DC Power box issue

Started by m_charles, August 21, 2023, 06:36:42 PM

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m_charles

Hi there.

Was wondering if anyone had an idea as to why I may be having this problem.

I built the schematic shown from Muziue. Works great, works exceptionally quite in my garage which shares an outlet with a computer charger and an iPhone charger. got rid of the hum that was coming through when using store bought power supplier for pedals. Only one problem. It powers fuzzers, drives, etc just fine. But if I plug anything that needs more mA, it won't work. Used an old analog boss chorus, Boss DD2, Boss DD3, all pedals which don't require extraordinary amounts of power, albeit more than a fuzz though of course.
Double and triple checked wiring, no errors.

Only changes from the schem:


- Used only 7809 IC's (the larger one's, not the smaller 78L09)
- made 6 outputs, each with its own 7809
- Didn't do the variable voltage output, don't need it
- Didn't add the direct 12v out, didn't need it.
- Added a 100ohm in series at the 12v input to provide additional filtering.[/li][/list]


Power supply is 12v DC 500mA
Tested all the higher draw pedals using only 1 of the outputs, no other pedals plugged in, they still won't work.

Could using all 7809 IC's be the issue?
Any advice is appreciated. Thx so much in advance!

-Chuck



PRR

Quote from: m_charles on August 21, 2023, 06:36:42 PM
- Added a 100ohm in series at the 12v input to provide additional filtering.

Hmmmm.
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Rob Strand

#2
Quote
I built the schematic shown from Muziue. Works great, works exceptionally quite in my garage which shares an outlet with a computer charger and an iPhone charger. got rid of the hum that was coming through when using store bought power supplier for pedals. Only one problem. It powers fuzzers, drives, etc just fine. But if I plug anything that needs more mA, it won't work. Used an old analog boss chorus, Boss DD2, Boss DD3, all pedals which don't require extraordinary amounts of power, albeit more than a fuzz though of course.
Double and triple checked wiring, no errors.

QuotePower supply is 12v DC 500mA
Is that what is marked on the label?

Power supplies built for pedals aren't 12V DC they are usually 9V DC. A 9V unregulated supply will measure 12V DC with no load but soon as you put a load on it the voltage drops.

Under load, what voltage measure at the inputs of the regulators?

A second issue is a regulator needs an input voltage 2V to 2.5V higher than the output voltage.  In fact the bottom of the ripple needs to be higher than that.   The input protection diode adds another 0.7V on top of that.  So you need about 12V min, and that means a supply with a 9V DC rating will poop out with little to no load.

If you have a 12VDC unregulated supply, as marked on the label,  it should measure about 16V with no load.   At the rated load current the average voltage could drop down to 11V and the bottom of input ripple with be a volt or more below that.   So your regulator will work but not at the full rating of the 12VDC supply.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

m_charles

Quote from: Rob Strand on August 21, 2023, 09:03:21 PM
Quote
I built the schematic shown from Muziue. Works great, works exceptionally quite in my garage which shares an outlet with a computer charger and an iPhone charger. got rid of the hum that was coming through when using store bought power supplier for pedals. Only one problem. It powers fuzzers, drives, etc just fine. But if I plug anything that needs more mA, it won't work. Used an old analog boss chorus, Boss DD2, Boss DD3, all pedals which don't require extraordinary amounts of power, albeit more than a fuzz though of course.
Double and triple checked wiring, no errors.

QuotePower supply is 12v DC 500mA
Is that what is marked on the label?

Power supplies built for pedals aren't 12V DC they are usually 9V DC. A 9V unregulated supply will measure 12V DC with no load but soon as you put a load on it the voltage drops.

Under load, what voltage measure at the inputs of the regulators?

A second issue is a regulator needs an input voltage 2V to 2.5V higher than the output voltage.  In fact the bottom of the ripple needs to be higher than that.   The input protection diode adds another 0.7V on top of that.  So you need about 12V min, and that means a supply with a 9V DC rating will poop out with little to no load.

If you have a 12VDC unregulated supply, as marked on the label,  it should measure about 16V with no load.   At the rated load current the average voltage could drop down to 11V and the bottom of input ripple with be a volt or more below that.   So your regulator will work but not at the full rating of the 12VDC supply.

Hey Rob.

I actually lucked out a tried bypassing that 1/2 watt 100ohm in series. I noticed while doing it that it was also getting really hot! I guess that was choking the current. I'll need to do some reading to understand why, but once I got rid of it, everything works perfectly...
Just for the heck of it, here's the power-supply info...

12V DC unregulated. Direct from the wall it measures approx 18.5V DC, so a bit more than your estimate.
outputs for each jack measure about 9.1V DC.
at the inputs to the regulator I'm getting approx 14.5v DC

Thanks!

m_charles

Quote from: PRR on August 21, 2023, 08:06:56 PM
Quote from: m_charles on August 21, 2023, 06:36:42 PM
- Added a 100ohm in series at the 12v input to provide additional filtering.

Hmmmm.

hmmmm is correct. once I got rid of it, it was fine, haha

FiveseveN

Quote from: m_charles on August 22, 2023, 02:36:49 AM
I guess that was choking the current. I'll need to do some reading to understand why
A resistor's only job is to resist the flow of current. 14.5 V / 100 Ω = you can only draw 145 mA.
Quote from: R.G. on July 31, 2018, 10:34:30 PMDoes the circuit sound better when oriented to magnetic north under a pyramid?

Rob Strand

#6
QuoteHey Rob.

I actually lucked out a tried bypassing that 1/2 watt 100ohm in series. I noticed while doing it that it was also getting really hot! I guess that was choking the current. I'll need to do some reading to understand why, but once I got rid of it, everything works perfectly...
Just for the heck of it, here's the power-supply info...

