60hz hum

[WhenBoredomPeaks]

I started a diy modular synth a while ago. I realized that i made some errors regarding grounding and now i pay the price. Basically i followed stompbox priciples but there are some differences in the synth world.

The problems: My panels of the different modules are made of metal and they are touching. The jacks are not isolated from the panel. The panels are connected to the circuit ground. Every module have it's own ground cable from the supply. So basically there are ground loops everywhere.

2 of 3 filters have a loud 60hz-ish hum, (Moog, Steiner) the MS20 is dead silent. There is an audio to gate detector which triggers by itself once in a while if i touch some metal things here and there.

Despite this i am still thinking that the 60hz noise is actually power supply ripple. I used 1000uF smoothing caps on the power supply. (+/- 15V supply, i think i draw around 200mA)
I see way bigger smoothing caps on the internet regarding synth PSUs, is it possible that the 60hz hum i hear is power supply ripple instead of grounding problems?

(if it is ripple then i just have to replace the smoothing caps, but if it is something ground related then i should remake the panels from some non-conducting material )

ps: i can add images once imgur is up.
i used a supply like this but with 1000uf caps: http://www.cgs.synth.net/modules/pic/schem_cgs14_psu.gif

[brett]

Hi
the power supplies should be regulated (7815, 8915).
A smoothing electro cap makes the job easier for the regulator, but it's unrealistic to rely on it alone. The film caps associated with the regulator play an important role, too.
cheers

[WhenBoredomPeaks]

Hi
the power supplies should be regulated (7815, 8915).
A smoothing electro cap makes the job easier for the regulator, but it's unrealistic to rely on it alone. The film caps associated with the regulator play an important role, too.
cheers

i used this circuit:



but i used 1000uF smoothing caps only so maybe that is why i hear ripple

[Thecomedian]

Why would you need such high Farad?

[WhenBoredomPeaks]

Why would you need such high Farad?

I don't know the science behind it but i've seen here that it is common practice to put bigass electro caps inside power supplies to avoid ripple.
(so the caps here are not used for filtering power supply noise but to make the DC coming out of the rectifiers actually looking like DC)

Some calculations: http://en.wikipedia.org/wiki/Ripple_(electrical)

I've seen a calculation there, let's say we want to get the ripple down to 0.1Vpp (i don't know if this is actually too high or low but it seems reasonable to me)

Let's say my circuit will need max 1 amper. (actually it draws around 0.2 amper) The electricity is 60hz here. Then i would need a 83000uF cap! (i used Amper, Farad and Hz in my calculation, hope those units were right)
Someone could check my calculations if they were right.

[defaced]

Power supply ripple would be 120hz because it is rectified.  Ground loop hum is 60hz.  Not to say you don't have both, but if you have 60hz hum, at a minimum I'd be looking real hard at ground loops.  I'd recommend checking out the first chapter of Tim William's Circuit Designer's Companion for some background and ways this is dealt with. 

[WhenBoredomPeaks]

Power supply ripple would be 120hz because it is rectified.  Ground loop hum is 60hz.  Not to say you don't have both, but if you have 60hz hum, at a minimum I'd be looking real hard at ground loops.  I'd recommend checking out the first chapter of Tim William's Circuit Designer's Companion for some background and ways this is dealt with. 

I run a frequency analyzer on the hum and it is actually 100hz
We probably have 50hz mains frequency here. (im in central-europe)

Is that confirms that it is ripple noise?

[defaced]

That does seem to support that conclusion.  If you have a scope, just scope the power supply rails.  If you see ripple, then you can definitely confirm it.  You can then also isolate which rail is giving you problems. 

[R O Tiree]

Check that you're getting at least +17V at the 7815 input and -17V at the 7915.

Check that the input and output jacks from those modules have signal and GND connected the right way around - I had terrible hum from a BMP once because I got the output wires reversed.

[WhenBoredomPeaks]

Check that you're getting at least +17V at the 7815 input and -17V at the 7915.

