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Please ... Help needed with 7805 regu...

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bob9999
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Post Number: 123
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Posted on Sunday, 16 October, 2011 - 07:20 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I have been building, using and designing power supplies for ages now. To date only a couple of faults, (usually my error)!
This latest PSU will only output about 4.1 volts from a 7805.
I've changed the 12 AC transformer, changed the small bridge rectifier, changed the 7805. Still low output voltage.
I'm at my wits' end here. I can't figure it out. I've isolated everything on the stripboard.
No shorts anywhere. Still the low output voltage.
I know I can ramp it up with a couple of diodes in series with the earth lead but thats not the point.
Has anybody had this error before please? Just knowing where the fault lies would be enough.
By the way, the PSU is for another PIC project.
Hoping you can help me,
Kind regards
RGR
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joe
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Posted on Sunday, 16 October, 2011 - 07:37 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Two things immediatly spring to mind assuming that you've measured the regulator input and it's at least 7v.

First, check the datasheet for the maker of the actual regulator. There may be a minimum load requirement; there is for the 7905 as without a 5ma load, it's voltage will be too high.

How about the regulator just being faulty, even if it's brand new. I can remember a Texas Instruments regulator putting out 3.6v and I thought that perhaps it had just been labeled incorrectly.

Regards,
Joe
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pebe
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Posted on Sunday, 16 October, 2011 - 07:39 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Have you got decoupling capacitors on input and output terminals as detailed in the datasheet? Without the caps, the 7805 can oscillate and that can give the effect you have.
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twintub
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Posted on Sunday, 16 October, 2011 - 10:26 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

....Also monitor the regulator input ripple with an oscilloscope: Make sure the LOWEST part of the waveform does not drop below 7V. If it does, suspect the main smoothing capacitor is faulty or that the capacitor value is too low. You must do this when the load is connected of course!
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phonoplug
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Posted on Sunday, 16 October, 2011 - 10:28 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Is its output low because the loading on it is too high?
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bob9999
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Posted on Monday, 17 October, 2011 - 10:15 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Gentlemen,
Many thanks for the kind assistance. I can go forward now. The input voltage is 13.6 volts DC and I've swapped regulators, so no fault there.
I didn't realise the decoupling capacitors were so critical. On measuring the Cap'ce the output (0.1u) is OK, the input, even though marked as 0.33u is in fact 0.02u!
Hopefully the fault lies here. I'll get my trusty scope out to check though.
I don't normally use a smoothing capacitor. Is it really necessary? If so, will 100u or thereabouts do the job? I assume the capacitor comes after the bridge rectifier?
I have isolated the load, so the drain is zero. I measured 4k7 across the live and earth output and assumed that would be enough of a test load.
I'm just driving a PIC, 3 LED's and a relay coil, not much load there anyway.
Will revamp the input capacitor and try again. I suppose the oscillation will stop the 7805 from functioning properly? Hopefully!
Kind regards,
RGR
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joe
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Posted on Monday, 17 October, 2011 - 10:28 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

You really do need the main smoothing cap, and don't be tempted to put it on the regulator output; always on the input side else you can blow the regulator and make sure it's rated correctly.

Check this out for help with calculating the value of your smoothing cap:
http://www.kpsec.freeuk.com/powersup.htm#smoothing

I would guess that 100uf is too small for all but the lightest of loads and the fact that your driving a relay makes me think that 470uf would be more appropriate; but that's an off my head gut feeling without seeing your circuit.

I personally don't ever use anything less than 220uf to 470uf.

One other thing to watch out for; your input voltage of 13.6 is quite high for a 5v regulator - your asking it to drop nearly 9v with is dumped as heat. If your regulator is running too hot it's thermal protection can start to kick in and cause all sorts of problems.

Also, as you increase the smoothing capacitor value, your DC input voltage may rise compounding the heat problem.

Regards,
Joe
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atferrari
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Posted on Monday, 17 October, 2011 - 11:22 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Hola Joe,

As far as I can recall I have always seen the 780XX fixed regulators suggested in typical applications with two capacitors: one at the input and one at the output.

I use them always like that.

And since I prefer not to rely in memory I have to check the data sheet for values, every time.

Did I, maybe, understood you wrong?
Agustín Tomás - Buenos Aires - Argentina
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joe
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Posted on Monday, 17 October, 2011 - 11:44 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Hi Agustin,

with 78xx regulators you always have to have the two small 100nf or so value capacitors connected from the regulator input to gnd, and from the output to gnd, and located as close to the regulator as possible (at least with all the 78xx regulators I've used).

What I was trying to say was that the larger value smoothing capacitor must be on the regulator input, and not it's output.

I've seen people add large value caps on the output thinking that they could use lower voltage rated capacitors to save money/space, but this is dangerous. These regulators can be damaged if the voltage on their output pin goes higher than the input pin voltage as happens when you switch the device off. The regulator input voltage immediatly falls to zero, but the large smoothing capacitor on it's output is still charged; this is often why you see a diode connected from the regulator output back, to the input to try and prevent this.

I've also heard that the regulator isn't as efficient if the smoothing capacitor is on it's output instead of it's input; but I've never checked this.

