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DVM meters with Square waveform mode

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(*steve*)

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The square waveform output of the DVM meters is only 2volt p/p so i don't think a line transformer would do anything

Such a transformer will work perfectly well with 2V input. for a mains transformer, just as 2V is 1/125th of my mains voltage, the secondary will be 1/125 of what it is marked as.

If the primary voltage is rated at p V, and the secondary at s V, if you present 2V at the primary, the secondary will produce 2 * s / p V. (the ratio of your mains voltage to 2V will vary, but the transformer ratio will remain the same, that is s/p).

But note what Hevans1944 and I said earlier. The function may not be useful. It may have an impedance too high to be useful, and you don't have the option of changing the frequency or the amplitude (not to mention the waveform).

It would certainly NOT be able to provide sufficient power to operate a fuel injector.

You also might want to investigate the mains frequency where this device was manufactured.
 

hevans1944

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... This goes back to maybe it's ment for a cars fuel injection or timing PWM circuits for cars

It's meant as a "feature" to sell cheap digital multimeters. It's up to the buyer to figure out something useful for it.
 

KrisBlueNZ

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The transformer will change the waveform too. A transformer that's designed to operate at 50/60 Hz is designed to pass a sinewave, not a squarewave. A squarewave has lots of harmonics, which (by definition) are at higher frequencies than the fundamental, and a mains (line) transformer will not reproduce these very well.
The square waveform output of the DVM meters is only 2volt p/p so i don't think a line transformer would do anything
Why don't you try it?

Try it both ways - put the signal from the meter into the primary of the transformer and measure the secondary, then try it the other way round, so the transformer is being used as a step-up transformer. Be careful with the connections because there could be some significant voltage there.
Yes it passes a sine waveform at 120VAC but a square wave at 2 volts p/p it might pass because the voltage is so low?
Why don't you try it and see. That's quicker and more reliable than asking!
This goes back to maybe it's ment for a cars fuel injection or timing PWM circuits for cars
Steve and Hop have both told you "what it's for". It's there as a "USP" - a "Unique Selling Point" - for that multimeter. The manufacturer has included a cheap, simple option that sounds like it might be useful for various things, in the hope that people will think that it's an important and valuable feature, which is unique to that multimeter.

They are hoping that people will choose that model because it has this unique feature that other multimeters don't have. It's designed to make the meter stand out among the competition. Because there is a LOT of competition in the cheap nasty Chinese multimeter market.
 

willwatts

Nov 15, 2014
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It may have an impedance too high to be useful,

The impedance on the output of the square waveform generator? if the impedance is too high why can't it be useful? why do you want the impedance to be low on a square waveform generator?
 

willwatts

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auto injectors use a square wave that has a fixed voltage the fuel is metered by the pulse with , hence PWM

the 30hz I think is for diesel injectors
 

(*steve*)

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auto injectors use a square wave that has a fixed voltage the fuel is metered by the pulse with , hence PWM

the 30hz I think is for diesel injectors

That's irrelevant because this meter:
  1. operates at a different frequency
  2. does not allow the mark/space ratio to be changed
  3. will not be able to provide enough power to operate the injector.
 

willwatts

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ok what about the impedance issues you're talking about

It may have an impedance too high to be useful,

The impedance on the output of the square waveform generator? if the impedance is too high why can't it be useful? why do you want the impedance to be low on a square waveform generator?
 

(*steve*)

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The impedance on the output of the square waveform generator?

Yes

if the impedance is too high why can't it be useful? why do you want the impedance to be low on a square waveform generator?

Those are great things for you to google to see if you can find an answer for yourself.
 

hevans1944

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ok what about the impedance issues you're talking about

It may have an impedance too high to be useful,

The impedance on the output of the square waveform generator? if the impedance is too high why can't it be useful? why do you want the impedance to be low on a square waveform generator?

Ideally a waveform generator has a low impedance so that whatever load it is driving does not appreciably attenuate the waveform, or if appreciable attenuation does occur, the attenuation is predictable. For example some waveform generators have 50 Ω output impedance; when connected to a matched load of 50 Ω impedance the output amplitude will be half the un-loaded output amplitude.
 

willwatts

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but other signal generators are at 600 ohms , most oscilloscopes or counters inputs are at 1 meg
I would need a 600 ohm termination plug for the Oscope and Counter?

So a load on a high impedance output generator will attenuate the waveform?
A load on a low impedance output generator will not attenuate the waveform?
 

willwatts

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but what is the impedance of an injector or a transformer at 50Hz

Transformers primary impedance should be very low resistance

So a signal generators output is High impedance or low impedance shouldn't attenuate the waveform since the load is very low resistance

High Impedance outputs don't like low resistance loads?
 

hevans1944

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but other signal generators are at 600 ohms , most oscilloscopes or counters inputs are at 1 meg
I would need a 600 ohm termination plug for the Oscope and Counter?
Probably not. Matched terminations are used to prevent signal reflections on transmission lines (a coaxial cable is a transmission line with a characteristic impedance).
So a load on a high impedance output generator will attenuate the waveform?
A load on a low impedance output generator will not attenuate the waveform?
Unless an output generator has zero source impedance, the signal across any load will be attenuated by the ratio of the output impedance to the load impedance. This is simply Ohm's Law. Obviously for lower output impedance there will be less attenuation when the load is connected. Zero output impedance requires negative feedback and can difficult to obtain in a stable circuit that does not oscillate for some load impedances.
 

hevans1944

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I think we are done here. Why don't you ask a moderator to close this thread and allow you to open another?
 
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