Check that the input and output jacks from those modules have signal and GND connected the right way around - I had terrible hum from a BMP once because I got the output wires reversed.

i rectify 12v ac which should be ~16,97V dc into the regulators

i'll check them

[WhenBoredomPeaks]

This kept bugging me while i wait for the bigger caps and stuff:

Is it possible that the MS20 filter have a better Supply Voltage Rejection Ratio than the other two? The MS20 filter uses opamps, the others are using transistors.

The other thing i thought about is that maybe my DC voltage into the regulators is too low. (~16.96V into 15V regulators)

[PRR]

RO>> Check that you're getting at least +17V at the 7815 input and -17V at the 7915.

WBP> i rectify 12v ac which should be ~16,97V dc into the regulators

As RO asks: Check that you're getting at least +17V.....

Calculation is not reality.

[PRR]

> say we want to get the ripple down to 0.1Vpp (i don't know if this is actually too high or low but it seems reasonable to me)

No. 5% ripple is reasonable at first cap. <1% is un-reasonable.

(In fact stray resistance means you "can't" get 0.5% ripple at your first cap, not without heroically fat busbars and caps.)

If you need less than several percent, you use an additional L-C or R-C filter stage, or.... a regulator.

IIRC, the PAiA synth was metal panels in wood cabinet, and it did help to make sure the panels could not touch (I have dim memory of taped edges).

> better Supply Voltage Rejection Ratio than the other two?

Without knowing what "other two", who can say?

[WhenBoredomPeaks]

RO>> Check that you're getting at least +17V at the 7815 input and -17V at the 7915.

WBP> i rectify 12v ac which should be ~16,97V dc into the regulators

As RO asks: Check that you're getting at least +17V.....

Calculation is not reality.

ok! will update the thread with results.

(btw i can try it with a bigger transformer but the reason i did not tried is that then my regulators would be too hot (it' a 2x19VAC transformer i guess)) (now they run at room temperature without heatsinking)

[WhenBoredomPeaks]

> say we want to get the ripple down to 0.1Vpp (i don't know if this is actually too high or low but it seems reasonable to me)

No. 5% ripple is reasonable at first cap. <1% is un-reasonable.

(In fact stray resistance means you "can't" get 0.5% ripple at your first cap, not without heroically fat busbars and caps.)

If you need less than several percent, you use an additional L-C or R-C filter stage, or.... a regulator.

IIRC, the PAiA synth was metal panels in wood cabinet, and it did help to make sure the panels could not touch (I have dim memory of taped edges).

> better Supply Voltage Rejection Ratio than the other two?

Without knowing what "other two", who can say?

i still need bigger caps to get the ripple down to 5%
i will move onto grounding issues once i remade the supply and the hum is still there
the other two is a yusynth moog vcf and a cgs steiner but i stated that in the first post

[Paul Marossy]

I run a frequency analyzer on the hum and it is actually 100hz
We probably have 50hz mains frequency here. (im in central-europe)

I find that it's very easy to confuse one for the other. Power supply ripple sounds a little more like a buzz than a hum to my ears.

[WhenBoredomPeaks]

I added a bigger transformer and it seems like it solved the hum problem. After some measurements on the negative rail the voltage was something like 16.5VDC before the 15V regulator which was not enough. So the guys who said that i should check my voltage into the regulators were right. I am now about to rebuild the supply.

Now the bigger transformer is a good quality 2x19V toroid, 4 Amps/secondary. It's an expensive leftover from a DIY high quality music amplifier project.

Now the voltage after rectifying is something like 27V into the 15V regulators which made them super hot in 30 seconds.

I don't want to throw away this expensive toroid, can i divide down it's voltage with resistors? Is it better to divide before rectifying or after? I would like to get down the DC voltage around 18V.
I would draw 200-300mA from the supply @ +-15V. Would something like an 5W resistor be enough for the divider?