Regards,
Joe
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muskrat
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Posted on Monday, 17 October, 2011 - 12:40 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Well THAT is interesting, Joe. Like Agustin, I also have used a small elco on the output of the regulators, as well as 100nF caps to avoid oscillation.
Looking at the National data sheet, they show a small (10u to 22u) elco at the output.
I haven't experienced problems with these regulators in the past, but I'll watch carefully after reading your comment. Thanks Joe.
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joe
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Posted on Monday, 17 October, 2011 - 12:53 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I just checked to make sure I'd not had some form of brain-fart and was dreaming this as I thought this was common knowledge.

http://cva.stanford.edu/classes/cs99s/datasheets/LM340.pdf

Read the text at the bottom of page 9.

This is for the LM340 fixed voltage regulator, but as far as I know, it applies to all fixed voltage regulators.

I think your ok with small value electrolyitics as the circuit load discharges them very quickly, but larger values on the output should be avoided.

Regards,
Joe
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dselec
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Posted on Monday, 17 October, 2011 - 01:34 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

TRy adding 1 ohm resistor in series to regulator input and check calculate the current.
should not be more than 400ma .
are you using the small plastic regulator or the big with heat sink model ?
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vlf
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Posted on Monday, 17 October, 2011 - 05:52 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

bob9999... all the above comments are sound advice; first fit the two decoupling cap's, if the regulator is still giving low volts without a high output load then; please do-not take offence but; check the correct three pin connections: ST pinouts = (Left to right top view “Input-Gnd-Output”) !.. as silly it might sound, I have in hast installed these regulators the wrong way around, without using any heat-sink, and had low voltage problems.

The other possibility; if your not using a heat-sink for the 7805, where no substancial screening to supply rail Ground is available... and if the regulator is close to any R.F. clock or Mhtz oscillator, “even sharing the same supply-rail” then; place a few ohms in series with the regulator input, and decouple to ground o-volts rail, with a 100nF cap at the resistor junction, and at the regulator input pin, with a 100nF cap and a 22Mfd to ground o-volts... keep these component leads and tracks short.

It would-not go amiss, to provide additional decoupling caps around any R.F. source, on the supply-rails.

Regards, Brian L.
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bob9999
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Posted on Monday, 17 October, 2011 - 06:34 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Gents,
Many thanks for the further information. I have taken it all onboard.
I have a correct 0.33u capacitor for the input now, and won't bother with a smoothing cap,(as usual).
All being well, the thing should work now. If not, I'll be back.
Kind regards,
RGR
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dselec
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Posted on Monday, 17 October, 2011 - 09:43 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Smoothing capacitor for 10% ripple, C = 5 × Io
Vs × f = 800uf (for 500ma and 12v) so how did we get 0.33uf ?
and more i 10uh coil at reg output will help .
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pebe
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Posted on Monday, 17 October, 2011 - 10:53 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

The 0.33uF cap is required for decoupling - not smoothing.
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epithumia
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Posted on Monday, 17 October, 2011 - 11:14 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

I'm gobsmacked at the idea of NOT using a smoothing capacitor.

Without that you won't get a remotely regulated output. It might give you 5V at no load, but at any real load what you've made is a 100Hz pulse generator...

Epi
If you need me, Neil and me will be hanging out with the Dream King. - Tori Amos
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zeitghost
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Posted on Tuesday, 18 October, 2011 - 11:50 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

How interesting.

I discovered that a 7805 will oscillate nicely at about 40MHz if you forget to fit the 100nF caps on input & output.

It was quite happily singing away to itself.
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bob9999
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Posted on Tuesday, 18 October, 2011 - 03:05 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Gentlemen,
I have foung the error, and it was my mistake. Your promptings about smoothing capacitors did it.
To date, most of my PIC projects run from an ssortment of variable DC supplys I've previously built.
Note the DC bit! The current device is an AC device and, it's been so long since I last built one, I just relied on the decoupling capacitors!!!
(This by way of an excuse for ignoring smoothing capacitors).
Anyway, I got my scope out and was horrifiied. Max ripple. Then it clicked. I stuck in a 250u and suddenly the output stabalised at 5.04 volts.
Still a lot of input ripple though so I substitued for a 2200u cap.
Now there is no ripple, stable voltage but hot 7805, (input 18volts AC). Too much. I'll dig out a smaller transformer, that should do it.
Many thanks for all your much needed comments, without them I would still be stuck.
It's always the little things that make all the differance isn't it?
Kind regards,
RGR
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dselec
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Posted on Tuesday, 18 October, 2011 - 04:07 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

"(input 18volts AC) " are u using a secondary center tapped transformer 2x 9vac if so use half way rectifier .
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joe
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Posted on Tuesday, 18 October, 2011 - 05:50 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

"input is 18v AC"
You actually mean 18v DC input to regulator I hope :-)

If your using a full wave bridge you can use a half wave instead; that will drop the voltage a bit. Also, you can slip a few diodes in series before the regulator to drop a couple of volts as well.
It may be enough depending on your hestsink and circuit load.

If you don't have a better transformer you could use a 12v reg to drive the 5v one if you have one kicking around. Ok till you can build something a bit more perm.

