# need some transformer schooling....

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#### strokedmaro

##### New Member
Im building a cnc machine and just got done building a controller kit. The kit is rated at a max voltage of 42Vdc. I purchased a toroidal transformer with dual 35Vac secondaries (wanted to parallel both 35vac secondaries together for approx 10 amps) However after asking a few "pros" some questions on construction I was quickly told that this was going to blow my driver chips.

Their reasoning was that AC voltage when rectified to DC increases the voltage by 1.414 times and that my 35vac when rectified would be roughly 49.49Vdc. (35 * 1.414=49.49) I plugged the tranny in and measured 34.89Vac before the bridge rectifier and about 30.5VDC output of bridge rectifier. these readings were with no load.

I can understand how amperage is increased and decreased but how can rectification cause a voltage increase?? Quite a few "pros" confirmed this but no one could explain why...so I turn to you guys (my experts )

If this is indeed true than is this example also true?:

a 12vdc power supply has a transformer output of 8.48vac (12 / 1.414) which when rectified produces the 12vdc??

I cant wrap my head around this one so any transformer guru's out there please give me a hand.

THANKS!!

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#### Hayato

##### Member
Well, when we say that a xformer has a 35 V AC output, we are talking about RMS values.

RMS value is a mean value, it does not represents the true value of the sine wave present on yout xformer output/input.

When you rectify and filter, the filtering capacitor will be charged with the peak value of the sine wave signal.
In this case is Vpk = Vrms*sqrt(2). -> 35 * sqrt(2) =~ 49,50V

But if you does not filter the outputwave, your multimeter will only show the Average value, that is the integral of the |sine wave| over the period, that is Vpk*2/Pi -> 49,50 * 2/3,1415 =~ 31,50 V

#### Hero999

##### Banned
Im building a cnc machine and just got done building a controller kit. The kit is rated at a max voltage of 42Vdc. I purchased a toroidal transformer with dual 35Vac secondaries (wanted to parallel both 35vac secondaries together for approx 10 amps) However after asking a few "pros" some questions on construction I was quickly told that this was going to blow my driver chips.
Yes, if the maximum DC rating is 42VDC, then 35VAC is too high, I'd recommend a 24DC transformer assuming it can operate between 30V and 40VDC.

Transformers also tend to give a higher off-load voltage (typically by a factor of 20% but varies widely depending on the size of the transformer) so always bear that in mind when desining things.

Their reasoning was that AC voltage when rectified to DC increases the voltage by 1.414 times and that my 35vac when rectified would be roughly 49.49Vdc. (35 * 1.414=49.49)
That makes sense to me.

I plugged the tranny in and measured 34.89Vac before the bridge rectifier and about 30.5VDC output of bridge rectifier. these readings were with no load.
It should have measured higher than this, about 50V as you calculated.

What size was the filter capacitor?

If you didn't use one then the meter will try to measure the pulsed DC waveform and more than likely fail to produce an accurate reading.

I can understand how amperage is increased and decreased but how can rectification cause a voltage increase?? Quite a few "pros" confirmed this but no one could explain why...so I turn to you guys (my experts )

If this is indeed true than is this example also true?:

a 12vdc power supply has a transformer output of 8.48vac (12 / 1.414) which when rectified produces the 12vdc??

I cant wrap my head around this one so any transformer guru's out there please give me a hand.

THANKS!!
There is no voltage increase.

AC voltages are normally measured in RMS, the peak voltage is always higher by a factor of √2.

An RMS voltage produces the same heating effect in a resistor as a DC voltage.

Your 35V transformer have a nominal secondary voltage of 49.5Vp

You could also say it has a secondary voltage of 99Vp-p. If you looked at the waveform on an oscilloscope and counted the number of squares between peaks multiplying by the appropriate scaling factor you'd get the peak-to-peak voltage which is twice the peak voltage.

#### strokedmaro

##### New Member
^^ once again you guys put it in terms anyone can understand! thanks for the info. I never connected the filtering cap but im going to do it now Just so I can see it on the meter. Thanks again...after your information I found a \$18 24vac, 10 amp transformer that should work a little better you guys rock!

EDIT: With the filtering cap connected I read 48.5vdc....nice to see it for myself. Thanks again for the information!

Strokedmaro

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#### Hero999

##### Banned
The filter capacitor needs to be very large.

As a general rule the capacitance in µ should be greater than the value calculated using the following formula:
$C= \frac{I \times 10000}{Vripple}$

EDIT:
Note that this formula only works for 50/60Hz supplies.

C is the capacitance in µF.
I is the current in Amps.
Vripple is the maximum allowable ripple.

To calculate V ripple subtract the rectifier losses from the peak voltage, is a generally used for bridge rectifiers. Now subtract the minimum voltage your appliance will work down to.

For example if you have a 24V transformer, you're drawing 10A, the rectifier looses 2V and your device can work down to 23V.

Vripple a 24√2 - 2 - 23 = 8.94V

$C= \frac{10 \times 10000}{8.94} = 11186\mu F$

Use the next standard value up, 12000µF or 15000µF

The capacitor must be rated to the peak voltage, 33.94V, unfortunately 35V is probably cutting it too fine, use a 40V or even 50V capacitor.

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#### strokedmaro

##### New Member
The filter cap that came with the kit I bought is quite large...like a mini coke can. 27000µF, 50V. Again, thanks for the excellent information...I learn something new every day here!

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