[R.G.]

i used this circuit:

It is critical for eliminating hum (along with having enough voltage for the regulators to regulate) that the CT of the transformer go to the junction of the two first filter caps, and nowhere else. This may not be too bad for opamp circuits, but any single ended circuits can have a hummy buzz from this that is otherwise impossible to remove. It's caused by the high currents in that CT lead.

I added a bigger transformer and it seems like it solved the hum problem. ...
Now the voltage after rectifying is something like 27V into the 15V regulators which made them super hot in 30 seconds.

I don't want to throw away this expensive toroid, can i divide down it's voltage with resistors? Is it better to divide before rectifying or after? I would like to get down the DC voltage around 18V.
I would draw 200-300mA from the supply @ +-15V. Would something like an 5W resistor be enough for the divider?

It's generally not a good idea to divide down power supply currents with resistor dividers. It can be done, but not without wasting massive amounts of power. Even so, I'll tell you how to go about it in this case.

A little math that's NOT available from online calculators will show the problem pretty quickly. If you're wanting 300ma at 15V out, and the raw DC power supply is producing Vraw, then the power wasted in the regulator is going to be (Vraw-15)*300ma. For Vraw=27V, that's 12V*300ma = 3.6W.

You have a few options.
(1) Heat sink the regulators; a TO-220 regulator can dissipate 4W if it's tied to an appropriate heat sink. This still wastes the same amount of heat, but spreads it out so the temperature is lower.
(2) Waste some in a resistor, some in the regulator. The regulator always needs more than 2V across it to regulate. If you insert a series resistor between the raw DC supply and the regulator, and ensure that at maximum current the resistor drops only enough voltage to leave the regulator its 2V, then you have still wasted the same amount of heat, but you have moved it to the resistor to waste. In this case, with 12V and 300ma to waste, and 2V for the regulator at least, you could waste 10V in a resistor. That resistance is 10V/0.3A = 33.333 ohms. The power it wastes is 10V*0.3A = 3W. The regulator wastes 2V and 0.3A = 0.6W, and the sum is again 3.6W.
(3) Use a resistor divider. Here's how. You need to understand Thevenin equivalent circuits to do it well. You need to supply the regulator with at least 300ma and at least 17Vdc. A Thevenin equivalent circuit is a circuit which at its output "looks like" a voltage source in series with a resistor to the rest of the circuit. For a simple setup with a larger voltage source and a two-resistor divider, the voltage source is the divided voltage from the resistor divider, that is Vin* R2/(R1+R2), and the resistance in series is R1||R2.

There are an infinite number of solutions to this. The single series resistor is one of these, with R1 = the series resistor and R2 = infinity. All the other solutions involve bleeding some power through the resistive divider, and tapping off a little of this to the regulator. If R1 and R2 are quite small, the current through the resistive divider gets huge compared to the current through the regulator (load, in general). However, the power wasted in going through the resistors instead of the regulator gets huge too. All of the solutions to this kind of resistive divider (that is, with R2 < infinity) waste more than the 3.6W of the original.

(4) Get a different transformer. In this case, you want about 30VCT at 500ma.

[WhenBoredomPeaks]

i used this circuit:

It is critical for eliminating hum (along with having enough voltage for the regulators to regulate) that the CT of the transformer go to the junction of the two first filter caps, and nowhere else. This may not be too bad for opamp circuits, but any single ended circuits can have a hummy buzz from this that is otherwise impossible to remove. It's caused by the high currents in that CT lead.

That CT issue is interesting. Now my current noisy supply have two rectifiers like on the bottom of the picture which suits my transformer with it's dual secondaries:



Now my "overpowered" transformer looks like this:



I want to rebuild my supply with a single rectifier, should i connect the two middle wires on my drawing (the middle blue and middle green) for a CT wire or i have to stick to the dual transformer setup?

I added a bigger transformer and it seems like it solved the hum problem. ...
Now the voltage after rectifying is something like 27V into the 15V regulators which made them super hot in 30 seconds.