Regards,
Joe
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bob9999
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Posted on Tuesday, 18 October, 2011 - 07:33 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Joe: You are right. I do mean 18v DC, (1.4 x 12v AC=16.8v DC).
My transformer is not centre-tapped.
I don't like the idea of using half wave as there is usually too much "noise", which for a microprocessor is murder.
Again, I could drop as you suggest, but that's more heat.
I canabilised my parts from a defunct washing machine, hence the use of 12v relays!
When all's said and done, better to use a 6v AC transformer with 6v relays. (I hate spending money though)!
Anyway, thanks again,
Kind regards,
RGR
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twintub
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Posted on Tuesday, 18 October, 2011 - 10:25 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

bob9999
A simpler method to cool down the regulator is to use an appropriate wattage & value of series resistor at the input to the regulator (placed just after the smoothing cap but before the decoupling cap). Assuming you have a 0.1A load (which is likely given that you're only driving a PIC, LED and a relay!) an 82ohm resistor will drop 8.2V and will dissipate 0.82W. For a 0.5A load the resistor could be 15ohm (dropping 7.5V) and would dissipate 3.75W.

These wattages will effectively be alleviating the regulator of the burden.
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bob9999
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Posted on Wednesday, 19 October, 2011 - 10:29 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Twintub:
Thanks. I know I could use a resistor but thats not the point. I always remember my mentor's teachings. His constant mantra was ... "If you've got heat, you've got a bad design"!
I've worked with that with all my designs.
My latest incarnation is to use one of my 9v DC supplies (as usual) then feed the mains in for the relay output. Bit Heath Robinson I know. I'll keep the other bits for "Later"!
I am however so chuffed to discover the fault in the circuit that it has all been worth it.
Too. It's given you guys somthing to think about ... for which I am most grateful. Without your help I would still be head-scratching.
Who would have thought the lack of a simple smoothing cap could produce such a headache!
Kind regards,
RGR
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muskrat
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Posted on Wednesday, 19 October, 2011 - 11:02 am:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

It's always the little things. The things you take for granted.
I always tell my students: If someone explains something with "because that's the way we've always done it", then question, question, question and don't give up until you find out WHY!!!
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bob9999
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Posted on Wednesday, 19 October, 2011 - 01:42 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

An excellent phillosophy Muskrat.
I only wish more subjects were tought with that in mind, what an improved society we'd generate.
That, and our reverting to teaching Logic as a stand-alone subject!
Kind Regards,
RGR
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dselec
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Posted on Wednesday, 19 October, 2011 - 02:30 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

"(I hate spending money though)!
Anyway, thanks again, "
Don't waste a penny i assure you a dalf wave rectifier will do use a ( pi=3.14 )filter for regulator input that's a 2200uf capacitor in parallel with a 10uf capacitor adding a 510-1000ohm resistor between the capacitors.replace coil with resistor or add resistor in series with coil see drawing
http://en.wikipedia.org/wiki/Capacitor-input_filter
http://www.google.com/search?q=pi+filters&hl=en&sa=X&biw=1024&bih=659&prmd=imvns&tbm=isch&tbo=u&source=univ&ei=U9CeTsK_FIva4QTZoaXbCQ&ved=0CFAQsAQ

(Message edited by dselec on 19 October, 2011)
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dselec
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Posted on Wednesday, 19 October, 2011 - 02:34 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

http://www.google.com/search?q=pi+filters&hl=en&sa=X&biw=1024&bih=659&prmd=imvns&tbm=isch&tbo=u&source=univ&ei=U9CeTsK_FIva4QTZoaXbCQ&ved=0CFAQsAQ
if you need help calculating the resistor ill need your total current measured .
you can also add a resistor and cheap 1$ 5watts 9v zener

(Message edited by dselec on 19 October, 2011)
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ant
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Posted on Wednesday, 19 October, 2011 - 06:08 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Hello Bob,

I have a lot of transformers which must go, some are in your voltage range, most but not all are s/h. I'm looking for postage and, if you're really pleased, a donation to my current favourite charity.

Regards Ant
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bob9999
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Posted on Wednesday, 19 October, 2011 - 10:09 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

Dselec,
That takes me back a bit! I'll explore the possabilities on breadboard for a later project. Thanks.
Ant,
Thanks for the offer but I'm finally sticking with a 9v DC supply and feeding the mains into the relay, (Heartbreakingly, I've had to lash out on a 6 volt relay).
I was thinking in terms of some form of diode pump to raise the voltage for the relay but enough is enough!
Kind regards,
RGR
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dselec
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Posted on Thursday, 20 October, 2011 - 03:52 pm:   Edit Post Delete Post Print Post    Move Post (Moderator/Admin Only)

"diode pump to raise the voltage "
OK you can use bridge rectifier for your 7805 and use the same ac volt from secondary for multiplier the relay and the bridge do not know each existence of the other.
this link will solve both your 9v and 12 volts relay problem.

http://www.allaboutcircuits.com/vol_3/chpt_3/8.html
did you see the filter link i posted ?

(Message edited by dselec on 20 October, 2011)

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