# Calculatin impedances of combined transformers.

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Would anyone try to educate me as to how the impeadance values are calculated when combining transformers?

As in this Ben Tongue example.

In the example, If pin 1 of the UTC 0-15 is connected to ground (only) the input is 1.35 Mohms. If it is as drawn above, with pin 1 connected to the 300 ohm tap the input is 1.0 Mohms What would the input be if pin 1 was connected to the 600 ohm tap? Or the 1.2k tap? A couple of examples may get me enough understanding to figure it out myself.

This comes up because sometimes we are looking for 200 to 1 turns ratios for matching to crystal radios

Thanks, Mikek

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Does that work? I mean, did UTC ever make a truly "1M" transformer? And over what bandwidth? It must not be much.

If you mean ideal transformers, ignoring actual characteristic impedances and matching, and being concerned only about the ratios between windings, then if you first convert impedance ratios to voltage ratios, and run through the loop equations, you'll find the overall voltage ratio, the square of which is the impedance ratio. Relate that back to whichever reference winding and there is your answer. Note that at least one tap has to be terminated for there to be a resistance to reflect.

Tim

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• posted

The UTC 0-15 spec's are Grid to Plate 10 to 1 Primary 15,000 ohms, secondary 1 Megaohm. (doesn't jibe with 10 to 1)

100 to 3000 hertz + or - 1db

Or 10,000 to 1 Megaohm

50 to 5k hertz + or - ?db

I found several UTC old (50's) catalogs that list it as 15k to 1 Meg.

Yes, ideal transformer.

Yes terminated in some resistance. Let's assume a 1.2k on the 1.2k tap.

I don't know if you looked at my diagram. To me it is not a simple look at ratios.

You can short the RC* in the circuit. and then try to figure it out.

Thanks, Mikek

*That's a "Benny" "Ben (Tongue) taught us that if a transformer is connected directly to the diode, as the signal increases, there will be a dc current flowing through the diode, tank and transformer. The dc resistance of the transformer is much lower than it's impedance. As the signal increases, the low dc resistance of the transformer will load the tank, causing distortion and bad selectivity.

The Benny will restore the high dc resistance to the detector circuit. The .1 uF capacitor across the resistor allows audio frequencies to pass with little loss. The resistor in this case is an audio taper (log taper) potentiometer."

From Dave at;

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I have made a simplified drawing and a "here's how I would have done it", and why didn't the educated designer do it that way.

Can a grown up look and see if they can explain?

Thank you, Mikek

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You have changed the transformation ratio by moving the bottom end of the UTC transformer from 300 ohm tap to ground.

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Good Catch, Ok I have the corrected version here;

So, I'm still in the dark. Thanks, Mikek

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• posted

There's a thumb-sized hermetic instrument input transformer from Triad, that has 50ohm and 1megohm markings, 2.5KHz to 30KHz. A 1:7070 turns ratio, 18mH primary.

Seems rather limited in application, regardless of the care that went into its construction. RL

• posted

When the 'primaries' are in series, there would be an inductive voltage division, so perhaps the phantom 1M040:10K030 'impedance' coupling preserves the relative voltage amplitudes, over a wider bandwidth, by compensating for this uncontrolled inductance ratio.

RL

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We want to try to have a bottom roll off of at least 300 Hz. Audio range 300Hz to 3000Hz. Bell Labs, found this was enough to give intelligible speech on the phone lines. Mikek

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I don't know about 30kHz, but Mikek's at least is apparently 3kHz, which is still only 53pF stray across at least 7000 turns. Care? I can hardly imagine!

Personally, I have some 50-250-600 ohm to 50k UTCs on hand, which are also quite nice. Half tempted to crack one open and look...except, not...

Tim

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Electrical Engineering Consultation ```
• posted

Yes as I stated before 300Hz to 3kHz is all I need. His starts at 2.5KHz.

Don't know what you mean by that.

UTC spec'ed the 0-15 transformer in their 1962/63 Catalog as 10 to 1, 15K to 1 Meg. Frequency response 100Hz to 3kHz. The graph shows a little better on the low end.

Here's the catolog. It shows the 0-15 transformer on page 17 of the catalog, page 19 of the pdf. With graphs at the bottom of the page.

(I don't know why the listed ratio is 10 to 1 and the impedance ratio calculates to 8.16.)

Mikek

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I used to work with Ben. Did you know him?

Mark

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No. He has a great site that he placed a lot of research he did on optimizing crystal radios.

My transformer question must me very difficult, no one has offered a solution.

I have another thread going about resurrecting a computer to get my impedance meter going. When that happens I'll measure it to get the answers. I'm waiting for a part that's on a slow boat from China.

Mikek

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What makes you think you can remove the coupling capacitor and 'fudge' resistor used in the original cct?

RL

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The impedance reflects the turns ratio only and assumes the nominal bandwidth of the part's design. It seems rather elaborate for interface to a simple crystal radio, which doesn't have a regularly defined impedance.

RL

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I just want to learn how to calculate the transformer driving transformer. And I tried to simplify the circuit. Your welcome to analyze the circuit with the "Benny" in place. Here is an explanation of the "Benny".

"The circuit is straight forward. The only unusual part of this circuit is that variable resistor and capacitor in parallel at the signal input. This is called a Benny, named after Ben Tongue, a fellow crystal set hobbyist that knows what he is doing. Ben taught us that if a transformer is connected directly to the diode, as the signal increases, there will be a dc current flowing through the diode, tank and transformer. The dc resistance of the transformer is much lower than it's impedance. As the signal increases, the low dc resistance of the transformer will load the tank, causing distortion and bad selectivity.

The Benny will restore the high dc resistance to the detector circuit. The .1 uF capacitor across the resistor allows audio frequencies to pass with little loss. The resistor in this case is an audio taper (log taper) potentiometer. The pot is wired so that there is a short when the pot is turned fully counter-clockwise. This allows for a better adjustment at the lower resistance values."

Info from here:

The only part I question (but only a little) is, "the resistor allows audio frequencies to pass with little loss" It's X sub C is 5300 ohms at 300 hz, maybe not a lot, if it's on the tank side and the tank impedance is 800k ohms.

Thanks, Mikek

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If the C value is intended to be part of the detector, I'm surprised that the circuit would work without it.

The output of a crystal radio is DC with a LF AC component. Ceramic headphones don't load the rectified DC - hence the load resistor inserted in most circuits to allow the AC output component to track the signal, in the presence of uncontrolled/unstable rectifier leakage.

The actual winding resistance of a transformer is designed to be much lower than the reflected load impedance and so can seriously load the DC biasing of the low-power crystal detector.

The DC resistance of the '1Meg' winding of the signal transformer mentioned earlier (13mH)is only 360ohms. DC impedance of larger parts intended for lower R match will be lower still.

RL

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Just a note.

On the small 50R:1Meg instrument transformer, it is the 50R winding that measured 18mH. It's resistance is just under 5ohms. Measuring inductance of the 1Meg winding (360R resistance) is out of the range of any direct-measuring meters on hand.

As inductance scales with reflected z (same n^2 relationship), it might be expected to measure over 100H, in the absence of DC. It would need to be to still look like 1Meg at the 2.5KHz lower frequency.

RL

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It's not, it's after the diode.

Mikek

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