I don't want to throw away this expensive toroid, can i divide down it's voltage with resistors? Is it better to divide before rectifying or after? I would like to get down the DC voltage around 18V.
I would draw 200-300mA from the supply @ +-15V. Would something like an 5W resistor be enough for the divider?

BTW do i have to connect the earth/ground in the mains connector the the CT/ground in my supply? (my amplifier is connected to that ground so when i connect my modular synth to it with a patch cable the my synth's ground will "meet" with the grounding in my house.

It's generally not a good idea to divide down power supply currents with resistor dividers. It can be done, but not without wasting massive amounts of power. Even so, I'll tell you how to go about it in this case.

A little math that's NOT available from online calculators will show the problem pretty quickly. If you're wanting 300ma at 15V out, and the raw DC power supply is producing Vraw, then the power wasted in the regulator is going to be (Vraw-15)*300ma. For Vraw=27V, that's 12V*300ma = 3.6W.

You have a few options.
(1) Heat sink the regulators; a TO-220 regulator can dissipate 4W if it's tied to an appropriate heat sink. This still wastes the same amount of heat, but spreads it out so the temperature is lower.
(2) Waste some in a resistor, some in the regulator. The regulator always needs more than 2V across it to regulate. If you insert a series resistor between the raw DC supply and the regulator, and ensure that at maximum current the resistor drops only enough voltage to leave the regulator its 2V, then you have still wasted the same amount of heat, but you have moved it to the resistor to waste. In this case, with 12V and 300ma to waste, and 2V for the regulator at least, you could waste 10V in a resistor. That resistance is 10V/0.3A = 33.333 ohms. The power it wastes is 10V*0.3A = 3W. The regulator wastes 2V and 0.3A = 0.6W, and the sum is again 3.6W.
(3) Use a resistor divider. Here's how. You need to understand Thevenin equivalent circuits to do it well. You need to supply the regulator with at least 300ma and at least 17Vdc. A Thevenin equivalent circuit is a circuit which at its output "looks like" a voltage source in series with a resistor to the rest of the circuit. For a simple setup with a larger voltage source and a two-resistor divider, the voltage source is the divided voltage from the resistor divider, that is Vin* R2/(R1+R2), and the resistance in series is R1||R2.

There are an infinite number of solutions to this. The single series resistor is one of these, with R1 = the series resistor and R2 = infinity. All the other solutions involve bleeding some power through the resistive divider, and tapping off a little of this to the regulator. If R1 and R2 are quite small, the current through the resistive divider gets huge compared to the current through the regulator (load, in general). However, the power wasted in going through the resistors instead of the regulator gets huge too. All of the solutions to this kind of resistive divider (that is, with R2 < infinity) waste more than the 3.6W of the original.

(4) Get a different transformer. In this case, you want about 30VCT at 500ma.

Let's go through the options in mixed order:

(4) I wanted to avoid that, that is why i started thinking about dividing down the supply voltage. ~25$ vs. 1-2$ (new transformer vs. some resistors)
(1) It is pretty problematic and/or expensive to get proper heatsinks in my country. And ordering from Ebay or some other place would be more expensive again than some resistors.
(3) That is what i wanted to do, i was able to found out the ratio and that i have to use higher resistance for less wasted current but i have never dealt with such "powerful" things as mains transformers so i had to ask about this before i blow up something. How high is practical? The reasonable sized 5W resistor's upper resistance limit is about 1000ohms in my preferred shops.
Would that be enough? (i think i would end up with something like in the 500-800 ohm region)
(2) Although it uses only one resistor i had to think about this one for a while then i realised that the value of the resistor depends on the current draw of the circuit.
Now sadly that current draw is not constant. Right now my actual current draw is around 110mA for the positive rail and 50mA for the negative rail. (according to the designers of those circuits) I guess that means that i should use different resistors in both of the rails. But if later i want to add some other modules then the increased current draw will result in increased voltage drop. Though i don't think my actual current draw will ever be higher than 200mA on any of the rails i am still unsure about this option.

