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Help me understand rectifier behavior

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(@cecil)
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While waiting for the transformer for my dual +12V -12V power supply I decided to test my circuit on a breadboard.  The circuit works but I cannot understand the voltage I measure out of the bridge rectifier.  The transformer puts out 15.5VAC on one secondary and 20VAC on a second winding.  Yet the DC voltage going into my circuit is either 21VDC or 27VDC.  I expected the DC voltage to be lower than the AC voltage by whatever the rectifier diodes consume.  What am I missing?


   
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robotBuilder
(@robotbuilder)
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It would be so much easier if you give more information.  Specs on the transformer would be a good start as well as an actual circuit diagram showing the connections to your circuit otherwise it becomes a guessing game. 

 


   
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Will
 Will
(@will)
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@cecil

The 20VAC is the root mean square of the variable voltage. The peak will be that times approximately 1.414 (the square root of two).

So 20*1.414 is approximately 28V.

Anything seems possible when you don't know what you're talking about.


   
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(@cecil)
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@robotbuilder Unfortunately I don't know much about the transformer.  It was salvaged from an APC 350VA UPS.  It was just convenient to grab it for testing.  I was testing the positive half of my dual PSU circuit and it worked as expected.  But the regulator circuit must be irrelevant because the voltage on the DC side of the rectifier is the same whether my circuit is connected or not.  The rectifier is a chassis mount, potted, full bridge, 50A, 1000V.

@Will says my DC voltage measurements are to be expected and my confusion is just an expression of my poor knowledge.  If I understand him the DC voltage rectified from a transformer will always be 1.414 times the AC  voltage on the transformer secondary, less the losses in the diodes.


   
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Will
 Will
(@will)
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@cecil 

Think of it this way ... you know that the voltage ranges up and down over the course of the cycle. The rms value is derived by taking the area between the voltage and the zero line (technically the integral of the curve) and divides this value over the length of the cycle.

That is, it calculates the "average" voltage over the entire cycle. The rectification diodes essentially turn the negative half of the wave into a positive voltage and produce a sort of double hump.

So the 20VAC represents the "average" voltage over the cycle, but since the voltage curves, the peak voltage must always be greater than the average.

I won't bore you with the mathematics but they yield the fact that the peak voltage on the sine curve will be the mean value times the square root of two (i.e. 1.414 app).

Hope this helps clear it up for you.

Anything seems possible when you don't know what you're talking about.


   
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(@cecil)
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@will  Thank you very much, you were very clear.  I've dabbled in electronics projects for 55 years without ever gaining much understanding.  I always get by learning what I need to know to get the project done but I've never taken the time to understand the fundamentals.  Now I have nore time so maybe I'll finally learn something.

I wish I had had your teaching sooner.  I just ordered a 30V center tapped transformer to make a +12V, -12V PSU.  I thought 30V was needed to provide sufficient headroom given losses at the diodes and voltage regulator.  I now see I could have used a 24V transformer which is more common and cheaper.  The 30V transformer I ordered represents wasted money, wasted energy every time I use it and will lead to a larger heatsink on the voltage regulator.


   
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robotBuilder
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@cecil 

Whew glad @will has sorted it for you I can delete my response 🙂

Here is an example of such a supply but I assume you already have a circuit example you are following.

https://wired.chillibasket.com/2020/06/dual-power-supply/


   
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(@cecil)
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@robotbuilder   I do have a screen capture of a chalkboard schematic from a YouTube video.  It is nearly identical to the full bridge example in the article you pointed out to me.  Great article, especially because it explains the choice of components.  I can't thank you enough for that resource.


   
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Ron
 Ron
(@zander)
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@cecil Is there a reason you can't return it? Also, what kind of project needs both +12V and -12V?

First computer 1959. Retired from my own computer company 2004.
Hardware - Expert in 1401, and 360, fairly knowledge in PC plus numerous MPU's and MCU's
Major Languages - Machine language, 360 Macro Assembler, Intel Assembler, PL/I and PL1, Pascal, Basic, C plus numerous job control and scripting languages.
My personal scorecard is now 1 PC hardware fix (circa 1982), 1 open source fix (at age 82), and 2 zero day bugs in a major OS.


   
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(@davee)
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Hi @cecil,

  @will has already given you a pretty clear answer, to which I would add that the 1.414 factor, minus two diode drops, is the peak voltage, observed when there is no load to discharge the capacitors.

