Hi,
I am currently working on a power supply that provides two independently adjustable and voltage-regulated outputs (one negative and one positive). It is a project from my electronic book ("Electricity and Electronics" by G. R. Slone) and I am self-studying it.

After I have installed two 24 volt transformers and a 6-amp 200 volt bridge rectifier module, along with a fuse and some wiring. (I have attached the schematic diagram) I connected the two transformers in series and used the middle as circuit common. Then, I've connected the transformer to the bridge. In the end, I connected a 10K ohms resistors (RT1) to the positive output of the bridge, and another 10k resistors (RT2) to the negative output. And the other ends of both resistors are connected to the circuit common. I tried to test the outputs under the instruction of the book.

I have first measured the DC voltage across RT1 and RT2, they are about 24 volts and -24 volts respectively as the book suggests. However, when I was measuring the AV voltage across RT1 and RT2, I got 50 volts and 0 volts respectively, while the book suggests 12 volts for both resistors. The book says the AC voltage is called the AC component or the ripple. But it didn't explain any situation like mine. So, is my measurement indicating anything wrong in my circuit? What's the explanation behind it?

Thanks a million.

 

You've made a mistake about where you placed your test leads when you changed the meter to AC.

 

First, from what you say it sounds like your circuit is actually working, so congratulations.

I don't get why RT2 measures 0V on AC so I'm ignoring that little part. It may read something f you swap the leads around the other way.

Assuming you have an inexpensive (under $100) meter you can't expect it to give you accurate results for the type of wave you are measuring. RT1 and RT2 are seeing rectified pulses that are of a fixed polarity (meaning + on one side and - on the other), but such "complex" waves do not get measured very accurately by a simple meter.

Based on you seeing "24 volts and -24 volts" means your circuit works. Do not expect your meter to give you anything accurate unless you change the wave. If you replace the resistors by capacitors (and the value really doesn't matter for this) the caps will charge to the peak voltage of the wave and your meter should read THAT accurately for DC.

An expensive meter will be able to do "True RMS" readings for AC, which should be accurate for any waveform. Inexpensive meters typically "assume" you are putting a sinewave in, so for any other wave they can lie and lie big.

If you can beg/borrow/steal an oscilloscope you can see what is really happening here.

And welcome to the forums!

 

Based on you seeing "24 volts and -24 volts" means your circuit works. Do not expect your meter to give you anything accurate unless you change the wave. If you replace the resistors by capacitors (and the value really doesn't matter for this) the caps will charge to the peak voltage of the wave and your meter should read THAT accurately for DC.

Thank you, I really appreciate your answer.

I was reading the older posts in the forum, something about the meter reading absolute value vs. true RMS. Unfortunately, I don't think I am able to get a oscilloscope.

However, since this is my first electronic project, I just wanna be super-cautious from electrically shocking myself. In the later part of this project I will actually install filter capacitors and an regulative circuit. Regarding the AC measurements and the DC measurements, am I safe to continue the project?

Also, I don't understand why the book suggests a 12 volt AC, shouldn't the turn RMS be 24 volts the same as the DC voltage?

 

You are safe enough as long as you make the 120 volt connections nice and steady, then insulate them so you never see them again.

Your last sentence has so many misspellings that I can't understand it. However, I can say that the differences in the meter readings are not a problem at this point in your life. It's just overcomplicating a simple chore.

 

Your last sentence has so many misspellings that I can't understand it. However, I can say that the differences in the meter readings are not a problem at this point in your life. It's just overcomplicating a simple chore.

Sorry for my English. and Thank you for your help.

Because the book suggests measurements of 12 AC across the RT1 as well as RT2, I was wondering where does the 12 AC come from? (or what's the calculation behind this?)

 

The transformer winding provides 24 volts. The rectifier makes sure only the positive half gets to RT1 and only the negative half gets to RT2. 24 divided by 2 = 12

 

The transformer winding provides 24 volts. The rectifier makes sure only the positive half gets to RT1 and only the negative half gets to RT2. 24 divided by 2 = 12

Sorry to keep asking about this, but I still don't quite understand. Since there are two 24 volt transformers in series, in total they apply 48 volts AC. And because I use the middle of the two transformers as the circuit common, both output terminals from the transformer will have 24 volts AC. After that, I used a bridge rectifier to convert the AC supply into pulsating DC, so one output terminal from the rectifier has positive 24 volt DC and another terminal has negative 24 volt DC. As far as I know, the rms value of AC is equivalent to DC (if not please correct me), shouldn't I get 24 volt true rms AC measurements from each terminal then? or the pulsating DC is different from the AC components?

Again, thanks for your help and patience.

 

I used a bridge rectifier to convert the AC supply into pulsating DC, so one output terminal from the rectifier has positive 24 volt DC and another terminal has negative 24 volt DC.

That's where your mistake is. Pulses of a single polarity do not make DC voltage.

 

If you connect the meter across the AC terminals of the bridge, Do you measure 48 volts (AC)?
Joe

 

The problem may be in the phase of the secondaries of the 2 transformers. Switch the leads comming out of one of them and see if that is what you are expecting.

 

That's what I was thinking. Joe

 

If you connect the meter across the AC terminals of the bridge, Do you measure 48 volts (AC)?
Joe

I measured the voltage across the AC terminals of the bridge, I actually got 100 volts...

And then, I accidentally remeasured the AC voltage across RT2 using different prongs arrangement of my meter (I connected the meter's COM prong to the negative terminals and connected mAV prong to the circuit common, instead the other way around which I used to do), it actually gave me 50 volts. So now, I have 50 volts AC across RT1 as well as RT2, if I use the prongs right.

However, I don't think I put the phase wrong, because I have checked the DC measurements and the book has explained the phase stuff very explicitly.

Right now, I think the most likely reason for I getting such reading is my cheap crappy meter Since, I don't think a AC measurement should depend on the polarity of the prongs, should they?

However, I still don't quite understand why the book suggests a 12 volt RMS.

Thank you guys so much, I really appreciate it, you guys are being very helpful!!!

 

I have just installed a pair of filter capacitors, the DC outputs from the capacitors were same as the book suggests (+ and - 36 volts). However, I am still measuring ridiculous large AC output from the terminals from capacitors (around 78 volts on each side)

 

The RMS value of a sinusoidal voltage function is Vmax/√2.

So a 24V RMS secondary has a Vmax=24*√2=33.94V

Ignoring rectifier diode forward voltage drops a full-wave rectified sinusoidal function has a DC equivalent [Vd] of 2*Vmax/∏.

So with Vmax=33.9V the ideal DC equivalent is 2*33.9/∏=21.6V

The raw rectified 24V RMS value would also ideally be 24V. If you blocked the DC component of the rectified waveform from reaching the measuring meter then you would only read the equivalent AC harmonics in the measurement.

The RMS value of the harmonic component of the rectified sinusoid would be ...

Vharm=√(RMS^2-Vd^2)=√(24^2-21.6^2)=√109.4=10.46V

Perhaps this is what the book is referring to but I very much doubt it is exploring such subtle concepts.

 

Perhaps this is what the book is referring to but I very much doubt it is exploring such subtle concepts.

This totally blow my mind off, because I never hear this formulas. Everyday there's something new It's good too know this.

Thank you so much!