[R.G.]

That CT issue is interesting. Now my current noisy supply have two rectifiers like on the bottom of the picture which suits my transformer with it's dual secondaries:
...
I want to rebuild my supply with a single rectifier, should i connect the two middle wires on my drawing (the middle blue and middle green) for a CT wire or i have to stick to the dual transformer setup?

Connect the two secondaries in the center and make a center tap. This is how all center taps really are anyway in a sense. Then use a single bridge.

(4) I wanted to avoid that, that is why i started thinking about dividing down the supply voltage. ~25$ vs. 1-2$ (new transformer vs. some resistors)

I figured that was the case, but I wanted to give you all the reasonable options.

(1) It is pretty problematic and/or expensive to get proper heatsinks in my country. And ordering from Ebay or some other place would be more expensive again than some resistors.

The proper heat sink may be as simple as a flat sheet of aluminum, perhaps 25-50mm by 50-75mm. 3-4W is not a huge amount. The to-220 package (if that is the package your regulators are in) can dissipate 2W in still air all by itself. Connecting it with heat sink paste to a piece of aluminum sheet that is several times bigger than the package itself can get you to 3-4W easily enough. If you do this, don't forget heat sink paste.

(3) That is what i wanted to do, i was able to found out the ratio and that i have to use higher resistance for less wasted current but i have never dealt with such "powerful" things as mains transformers so i had to ask about this before i blow up something. How high is practical? The reasonable sized 5W resistor's upper resistance limit is about 1000ohms in my preferred shops.
Would that be enough? (i think i would end up with something like in the 500-800 ohm region)

I really, really recommend you not do this. It can be done, but it's trickier than option 2.

(2) Although it uses only one resistor i had to think about this one for a while then i realised that the value of the resistor depends on the current draw of the circuit.
Now sadly that current draw is not constant. Right now my actual current draw is around 110mA for the positive rail and 50mA for the negative rail. (according to the designers of those circuits) I guess that means that i should use different resistors in both of the rails. But if later i want to add some other modules then the increased current draw will result in increased voltage drop. Though i don't think my actual current draw will ever be higher than 200mA on any of the rails i am still unsure about this option.

It does sound a little confusing, doesn't it?

I had to learn this one early in my career designing power supplies and regulators. The issue is not really whether the current varies or not - the regulator is there to take care of that. All you have to do is figure out the *maximum* current the load will draw, and size the series resistor to let the regulator circuit have just barely enough voltage when the current is that high. At that point, almost all the excess voltage is dropped across the resistor, and so is the excess heat. The regulator has just enough, and stays cool.

When the current goes down, the voltage lost across the resistor drops, and so the voltage across the regulator rises. But the current is dropping so the total power in the regulator plus resistor goes down, and the regulator and resistor stay cool enough.

I very much recommend this option. Here's an example design for 200ma.

Your DC supply is 27V. Your output is 15V. So there is 12V to be dropped across the resistor plus regulator. The regulator needs 2V minimum, even at maximum current. So at the maximum of 200ma, we need a resistor to drop 10V. 10V/0.2A = 50 ohms. The power in this resistor is 10V * 0.2A = 2W, so you need a 3W or 4W resistor for reliability.

The regulator power is 2V*0.2A = 0.4W. It's not particularly warm.

What happens if the current drops to 100ma? The voltage across the resistor drops by half, to 5V. It's power is now 5V*0.1A = 1/2W. The regulator has 7V across it and the same 100ma, and its power is now 0.7W. It's warmer at lower current! But that's still within the free air rating of the TO-220 package if you don't block airflow from it.

The action is that the resistor absorbs more and more of the excess voltage as the load current increases, so it shifts the waste power into the resistor, and away from the regulator.

I'd have to do some math to figure out the exact point where the regulator has the biggest power wasted, but you see the use of this technique, yes?