Voltage Regulators

What do you mean by 1mF ? It *should* mean 1000uF but popular US usage ( why are Americans *always* so bloody minded ? ) was to use mF = uF.

If you have 1uF, that's your problem.

Graham

cause it makes complete sense. m = *m*illi, u = micro cause u looks like the greek symbol from micro that is often used. I don't see a problem with it? Just cause someone started a bad trend should we keep on perpetuating it?

well, its not.

Jon

Capacitance units aside ;-) I would not expect that difference in reading between 200v and 20v range from any worth while VOM. It's not unusual to have the regulator wrongly connected and what you are saying sounds like it to me. Have another look :-)

--
Regards ......... Rheilly Phoull

Once in a while, I get flaky readings on a DMM. It usually means the battery is low.

Try replacing the battery in your DMM.

Regards,

Mark

Hi, Jon. Take a deep breath, and look at things one at a time.

Before anything else, check your meter. It shouldn't change readings between the 20V and 200V range. Get a known good 9V battery, and measure 9V on the 20V and 200V ranges. If they don't agree, replace the battery on the meter and try it again. If the meter still doesn't agree between ranges with a fresh battery, junk it and get another one for $10 or $15 at the hardware store. You need a known good meter to troubleshoot.

First, you want to look at your unregulated, filtered power supply. To see if it's working properly, try putting a load on the supply _before_ the regulator. As a rule of thumb, you should figure that, for a 60Hz supply (120Hz ripple), 8,000uF will give you 1V of ripple per amp of current. That means a 1,000uF cap will give you 8V of ripple per amp. On your DVM, that 8V will look like a 4VDC reduction per amp, because the ripple is roughly a triangle wave. So, get a power resistor, load up the power supply, and see what's going on.

Let's assume you have a 100 ohm, 20 watt resistor handy. Put it across the 35V peak supply. You'll have about 1/3 amp load, which means 1/3 *

8 or about 2.7V of ripple. On your DVM, that will look like your 35V will go down half that, or down 1.4V to about 33.6V. If your DC voltage measured across the cap doesn't go down by about 1.4V +/- 0.25V with that 100 ohm resistor load, then something's wrong. Your bridge rectifier may be hooked up wrong. Your cap may not be 1000uF. Your cap may be old and bad. Your cap may actually be 1uF (which would cause the problem you describe). Your transformer may be messed up. But you will _know_ that there's a problem in that part of the power supply. And if you have another value power resistor, work it out for that load. But if that part of the power supply performs properly, then you know that's where the problem _ain't_.

Once you get that taken care of, look at your regulator. Just get a

12VDC wall wart. Put the + at the input pin, with the - of the wall wart goijng to the middle pin of the 7808. Measure the output voltage. Yes, there might be some high frequency oscillations, but it _will_ give you 8V +/-10%, or it doesn't work. Then put a 220 ohm resistor between the output pin and GND, and see if the output voltage changes. If it changes more than several mV with a 36 mA load, it's busted.

If you look at the data sheet, you'll see that the Absolute Maximum input voltage to that device is 35V, and you've already come right up against that limit. You need to think this one out a little better, or lay in a good supply of spare parts. I'm figuring your 7808 is damaged (they act that way sometimes when they've been overstressed).

Putting 35VDC at the input of an LM7808 isn't doing it right, Jon. Not only that, but let's say you're pulling 1/2 amp from the 7808. With a

1000uF cap, you should have an average DC of 33V. Your power dissipation will be (33V - 8V) * 0.5A = 12.5 watts. Lotta extra heat, here.

The point of troubleshooting is to make good decisions based on available information. Read the datasheet, Jon. That's always important. And Absolute Maximum means that, if you exceed that, even by a millivolt, the manufacturer says you will have problems. But when you're doing something new, or even troubleshooting something that's known to have worked, you have to be open to the available facts. Try to make the observations that will make it apparent where the problem is, but more importantly, where it is _not_.

Make good observations, and believe your observations, even if they don't seem to make sense. The things you're seeing and measuring are trying to tell you something, if you can take the time to learn their language and are prepared to listen to them.

You also may want to take a minute and actually think out what you're doing here, and see if you can get there with the parts you have. Again, I'd really recommend just buying one or two open-frame linears, combine them with an LM317 for the variable voltage, and be done with it.

(And by the way, people stopped caling 1000uF "1 mF" about 35 years ago. If your cap says "1mF" on it, the cap is almost certainly a fossil, and shouldn't be used, anyway.)

