small transformer mistery

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paul_ma7

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when connected to the mains (125v) i can measure an output of 13v, so the ...?
ratio is 1 to 10, i'm building a power inverter with this transformer, so when i connect the 12.7v battery i would expect around 127v, or perhaps 110 considering the battery voltage drop, instead my multimeter reads a steady 222v, about double of what it should be,
what could be wrong?

could a higher or lower frequency than 60hz increase the voltage?
 
my guess would be the difference between ac and dc.
The ac transformer is 1-10 but with dc it will be a different ratio.
 
The output is higher because it is unloaded. Once it is properly loaded, I'm sure that the RMS output voltage will be too low due to the fact that the peak voltage will be around 125V. How are you driving the transformer?
 
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The output is higher because it is unloaded. Once it is properly loaded, I'm sure that the RMS output voltage will be too low due to the fact that the peak voltage will be around 125V. How are you driving the transformer?

i'm using an astable circuit, the transformer is between the collector and capacitor of one of the transistors. I'm planning on using two transformers, one attach to each collector and connecting both outputs together to double the current.
 
sure, i'm attaching it to this post
There are no component values on the diagram. Please fill them in. I fail to see how that circuit could generate much power to the load at all. Are you sure that is how it is wired?
 
What you have is a flyback type of circuit. That will generate a voltage much higher than the supply voltage times the turns ratio. Basically a flyback circuit stores energy in the transformer inductance and uses the inductive kick to generate a high voltage. This occurs because you are applying pulsed dc to the transformer, not AC (the applied voltage is only one polarity). See https://en.wikipedia.org/wiki/Flyback_converter for further explanation.

To use the transformer as a normal converter you need to alternately apply equal plus and minus voltage to the primary (bridge type circuit). You can also use a center-tapped primary and alternately apply a plus voltage to each side of the winding with the center tap grounded.
 
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Hi there,

Assuming you want an AC output...

If you have a transformer that you put in 120vac and you get out 12vac that's
a 10:1 ratio, so if you put in 12vac (secondary) you can get 120vac out of
the primary, at least approximately.
The only problem is that when we put in 120vac that has a 170v peak, and
the output has a 17v peak. Thus, if we put in a waveform with a 12v peak
(secondary) we would get a waveform out (primary) that is only 120v peak,
not 170v peak. The other thing is that it will be a square wave if it is
driven with a square wave multivibrator circuit. This means the rms output
will still be 120v, but it will lack the 170v peaks of a sine wave with the same
rms value.
The idea used in many low cost inverter schemes like this is to cut back the
waveform to a single pulse (rather than a real square wave) whos rms value
is equal to the original rms value, 120v. The only problem is, we cant get
a pulse like this out of the transformer primary without putting in a voltage
pulse that is higher than 12v.
One way around this is to remove turns from the secondary, but then we
run the risk that the transformer core saturates, so another way is to
add turns to the primary, which would boost the output enough so that
we can get the required peak for the modified pulse output we intend to use.
This can be hard to do too since the transformer might not have room for
more turns on it, so another idea is to find another transformer with the
right rating and ratio and connect the second transformer (primary or secondary,
as required for the turns ratio) in parallel to the secondary of the first
transformer, then connect the two outputs in series. This boosts the output
to a level that will indeed produce 170v peak, or perhaps a little lower.
If your app can stand 160v peaks you can try to boost the output by 40v
this way and that will get you there. This means you would require another
transformer that can put out 40v with 12v input, which could be a transformer
with 120vac input and 36vac output, also used in reverse. Since it would
be a good idea to have some regulation room (ie wiggle room) you might
want to think about going with a 120vac to 24vac and use in reverse and
also incorporate some feedback for output regulation.
 
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sure, i'm attaching it to this post

Sorry but that circuit is a pretty poor way of doing it.

For a start the transformer is in the wrong place, it should replace R4 and it'll only drive it with positive pulses so it will output an asemetrical waveform which most appliances won't like.
 
You are trying to make this lousy inverter:
1) Its capacitors have backwards polarity and blow up.
2) Its transistors have avalanche breakdown of their emitter-base junctions and the resulting high current is in the capacitors instead of in the load.
2) The base current of the transistors is far too low for an inverter.

It might work if the capacitors have the correct polarity, the transistors are high gain darlington type and their emitters have a diode in series.
 

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despite the drawbacks i decided to build this inverter, my question would be if i use an 18v power supply instead of a 12v, would that be ok ?
 

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despite the drawbacks i decided to build this inverter, my question would be if i use an 18v power supply instead of a 12v, would that be ok ?
The transistors have avalanche breakdown if the supply is higher than 7V. The avalanche breakdown uses most of the inverter's power to destroy the capacitors. 18V will destroy the transistors and capacitors. The simple circuit DOES NOT WORK!

If the circuit is powered by a 6V battery then the transistors will not have avalanche breakdown. If the backwards capacitors have their polarity corrected then the very low input current to the transistors will produce an inverter with an output power of about 16W. What can you power with only 16W?
 
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