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Creating A DC and An AC power Source With Approximately Equal Effective Voltage Outputs

Rory Starkweather

Nov 13, 2014
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This may take a little time because I am way outside my normal area of interest.

What I want to do is test an LM7809 with a pure AC input, then with a DC input. The thing I am primarily interested in is the % ripple on the output voltage. I won't be using anything but the bare LM7809. No capacitors or anything else.

I would like to use input voltages of 18 vdc and whatever AC voltage would be comparable to that.

I'm assuming that the AC source will need to produce 18 RMS volts AC. Is that correct?

I bought two 120 VAC to 18 VDC transformer/regulators for $1.58 each. I plan to use one as is, and cut the regulator circuitry off of the other one. I'm sure this will give me more than 18VAC, but I may be able to get it down to what I need somehow.

So does anyone see any gross errors in my methodology?

BTW, I realize that I could simulate the circuit, but I would like to wait until I have real physical data and then compare that to a simulation.

This is part 1. of a project to evaluate the effects of various support circuits on the LM7809. Primary data of interest is full circuit power dissipation, Output voltage and ripple and noise on the output.
 

davenn

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What I want to do is test an LM7809 with a pure AC input, then with a DC input. The thing I am primarily interested in is the % ripple on the output voltage. I won't be using anything but the bare LM7809. No capacitors or anything else.


you want to put AC into a 7809 ?
that's a no-no or did you really mean you wanted to rectify the AC first ?

there wont be any ripple with a DC source eg a battery

with no smoothing caps, you are likely to cause damage to the reg. chip
 

KrisBlueNZ

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Yes, as Dave says, if you feed "pure AC" into a 7809 you will damage it; it's not designed to have negative voltage at its input, and AC alternates between positive and negative.

If you want to see how the 7809 handles ripple, you could feed it from rectified AC without a smoothing capacitor - either half-wave rectified or full-wave rectified - which varies between 0V and some positive voltage but doesn't go negative. You will see a similar signal at the regulator's output.

An 18V AC RMS source will produce a peak voltage of 25.5V (calculated from sqrt(2) × RMS voltage). If you want an unsmoothed rectified voltage that peaks around 18V your transformer should be specified for about 13V RMS.

If your power supplies cost you $1.58 each they are probably switching power supplies. In that case you won't be able to remove the filter capacitors and get a rectified unsmoothed output voltage.

BTW, I have no idea why you would want to do any of this, and I can't see anything useful that will come of it!

If you tell us the ultimate aim of your experiments, we will be able to tell you what you would learn and what conclusions you would come to.
 

Rory Starkweather

Nov 13, 2014
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I had the impression, mistaken it seems, that I could put a really dirty signal in, and get something useful out. It appears now that an unfiltered full bridge rectified DC with ripple will be about the best I can do. Oh well. As I understand it, half rectification will just throw away half of the input signal and not tell me much.

The ultimate goal is a 'bullet proof' 9 volt DC supply. In the end it will have multiple filter caps, protective diodes, a PTC thermistor, current boost and variable current control. This power supply will be used for small calibration circuits.
 

KrisBlueNZ

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A regulator will turn a "dirty" signal into a much cleaner signal, but it will be damaged if the input voltage goes negative, and it cannot produce output voltage when there's not enough input voltage.

A regulator from the 78xx series has a dropout voltage of about 2V. That means that the input voltage needs to be at least 2V higher than the regulated output voltage, otherwise the regulator will "drop out" of regulation, i.e. the output voltage will drop below the nominal regulated voltage.

This requirement applies at every instant in time, because the regulator has no kind of energy storage. As soon as the input voltage drops below VOUT+2, VOUT drops.

Therefore if you put AC through a rectifier, but you don't have a smoothing capacitor, and you feed that signal into a regulator, the regulator's output voltage will have gaps at exactly the same time that its input voltage has gaps. To avoid the gaps, you need some kind of energy storage - a capacitor, connected at the regulator's input.

The following image is from http://macao.communications.museum/eng/exhibition/secondfloor/moreinfo/2_16_0_DiodeLab.html

2_16_0_12_eng.png

The grey dotted curves show the output of a full-wave rectifier without smoothing. The waveform goes all the way down to 0V and if you feed that voltage into a regulator, the regulator's output will also go all the way down to 0V during those times in the waveform.

Adding a smoothing capacitor provides energy storage to fill in the gaps, and the voltage across the capacitor will look like the red trace. As long as the lowest point on the red trace is at least as high as VOUT+VDROPOUT, the regulator output will be clean and stable.

You can connect smoothing capacitors across the output of the regulator, but the bulk smoothing must be done before the regulator's input, for various reasons - the main one being that this will allow the regulator to remove the ripple (the variations in voltage on the red trace) and produce a nice straight line at its output.
 

Rory Starkweather

Nov 13, 2014
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Actually, I know that, but the idea is to build the circuit incrementally. The second thing I am going to add to it is a 1000 uF capacitor from the input to ground. It will eventually be followed by two more caps there, and two across the output. One from output to sense. Diodes from output to input and output to sense. A thermistor across the input. A current booster, a current regulator at the output, and a few other things. I have a good idea what all of these will do, but I want to see the real world numbers.
 

KrisBlueNZ

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Okey dokey. BTW the thermistor will have to go in series with the input, not across it.
 
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