Good luck Chris

It simply isn't common practice to quote capacitance as milliFarads though. As I explained it's prone to being misunderstood due to prior bad US practice.

Graham

1000uF @ 200V is actually physically quite large !

Graham

A 1 milliFarad (1000 uF) 200 volt capacitor would be very large.

If the capacitor is marked "1mF", then I suspect that it is almost certainly 1 uF - you need a much larger value for your primary filter capacitor - 1000uF at 50 V would be a good start.

For historical reasons, "mF" = "uF" - in ancient times, a 1000 uF (or

1 milliFarad) capacitor would have been impossibly large - no-one would have a use for such a thing, so, as the Greek alphabet had not yet been invented, we used "mF" to mean "microfarad" - that practice is so deeply ingrained in the culture, that it is very undesirable to use "mF" for anything other than microfarads.

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When you finally get it sorted out, be aware that you will be pounding that chip pretty hard if you draw appreciable current through it. For example, at 200 mA (40% of the chip's rating) it will need to dissipate 6 watts. That's not a factor with your present load, but you really don't want to go over about 30 mA if the chip is not mounted on a heat sink. The problem is that the 30 volt Vin-Vout means the chip has to dissipate 30 * Iout.

Ed

--- Unless you have an auto-ranging meter you _do_ have to set the range yourself. If you don't know what range your reading will fall in, you should start with the highest range and switch down as required to get the precision you want.

-- John Fields Professional Circuit Designer

Hello Jon

First thing, Use the center tap of the transformer and 2 diodes to get the DC output of the base supply down to a reasonable level. 1000uf is a bit on the low side. Use a lower voltage cap, it'll be MUCH smaller.

Next, Make absolutely sure that the regulator is connected properly.

Front View ------- | o | | | |78M08| | | ------- input | | | Output | ground

Next, Put a 1-10uf capacitor between the output and ground pins of the regulator and if the regulator is more than a few inches from the main filter cap, put a .1-1uf on the input pin also.

Next mount the regulator on a small-medium sized heatsink.

Btw, An MC78M08 has a "peak" output current rating of 700mA. The data sheet says "Maximum output current with adequate heatsinking is 500mA".

Mike

--
Since:

         I dt
    C = ------
          dV


rearranging to get the ripple voltage due to a 500mA load gives us:


          I dt     0.5A * 8.3E-3s     
    dV = ------ = ---------------- = 4.15V
           C           1E-3F


Using the supply you suggest:


                              17VDC             8VDC
                               /   +-------+     /
MAINS>---+  +-------[DIODE>]--+----| 78MO8 |----+    
         P||S                 |    +---+---+    |
         R||E-CT-+--[1000µF]--+        |      [16R]  
         I||C    |            |       GND       |
MAINS>---+  +----|--[DIODE>]--+                GND
                 |
                GND

would put the input of the regulator at 17V.  With a 500mA load on
the output of the regulator, the ripple would cause the input
voltage to vary from 17V to about 12.8V.  

The regulator needs about 2 volts of headroom (typical, I couldn\'t
find worst case) to stay in regulation, so with a worst case
input-to-output differential of 12.8V - 8V = 3.2V there should be no
problem at all with a 1000µF cap.

But what about accounting for the cap\'s tolerance?

If it\'s + whatever, -20%,  then we\'ve got 800µF, minimum, and we
wind up with:

          I dt     0.5A * 8.3E-3s     
    dV = ------ = ---------------- ~ 5.2V
           C           8E-4F

of ripple, so with an 8V output and a 17V - 5.2V = 11.8V input
valley voltage, we wind up with an input-to-output differential of
11.8V - 8V = 3.8V.

Still no problem.

--
Rewrite:

The regulator needs about 2V of headroom to stay in regulation,
which means that for a fixed 8V output, the input _always_ has to
stay at 10V or higher.

With a 500mA load and 1000µF filter, the lowest the input will drop
to will be 12.8V, so that\'s no problem and 1000µF will be OK

It seems like it was the battery as I'm getting a consistant reading across the different voltage ranges.

Thanks for the help Jon

With no capacitance on the output of a 3 teminal regulator you run the risk of parasitic oscillations. Oscillations upwards of 1Mega Hertz. Detected by your DMM because it's not rated for that frequency. Detected by little pn junctions and filtered by stray capacitance, present a DC voltage to the DMM circuit. Switch ranges and the stray capacitance changes changing the reading. No? Get a good ole VTVM out preform a DMM in my opinion.