If you add a significant load, the voltage will vary with time, peaking when the incoming voltage is at its peak, and then falling as the load discharges the capacitor, and the incoming voltage is too low to recharge it. Thus, if your mains supply is 50Hz, the voltage will have a 100Hz 'ripple'.

The magnitude of this ripple .. i.e. variation between minimum voltage and maximum voltage increases with load current, and decreases if you increase the capacitance attached to the rectifier output.

Hence, if you added a significant load whilst measuring the rectified voltage, you would expect to see the average voltage fall and an oscilloscope would show increased ripple.

A linear voltage regulator will try to 'remove' this ripple, but it can only do this if the minimum voltage point in the ripple waveform (at the rectifier output) is equal to the expected output voltage, plus an allowance for the voltage drop across the regulator, which can be up to about 1.3V, depending on the choice of regulator.

Choosing the right transformer voltage is a bit of an "art", as it depends on lots of factors, including load current, transformer internal resistance, capacitance value, capacitance internal resistance, voltage regulator dropout voltage, etc.

So while you are probably right to suggest that some decrease in transformer secondary voltage was possible, picking a transformer that only achieved the minimum peak voltage under no load conditions could have been a worse mistake, as the final voltage might have had a high ripple variation!

Your approach may have been slightly naive, but it should have a good chance of meeting your expectations!

Best wishes and good luck, Dave


   
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(@cecil)
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@zander  I'm sure I could return it but from DaveE's comments it sounds like it wasn't a terrible choice after all.  It also has an advantage because of its small form factor.  It will fit into a small project box I already have.

My project is a milliohmmeter.  There are several designs freely available.  I chose one by a Brazilian who has a YouTube video about it in Portuguese and offers the PCBs at PCBWay.  It should be a usefull instrument, even though I don't need it, and should admit I'm just doing it for fun.

In addition, engineering a heatsink and perhaps a fan to fit in a small enclosure is well within my skillset.  I'm a pretty fair machinist and have a garage full of machine tools.  And I have 40 years experience assembling more than 500 PCs and the leftovers include hundreds of various heatsinks.


   
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Ron
 Ron
(@zander)
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@cecil Sorry, I can't be of any further help as I have no clue what a milliohmmeter(sp) is. I own a digital auto-ranging VOM, but it certainly doesn't contain a power transformer. Good luck!

First computer 1959. Retired from my own computer company 2004.
Hardware - Expert in 1401, and 360, fairly knowledge in PC plus numerous MPU's and MCU's
Major Languages - Machine language, 360 Macro Assembler, Intel Assembler, PL/I and PL1, Pascal, Basic, C plus numerous job control and scripting languages.
My personal scorecard is now 1 PC hardware fix (circa 1982), 1 open source fix (at age 82), and 2 zero day bugs in a major OS.


   
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(@cecil)
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@zander  By the way, I saw your post about needing the least expensive PC.  If you can get past "someone else's problem" I have a couple dozen you could pick from for free.  Otherwise, I would also encourage you to consider a refurbished Lenovo ThinkPad.  I've always found the original IBM design bombproof and it hasn't fallen off since Lenovo bought the design.  They were preferred by many corporations and many come off lease in little used condition and get sold as refurbs for $200-300.


   
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(@cecil)
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@zander  Can your VOM measure 0.010 ohm?  That's whar a miliohmmeter does.


   
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Ron
 Ron
(@zander)
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@cecil Thanks, but I already got a Mini PC, 500GB SSD. I only need it to run my USB scope and it joins the Raspberry Pi and Mac on my desk. $300 USD would be at least $500 with exchange and shipping, probably more since shipping across the border is a license to steal. Amazon and USPS or a company that takes care of all the extras and quotes prices to us Canadians in CDN all in shipping and taxes. Sorry, shipping is a sore point with most US companies.

First computer 1959. Retired from my own computer company 2004.
Hardware - Expert in 1401, and 360, fairly knowledge in PC plus numerous MPU's and MCU's
Major Languages - Machine language, 360 Macro Assembler, Intel Assembler, PL/I and PL1, Pascal, Basic, C plus numerous job control and scripting languages.
My personal scorecard is now 1 PC hardware fix (circa 1982), 1 open source fix (at age 82), and 2 zero day bugs in a major OS.


   
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