12V DC unregulated. Direct from the wall it measures approx 18.5V DC, so a bit more than your estimate.
outputs for each jack measure about 9.1V DC.
at the inputs to the regulator I'm getting approx 14.5v DC

Thanks!
Well it's good you sorted it out.   I wasn't sure if you had all the regulators after the 100 ohm or only on the 12V line.

The unloaded voltage varies quite a bit from wall-wart to wall-wart.  I have some 12V power supplies which are only 13.5V to 13.7Vno load and they drop somewhat below 12V under full load.   Back in the old days when these things replaced batteries the wall-wart manufacturers didn't want to fry a 12V battery driven devices with 16V to 18V.   You will also find the voltage drops quite a lot with a 10mA load (maybe 1.5V to 2V) then the voltage drops more slowly for each 10mA increase after that.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

PRR

500ma times 100 Ohms is 50 Volts.

Since you do not have 50V to spare, disappointment happens.

You have not said how much current your "more mA" pedals were trying to suck. Or what actually voltages were happening. And "doesn't work" is rarely a useful observation. It is good to know these things.
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m_charles

Quote from: PRR on August 22, 2023, 02:47:29 PM
500ma times 100 Ohms is 50 Volts.

Since you do not have 50V to spare, disappointment happens.

You have not said how much current your "more mA" pedals were trying to suck. Or what actually voltages were happening. And "doesn't work" is rarely a useful observation. It is good to know these things.

Perhaps you missed the replies above. I said I solved it. By more mA draw, it was anything above 20mA or so.

m_charles

Quote from: Rob Strand on August 22, 2023, 03:54:53 AM
QuoteHey Rob.

I actually lucked out a tried bypassing that 1/2 watt 100ohm in series. I noticed while doing it that it was also getting really hot! I guess that was choking the current. I'll need to do some reading to understand why, but once I got rid of it, everything works perfectly...
Just for the heck of it, here's the power-supply info...

12V DC unregulated. Direct from the wall it measures approx 18.5V DC, so a bit more than your estimate.
outputs for each jack measure about 9.1V DC.
at the inputs to the regulator I'm getting approx 14.5v DC

Thanks!
Well it's good you sorted it out.   I wasn't sure if you had all the regulators after the 100 ohm or only on the 12V line.

The unloaded voltage varies quite a bit from wall-wart to wall-wart.  I have some 12V power supplies which are only 13.5V to 13.7Vno load and they drop somewhat below 12V under full load.   Back in the old days when these things replaced batteries the wall-wart manufacturers didn't want to fry a 12V battery driven devices with 16V to 18V.   You will also find the voltage drops quite a lot with a 10mA load (maybe 1.5V to 2V) then the voltage drops more slowly for each 10mA increase after that.

Thx man. One interesting thing is the supply is one I found in a drawer, prob 25 years old and I have no idea what it was originally used for. I tried some newer supplies with enough volts and current, but they were leaking line noise into the signal. The old larger size guy did not. This may have been used originally for some piece of gear I had, so maybe it was better filtered t begin with, who knows...

Rob Strand

QuoteThx man. One interesting thing is the supply is one I found in a drawer, prob 25 years old and I have no idea what it was originally used for. I tried some newer supplies with enough volts and current, but they were leaking line noise into the signal. The old larger size guy did not. This may have been used originally for some piece of gear I had, so maybe it was better filtered t begin with, who knows...
You can measure the ripple with different loads and deduce the capacitor inside or even measure the cap directly from the output terminals.   A typical capacitance is 1000uF but you might see 470uF to 2200uF on some units depending on current and voltage ratings.

The cap ESR is a good thing to check on the output terminals.  Sometimes the ripple can be high but the capacitance measures OK.   The long wall-wart leads will increase the ESR measurement.

The internal electrolytic capacitors on these things can dry-up like any electrolytic caps.

If you put a 1000uF or 2200uF cap inside your power supply box it can help cope with crapped-out caps in the wall-wart.   It's not a bad idea anyway because it reduces ripple and prevents the regulators from dropping out.

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

amptramp

I have one wall wart where the top cap came off and inside, there was a fullwave rectifier followed by a capacitor with a 1 ohm resistor in series with the capacitor!  ESR was going to be more than one ohm.  I presume this was to reduce inrush current through the diodes, but it wasn't all that good for filtering.

Rob Strand

Quote from: amptramp on August 23, 2023, 09:04:31 AM
I have one wall wart where the top cap came off and inside, there was a fullwave rectifier followed by a capacitor with a 1 ohm resistor in series with the capacitor!  ESR was going to be more than one ohm.  I presume this was to reduce inrush current through the diodes, but it wasn't all that good for filtering.

I've never seen that.  There's some crazy stuff out there.

The full-wave configuration is very common.

Those small transformers have quite a bit of series resistance so the resistor isn't a big help to inrush -  it might help a low voltage unit like 3V or 4.5V.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

amptramp

Quote from: Rob Strand on August 23, 2023, 08:34:32 PM
Quote from: amptramp on August 23, 2023, 09:04:31 AM
I have one wall wart where the top cap came off and inside, there was a fullwave rectifier followed by a capacitor with a 1 ohm resistor in series with the capacitor!  ESR was going to be more than one ohm.  I presume this was to reduce inrush current through the diodes, but it wasn't all that good for filtering.

I've never seen that.  There's some crazy stuff out there.

The full-wave configuration is very common.

Those small transformers have quite a bit of series resistance so the resistor isn't a big help to inrush -  it might help a low voltage unit like 3V or 4.5V.

This one was a 6 volt unit.