kHz, 0dBu

Square Wave @ 100Hz, 0dBu

Cheers

 

[Jean Clochet]

That's good to hear.
Since your board looked so nice, I figured it was going to be fine

Thanks for posting measurment graphs. 
What is the Cinemag 1:5? Is it a mic in transformer that you raised the secondary load to 150K for a 600R source  rather than a 50K load for a mic Z source?  Or a line in 1:5?
I am interested in a 600 ohm input 1:4, split secondary myself at the moment.
You would get some additional loading loss, but you could try some lower load on the secondary to see if the square waves get any better?
And only if if it bothers you, but you could also play with values of an  r & c network on the secondary to clean the leading edge? 
But if it sounds good, it is good.

Glad you reconciled theory and practice.

 

[ej_whyte]

Its a CM-75101APC with the primaries wired in series: http://cinemag.biz/mic_input/cm-75101APC-schematic.html

I talked to David at cinemag about zobels & load, and he recommended the 150K load and no zobel, but i've left spaces for a zobel so i might just try something out to see what happens.

I have more graphs and stuff if your interested

Cheers

 

[TomWaterman]

Ace!

A teeny tiny rise in the HF but play music through it for a while before you go nuts!

Good work!

-T

 

[Jean Clochet]

i've left spaces for a zobel so i might just try something out to see what happens.

I have more graphs and stuff if your interested

Cheers

OK thanks for info on the Cinemag, I'll download the datasheet. 
I don't think the response you have is bad at all and I think Tom from "The Shire"  ( ) is right that you should listen to it and live with it for a while before doing anything.   
Although you could make a good case that every transformer should have a zobel or damping mechanism, practically no recording equipment in 1950's and 1960's had one.  I don't believe the U.S. distributer for Lundahl uses one when he builds stuff despite Lundahl having recommended networks on the datasheets.

If you're OK with sharing the graphs, then, yes  I'm interested.

Thanks. 

 

[ej_whyte]

Yeh -17.7dB is the baseline on that frequency plot, so its only +0.2dB on the HF and -0.4dB on the lows, and this is only a small part of the whole project so it will change by the end. Besides, if i really wanted it completely flat i may as well scrap this project and do it all digital.

Heres all the other plots as well, bear in mind that they include the line driver but like i said earlier, the frequency response of that never goes more than +/- 0.04dB (i didnt do all the other tests on it)

THD+N @ 20dBu

THD+N @ 0dBu

Square wave response @ 50Hz, 0dBu

IMD @ 20dBu

IMD @ 0dBu

Freq Response @ 20dBu

Veroboard Line Driver

Testing

Cheers

 

[ej_whyte]

I've been trying to measure the output impedance using this method: http://www.sengpielaudio.com/calculator-InputOutputImpedance.htm but it is coming out with funny results :s Its saying -109.2k, because my unloaded voltage is lower than the loaded voltage. I tested it without the 150k load resistor and it was 37.6mV, then with it it was 138.3mV.

I them took the resistance across the 680Ω shunt resistor and it showed 51Ω :s As far as i can tell it should be 599Ω?

1/( ((1/20050) x4) + (1/680)) = 598.77Ω

Anyone have any ideas as to what i'm doing wrong?

Cheers

 

[Jean Clochet]

Anyone have any ideas as to what i'm doing wrong?

Cheers

What level are you putting into one of the inputs, +4dBu?
138mV is -15dBu which is about right (- 19dB down from +4 dBu).

When you test the output Z, you want to leave in the 150K secondary load resistor.  You are expecting the output Z to be about 15K and so another 15K resistor across the secondary should drop the voltage by half or 6dB.  So start with a 20K pot and wind it down until the voltage drops by half. In other words, your 138mV is 69 mV with the extra circa 15K R.  Or whatever value your pot ends up being.
If your meter can't read such low voltages too well then raise the input V by twice or three times.

Don't forget that when you try to measure the shunt resistance value in situ, you are measuring every resistance in //.  Including the transformer dcr.  The reason it isn't 50R that is reflected to the secondary is because the DCR will add in series to the source Z and get reflected up by 25 X impedance or 5 turns. 

Thanks for the extra plots you posted

 

[ej_whyte]

Yeh it was 4dBu this time. I was following the method from the website posted above, where you measure the output voltage with and without the load resistor and put the 2 results in to an equation to get the output Z.

I'm confused what you mean about another 15k resistor across the secondary though, did you mean to put 150k? At the moment it is just the 15k output of the transformer with the 150k load across it, no 15k resistors :S Or do you mean use a pot as a means of finding out what the output Z is? So when the pot is applying 6dB attenuation, take it out the circuit and measure its resistance, and whatever that is should be the output Z? Luckily i found a really oldschool meter at work that does down to 1mV and measures dBu as well

With regards to taking the resistance across the shunt, shouldn't it be the 599Ω from all the input resistors and shunt in parallel, also in parallel with the 150k divided by 25 as it is reflected back to the primary?

150k/25=6k
1/ (1/599 + 1/6000) = 544.6Ω

Cheers