7805 - noisy rails

[westhomas]

I hooked a speaker to the positive and negative rail on my breadboard and there is very substantial hum, kind of low pitched. Does this constitute an oscillation, and if so, how can I fix it?

I;ve got a 10µF cap on the input voltage, and a 1µF, and 0.1µF across the output of the 7805. I'm worried about damaging components so I don't want to mess around too much.

thanks for any input,
wes

 

[audioguru]

A 10uF filter capacitor following the rectifier bridge is far too low for a power supply. It should be 220uf for low current or 3300uF for up to 1A of current.

The datasheet for the uA7805 shows a 0.33uF capacitor mounted very close th the pins of the regulator.

The lowest part of the input voltage to a 7805 regulator is 8V. Some of them work with an input as low as 7V. Yours was probably 0V.

 

[Ubergeek63]

there should be high frequency bypass on both the input and output as well.

 

[westhomas]

also, I'm using a wall-wart for converter - i don't know what's inside but it's rated 12V at 200mA. I've tried some other converters as well. same problem.

I've got a 555 and a 556 on my board but am running 100uF caps across their power rails. I don't think that's the problem

there should be high frequency bypass on both the input and output as well.

you mean a 0.1uF cap?

 

[Speakerguy]

What's the impedance of the speaker? You could be drawing way too much current from your wall-wart. You might be listening to 120Hz ripple from an input voltage that is sagging badly.

 

[Hero999]

What power rating is the speaker?

It'll probably smoke if you leave it connected for too long.

 

[westhomas]

it's an 8Ω rated for 50W (RMS) and it's a beater. I'm not so worried about damaging it so much as I am about damaging my circuit. I think I already degraded the quality of one of my sensors just while testing it.

Hmmm...I just tested the output from this thing and it's giving me 19V and 1 amp though it's rated for 12V and 220mA. But regardless, the 7805 is rated for up to 30V input, with a 1A output. And now I've tested a 9V battery - no noise. It must be a ripple from the bridge of the wall wart. Can you suggest a solution for this?

 

[Speakerguy]

Eight ohms on 5V = 5/8 amp = 625mA. If your wall wart can't deliver at least that, you'll have trouble. You need more current to supply the regulator. It can't output more than what's it getting in. You can also try a bulk electrolytic on the input side to reduce ripple, but a more stout wall-wart is the first order of business.

If you have good DMM, just look at the AC voltage on the 5V rail without the speaker connected. It should be negligible, although you have to have a good DMM for it to have a wide frequency response band.

I have only ever seen oscillation on a 7800 series regulator once, when used without output capacitors, and it had an oscillation frequency over 1MHz. A cheap electrolytic on the output fixed that. I seriously doubt you have any oscillation problems, and all of your problems are caused by hooking the speaker to the 5V rail.

 

[Willbe]

How 'bout posting a schematic?

 

[crutschow]

If you want to check the hum on a power supply with a speaker you need to put a capacitor (at least 470µF) in series with the speaker to block the DC.

 

[westhomas]

when I put a 470µF cap in series with the speaker the hum is gone. My multimeter has an AC function, on 200 it reads 10.2 across the rails. I have no idea what to make of this figure.

I think you've all demonstrated the limits of my understanding here. Should I follow the advice of Audioguru and put a 3300uF cap across the input of the 7805, considering it has a 1A output?

Here's what I have, plus add a 0.1uF across the output -

Thanks for all the suggestions

 

[audioguru]

Your 8 ohm speaker was shorting the wall-wart so it produced very high amount of hum.
It is normal for a cheap low current wall-wart to produce a much higher voltage with no load. Its voltage (12V) should be correct when the current is at its rating (200mA).
The wall-wart is not supposed to be shorted with a current meter (you shorted it and measured 1A).

For only 200mA then the main filter capacitor should be 660uF and it is probably inside the wall-wart. The regulator cannot produce an output of 1A when its input is only 200mA. Its output must not be loaded with more than 200mA or the wall-wart might blowup or catch on fire.

 

[crutschow]

My multimeter has an AC function, on 200 it reads 10.2 across the rails. I have no idea what to make of this figure.

That could indicate a high frequency oscillation.

Is the output capacitor directly across the 7805 pins?

 

[Willbe]

it's rated 12V at 200mA

C = I ΔT/ΔV.

A 12v dc value means 12(PI/2)=19v peak value for full wave rectification
and you want >7v at the input to the regulator
so ΔV = 19-7 = 12v, I = 0.2A, ΔT = 8 mS
so I get C at the regulator input should be >130µF.
Use a 30v cap rating because with no load the V can climb another 20%.

 

[westhomas]

The regulator cannot produce an output of 1A when its input is only 200mA. Its output must not be loaded with more than 200mA or the wall-wart might blowup or catch on fire.

yes, that makes sense. The 9V had enough current to satiate the regulator, and so no hum.

it's rated 12V at 200mA

C = I ΔT/ΔV.

A 12v dc value means 12(1/0.636)=19v peak value for full wave rectification
and you want >7v at the input to the regulator
so ΔV = 19-7 = 12v, I = 0.2A, ΔT = 8 mS
so I get C at the regulator input should be >130µF.
Use a 30v cap rating because with no load the V can climb another 20%.

Thanks for the equation. Absorption is best when you've reached a point where it critically affects what you're trying to do. I'll remember this one.

Thanks for the lessons. As you can tell I'm just teaching myself as I go. I hope someone else will find these posts useful.

 

[Willbe]

On second thought

The 7805 was in pretty short supply back in 1943 but if you can get your hands on the work of Schade
**broken link removed**
and, even better, an app. note from GE in Auburn, N.Y. by E.E. Von Zastrow dated 8/67 using Schade's graphs, you can figure this cap value pretty close.

The problem is Schade's graphs are for a fixed load resistor while the input to the IC regulator looks like an ideal current sink. Approximate, approximate!

 

[audioguru]

I use this graph for determining the value of a filter capacitor in a full-wave rectified circuit with a 60Hz mains. It shows the current and the resulting amount of p-p ripple voltage.

 

[Hero999]

I use a spreadsheet which uses this complected formula which I derived when I was bored. It's not 100% accurate as it doesn't account for the diode drops which I add on separately but it's accurate enough, maybe I'll modify it to account for this some day.

[latex]C = \frac{I}{V_R} \times \left(\frac{1}{4f} + \frac{arcsin{\left(\frac{V_{IN}-V_R}{V_{IN}} \right)}}{2\pi f} \right)\\
C=\text{Minimum capacitance required}\\
V_R= \text{Maximum ripple}\\
V_{IN}= \text{Input voltage, excluding diode losses}\\
I=\text{Current drawn by regulator}\\
f= \text{Frequency}\\
[/latex]

When I don't have the above formula handy I use this simple one, which safe as it oversizes capacitors and is easy enough to remember.

[latex]C = \frac{I}{2fV_R}[/latex]

 

[westhomas]

Thanks all,

It will be good to have these references in the future. I believe what was causing the problem was what audioguru described...I probably shouldn't trust a 50W speaker as an oscilloscope, heh?

...thanks again