Hi Guys,
I bought QA-402 quite some time back but didnt really have enough time tinkering with this great device.
Would like to measure bandwidth of my DIY input transformer. It’s 600:600R, 1:1, around 40R resistance in Primary/Secondary.
Do I need to add R in series when injecting the signal to its primary? I assume 2Vrms will be sufficient?
Any suggestion welcome.
Thanks.
Hi @Jimmy168, driving 600 ohms shouldn’t be an issue, but remember there’s 100 ohms of output Rin the QA402 outputs. And if you drive balanced that will be 200 ohms total. So, all up, you are probably looking at 800 ohms of load which isn’t an issue for the output opamps (OPA1612).
If your transformer wants to see specific input/output impedances, you will need to add R to get those numbers right. On the output side, it’d be series R. On the input side, it could be a parallel R across the inputs of the analyzer (or outputs of the transformer, depending on how you want to look at it).
And the, you can do a quick chirp at low levels (-10 dBV or so) to learn the frequency response, and then start some automated tests where you step up in voltage to learn more about the transformer saturation etc.
Feel free to post what you are measuring. I think a lot of folks would enjoy learning what you are up to.
Hi Matt, thanks for your reply.
I dont think there will be an issue with the 100R R output as I’m not really measuring on specific impedance.
What my worry was the primary/secondary DCR resistance is around 40R each. Not sure if it will be too low or not?
Noted on the R on input and R load at output.
My first time measuring Input Transformer - as it’s small and low primary/secondary resistance (40R), I really dont warn to damage either the transformer or my QA402… 
Thank you very much.
@matt , may I know the input impedance of QA402? Example if I remove 1K5 load from my Input Transformer output and direct to QA402 input, what load that my Input Transformer output side will see?
Thanks.
Jimmy.
Hi Jimmy, the input impedance appears as follows: The BNC input goes to a 4.7uF cap in series with a 470 ohm, followed by a shunt of 100k ohms. Overall, we consider it a 100k series input.
Thanks @matt .
Any idea if QA404 will do 384k (so we can measure 200 kHz)?
Hi @Jimmy168, no, the QA403 won’t do 384K.
How did you measure the 40 Ohm resistance? With a DVM?
Best regards?
Yes. Any recommendation how to measure?
That´s not the right way. Due the using of dc voltage you magnetize the core permanently. The core is made with highly permeable metal. Due the using of DC you can damage the transformer permanently. There is a exists a procedure made by Studer to undo this magnetization. The only way is to use a LCR-Bridge or impedance analyzer. Then you can measure the impedance of every winding. The DC-resistance is not the right value.
Best regards!
I understand that measuring impedance is different vs resistance. I just want to know the DCR of each winding.
So you mean using DMM to measure DCR might damage the transformer?
Thanks.
A transformer is a complex resistor made up of several components. The DC resistance is irrelevant because DC cannot be transmitted by a transformer. Only changing voltages can be transmitted. The impedance describes the behavior of the transformer in its typical use. Superimposed DC voltage increases the distortion factor of the transformer and also leads to increased microphonics due, among other things, to the permanent magnetization of the core by the DC voltage. Input transformers are only designed for low voltages and use highly permeable metal sheets to enable the required frequency range. That’s why input transformers are always DC decoupled. Voltages such as those from a multimeter can permanently damage an input transformer. What kind of multimeter do you use? The decisive factor for the induction, which is what we are talking about here, is the change in the potential from zero to the voltage for the test object, which is output by the multimeter, because without this no resistance can be measured. Recommended literature: Domsch: Domsch: “Der Übertrager der Nachrichtentechnik”. Sorry for my english. I´m not a native speaker.
Best regards!
Ah got it. I guess if the DMM delivers enough current to magnetize the core, the it might damage it…
So we need to demagnetize it then.
Thanks.
It’s not the current damage the transformer. It is the DC voltage that is applied.
Best regards!
That’s not what I understand… Thanks anyway.
I have tested the transformer and all good. No missing bandwidth. Was ready to demag if needed.
Thanks.
Only current in a winding can magnetize the core, magnetic flux is proportional to both current and turns-count. Voltage merely affects the rate of change of current. For sinusoidal signals lower frequencies mean a given voltage drives more current through the winding, ultimately limited by the DC resistance of the winding.
A DC voltage would create a DC current (in a circuit of course) and multimeters measure resistance by injecting a DC current into whatever is connected between their probes. It is the DC current that will both magnetizes the transformer’s core and reduces it’s inductance - unless of course said transformer is gapped or designed for some DC current.
We are talking micro amperes here.
On my Fluke 183’s internal current shunt, my Fluke 87 drove 337.6uA (micro amps) for a 50R measured reading on the 87.
Yes, I have same understanding. The current is too small to create issue. I have checked both Inductance and Bandwidth. No issue so far. Sound test also not indicating any issue on low frequency.
Thanks.
