N00b power supply questions

[SirNickity]

I'm trying to understand transformer-based power supplies as an introduction to DIY circuit design. One thing I'm stuck on is how the center tap works with respect to ground.

For example, my first project will be to convert two bass (guitar) effects pedals to a single rackmount processor. The catch is, one needs 9v, the other 18. So, my first thought is: simple, get a center-tapped 18v transformer and derive 9v from the CT to one leg. BUT, then doesn't that become an issue when the output of one is fed into the other? Potentially, this means the signal ground could be 9v higher on one, right?

Now, from what I know about op-amps (which isn't much yet), the signal output is usually bypassed to ground via a capacitor, presumably to shunt any errant DC, but what are the limits here?

Please set me straight here - spare not my feelings if I'm way off. :-)

[gain-wire]

Hi sir Nickity (that's a neat little name!)
obviously you're not as "noob" as you think, because you seem to understand that DC coupling issues can exist. However in my experience with guitar effects (actually any audio device that isn't balanced) I've never seen anything that wasn't coupled, input or output, with a capacitor, polarized or not. And I would take a chance in saying that most electrolytics will be rated at least 10volts. You'd be best to take those pedals apart and see for yourself.

HOWEVER, you need not be concerned with what voltage the power supplies are because they are not reflected on the output. Nobody wants DC on the output of their amplifier/effect/audio device! And especially not on a potentiometer, although I've never fully understood why. It just makes tons of noise.

Go ahead and make your bipolar supply, power up your pedals, and to be sure (just to show you) measure the DC voltage on the output and input of each pedal. You will see that there is no DC present. If there is then that could mean other things (no internal grounding resistor or poor design) which we'll tackle once we get there.

Also: the output signal is never bypassed to GROUND via a capacitor (that would short out your signal) it is COUPLED to the load with the capacitor. That's why they are called coupling caps. But yes it is to isolate any DC from the next device. But this only works IF there is a load connected to it.

cheers!

[SirNickity]

Haha -- thanks!!

Your clarification on terminology led me to research more on coupling. After reading for some hours, I've managed to confuse myself a little more.

On one hand, I'm beginning to see that DC on the output of the signal is fairly common and often to be expected. For example, some posts regarding tube circuits mention a couple hundred volts of DC bias on the output. I see now why presenting an (ideally) infinite impedance via the cap makes more sense than shorting it to ground.

But that only explains the signal's hot (tip) lead.

At the moment, I only have one of the effects pedals I plan to use. The Trace Elliot compressor is a rare breed, running on 18VDC. Pedal #2 will be an EQ, but I haven't picked one yet. Almost certainly it will be 9V.

Assuming I traced the compressor's output correctly, it does indeed run through a 2.2uF / 35V cap. Naturally, it's based around op-amps (and a VCA), which means there's probably a voltage divider somewhere to create +/- 9V. I will assume the EQ will have a similar arrangement. (Ironically, it would be easier to build the circuit to take advantage of my bipolar supply, but since it's already done, it's easier for me to provide asymmetric.)

I don't really know enough to guess whether the signal ground is dervied from the voltage divider's midpoint, or the DC negative point -- with the understanding that the signal output will be biased by 1/2 Vin (which the coupling cap will block.)

Either way, won't the signal ground be at different potentials, since the 0V reference is different between the two pedals? In fact, isn't that possible with "typical" power supplies, like e.g., a 9V battery in one pedal, and an amp running off 120VAC? Once the interconnect between them is established, then what?

I really appreciate the help... I don't like building something I don't understand. :-)

[Spiny]

A 9-0-9 transformer can be treated as a 0-9-18 transformer

the attached circuit shows how two voltages with a common ground can be provided (this is only one way) The only thing to note is the total current drawn must not exceed the rating on one half of the transformer.

For example using a 9-0-9 with 18VA is 1 amp current (9va) in each half, if you draw 0.75 amp on the 9v tap then only and extra 0.25 amps may be drawn from the 18V tap as this brings the total for the 9 volt section to 1 amp.

This is a bit of an over simplification but you get the idea

[SirNickity]

Ohhh! Duh.. That arrangement hadn't occurred to me yet. Makes perfect sense. Thanks!

As an aside, I've been wondering whether it would be best to use linear regulators - not sure how much difference there will be. The signal levels are low, which means their current draw is minimal (thus easier to filter the ripple from a trans), but any variance will be greatly amplified later in the high gain preamp. I'll read up on the PS rejection attributes of the op-amps used, but I would appreciate insight from anyone here as well.

On that note, I am assuming the 100nF caps are to keep the regulation swings under control (oscillation, right?), and the large caps are to provide low-impedance sources for better transient response, right? As far as I understand, none of them are strictly necessary, but regulation can suffer from the feedback circuit over-correcting when subjected to turbulence in the current draw. Also, I don't have experience enough on transformer output to know for sure, so this may be an obvious question, but even under light loads, an 18V peak-to-peak trans might not drive a 7818, right? A 20V might, depending on the tolerance of the trans and minimum in/out differential on any particular regulator, but a 24V would probably be better... yes?

Finally, might it be better to drive the 7809 via the 7818 so the voltage drop on the '09 is somewhat less drastic? (Assuming total load on the '18 would be less than 1A...) This could also alleviate imbalanced loads on the taps, and reduce the need for a second bridge. Or does cascading regulators bring its own challenges?

Oh, and while this solves the connection between the two devices, I'm still not clear on the theory behind signal grounds from post #3. If someone has the time to mentor a newcomer, I would be quite grateful. :-)

Thanks everyone, you guys have a saint's patience. Hopefully I can pass on the favor some day...

[star882]

Originally Posted by Spiny

A 9-0-9 transformer can be treated as a 0-9-18 transformer

the attached circuit shows how two voltages with a common ground can be provided (this is only one way) The only thing to note is the total current drawn must not exceed the rating on one half of the transformer.

For example using a 9-0-9 with 18VA is 1 amp current (9va) in each half, if you draw 0.75 amp on the 9v tap then only and extra 0.25 amps may be drawn from the 18V tap as this brings the total for the 9 volt section to 1 amp.

This is a bit of an over simplification but you get the idea

Remove the second bridge and just connect the input of the 9v regulator to the center tap.

[Spiny]

You could use a pair of 7809's of the 18volt supply. 18V rms is plenty - the peak is 25v, for a moderate current draw 22+ volts will be available, this could cook the 9v reg dropping 13+ volts, the stacked 9v will only drop 4 v so thats ok. In this configuration calculate the heat dissapated in the 9volt reg. I think a heat sink is needed.

The 100nf (connect close to the leads) is to deal with noise from the regs - they are not silent - the electrolytics also help control low frequency noise as I understand it as well as act as a reservoir.

[SirNickity]

Noted, but let's say... hypothetically... that someone jumped the gun and has a 7818 lying around.

Does this look reasonable?

[Spiny]

Originally Posted by SirNickity

Noted, but let's say... hypothetically... that someone jumped the gun and has a 7818 lying around.

Does this look reasonable?

That will work a treat - just calculate the dissapation in the 9volt reg and remember the 18volt is passing both output currents so check it can handle it.