#### RSRecords ##### Determining Turns Ratio of Transformer
« on: June 17, 2019, 04:37:48 PM »
Can you determine the ratio of transformer by measuring the nominal impedance of the Primary vs secondary?
For example:
10 ohms across primary
42 ohms across secondary

Is there a formula or is it a 1:4 transformer?
I recall it not being that simple but I couldn't find the thread discussing it. There's no way to determine reflected impedance without destroying it correct?

Thanks

#### squarewave ##### Re: Determining Turns Ratio of Transformer
« Reply #1 on: June 17, 2019, 06:14:22 PM »
Yes. The impedance ratio is just the square of the turns ratio (or in other words the turns ratio is the square root of the impedance ratio). For example, if turns is 1:4 then impedance is 1:16.

But 10 ohms and 42 ohms is pretty low. Is this a transistor transformer? Those sound like DCR measurements which is not useful. You need AC impedance.

Rather than trying to measure impedance, just measure the voltage ratio. The voltage ratio equals the turns ratio. Then square that to get impedance.

Update: I had the square / square root relation reversed. Fixed.
« Last Edit: June 17, 2019, 08:28:56 PM by squarewave »

#### RSRecords ##### Re: Determining Turns Ratio of Transformer
« Reply #2 on: June 18, 2019, 11:37:11 AM »
Yes. The impedance ratio is just the square of the turns ratio (or in other words the turns ratio is the square root of the impedance ratio). For example, if turns is 1:4 then impedance is 1:16.

But 10 ohms and 42 ohms is pretty low. Is this a transistor transformer? Those sound like DCR measurements which is not useful. You need AC impedance.

Rather than trying to measure impedance, just measure the voltage ratio. The voltage ratio equals the turns ratio. Then square that to get impedance.

Update: I had the square / square root relation reversed. Fixed.

Ok thanks! It's an mci jh110c output transformer.

So i applied a 1k sine wave at 1Vac RMS to the primary. Secondary reads 1.3vac.
1:1.3

so impedance ratio 1:1.69.

so if the output is supposedly 600 ohms. would that be
1/1.69=x/600
.59=x/600
354=x
354:600

Does that look right?

#### ruffrecords ##### Re: Determining Turns Ratio of Transformer
« Reply #3 on: June 18, 2019, 06:54:00 PM »
Ok thanks! It's an mci jh110c output transformer.

So i applied a 1k sine wave at 1Vac RMS to the primary. Secondary reads 1.3vac.
1:1.3

so impedance ratio 1:1.69.

so if the output is supposedly 600 ohms. would that be
1/1.69=x/600
.59=x/600
354=x
354:600

Does that look right?
Yes. The primary impedance is just 600/1.169 =  355.

Cheers

Ian
www.customtubeconsoles.com
https://mark3vtm.blogspot.co.uk/
www.eztubemixer.blogspot.co.uk

'The only people not making mistakes are the people doing nothing'

#### Marik ##### Re: Determining Turns Ratio of Transformer
« Reply #4 on: June 19, 2019, 03:32:52 AM »
There are a few things to be clear here. You have 10Ω and 42Ω. I assume it is rather DCR of the Pri and Sec. Doesn't tell you much--often those wound with different wire diameters...

Then you measured 1:1.3 voltage ratio--already much better. Once you know inductance of one of the Pri then it gives you much better idea what's going on. Say, you assume -3dB at 30Hz and measure Pri inductance as 1H. Then from the equation:
L = X / ( 2 × π × f )     Where X is impedance and f is frequency (another way of thinking about it is the low end response of the transformer is its inductance for given low end response)
you find that
X= 1Hx (2 × π × 30Hz )=188.4Ω (so in this case you would have a 200Ω transformer, which is pretty close)

Of course, your Pri inductance will be different. Once you are able to measure it you will know exactly what your transformer is.

Best, M

Samar Audio & Microphone Design

www.samaraudiodesign.com

The Art of Ribbon Microphones

Marcelland
• Posts: 9695 ##### Re: Determining Turns Ratio of Transformer
« Reply #5 on: July 29, 2019, 12:25:58 PM »
There's no way to determine reflected impedance without destroying it correct?
Absolutely not! Evaluating the reflected impedance is quite simple; once you knopw the voltage ratio, thus the impedance ratio, you apply that ratio to the load impedance. EG if you load your 1:1.69 xfmr to a 600 ohm load, the reflected impedance at the primary is 600/1.69 or about 355 ohms. That is useful to assess the drive capability of the driver; it must be capable of driving 350 ohms. This does not take into account losses or magnetizing current, but usually it does not make a big difference. A good designer (old-fashioned) will design the driver stage for being capable of driving 200 ohms.

Now if you want to assess the reflected impedance at the output, the quick n' dirty formula doesn't work, because of the losses and DC resistance of windings, that are much higher than the driver stage's output impedance. I'm talking about solid-state here, vacuum tubes are different.
Who's right or wrong is irrelevant. What matters is what's right or wrong.
Star ground is for electricians.

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