Regulated power supply questions

[juan_felt]

Hi everyone!

I have a few questions regarding the following power supply from tonepad: (http://www.tonepad.com/getFileInfo.asp?id=111).

1- I would like to add an LED in each DC output to indicate which one is in use, and also to indicate if there is an overload. (I understand I'd have to use a two color LED). What do I have to do to accomplish this?

2- If I want to use a 7809 or 78L09 voltage regulator instead of the LM317, do I have to change anything, or just swap them, taking care of properly connecting them?

Thank you very much!

[Keeb]

1. I personally don't really see the need for this... But a using the switch (normally used for the battery connection) on the DC jack to connect a LED when a power cable is inserted? I can't answer your overload question, sorry.
2. The LM317 has a higher output current (1.5A) compared to the 7809 (1A). A 7809 might do the job depending on how many outputs you're planning on having and what kind of effects you connect.

[slacker]

For a way to do indicator LEDs have a look at this thread http://www.diystompboxes.com/smfforum/index.php?topic=113132.msg1046296#msg1046296 this gives a LED that lights up when a cable is plugged in. The LEDs will dim or go off if the power supply is overloaded but they won't show which output is causing a problem just that the total current draw is too much.

Why do you want to replace the LM317?

[balkanizeyou]

2. The LM317 has a higher output current (1.5A) compared to the 7809 (1A). A 7809 might do the job depending on how many outputs you're planning on having and what kind of effects you connect.

Also, note that if you're planning on getting close to maximum stated output current, you're going to need a heatsink.

EDIT: ^well, duh, that's stated in the build document so my reply was kinda reduntant.

[R.G.]

1- I would like to add an LED in each DC output to indicate which one is in use, and also to indicate if there is an overload. (I understand I'd have to use a two color LED). What do I have to do to accomplish this?

You'll have to construct some kind of circuit that detects when the voltage is regulation, versus falling out of regulation. There are quick and dirty solutions as simple as watching how bright the "on" LED is, all the way to detecting the output voltage falling below X voltage with some accuracy. Everything else will be various levels of compromise. The regulator itself has no external signal of an overload, just that it limits current flow, and that means the output voltage falls.

2- If I want to use a 7809 or 78L09 voltage regulator instead of the LM317, do I have to change anything, or just swap them, taking care of properly connecting them?

The LM317 is fundamentally different from the 78xx series. It is a floating regulator - there are no ground pins on it, only in, out, and adjust, for adjusting voltage. On its own, it only produces an output voltage of about 2.4V. You must set the output voltage by setting the ratio of at least two resistors.

They are NOT interchangeable.

Datasheets are your friend. Always look at the datasheet for new parts you're not familiar with.

[PRR]

> indicate which one is in use, and also to indicate if there is an overload]

Define "in use". Define "overload".

You might think "in use" is "non zero current". But current is never dead zero. There's always leakage across insulators, even through air. This may be smaller than "normal loads"-- but what about a passive switchbox with an LED. In the off position it draws "no current" (small leakage"..... but it is IN USE in the sense that when you stomp the switch that LED better light up.

Overload: the power supply is self protecting. In general, in practice, it either works or it doesn't. A simple resistor and LED will show that it has output. There is a marginal case above 1 Amp where the regulator will drop its voltage trying to stay not-much-over 1 Amp. However dropping output voltage means more voltage across the regulator, more heat in it. With a "small" heatsink and these circuit values, regulator temperature quickly reaches 151 deg C, and it will shut OFF to protect itself.

But while 1 Amp, 1,000 mA, may not be "overload" for the regulator, individual pedals may be "in overload" at much smaller currents. Say that passive switch but instead of a 4,700 Ohm resistor to the LED your helper puts in a 47 Ohm. Current is 100 times more than expected! Resistor too hot to touch, LED glowing like an incandescent. Meanwhile the regulator is utterly happy at maybe 1/5th of its maximum load.

Use good cables. Label them. Don't abuse them. An LED + resistor on the regulator gives first-order assurance that all is well.

[juan_felt]

Hi!

Thanks to everyone for the replies.

I'm sorry if I didn't explain properly what I meant by "in use".

My idea is to have many regulators, one for each output, and what I wanted to accomplish is to have an LED that indicates when power is being drawn from that particular output. Is this possible without adding anything else than an LED + Resistor?

The "overload" LED is not really important, since the regulator is self-protecting (now I know, thanks! PRR).

Regarding using 7809, or 78L09, I will refer to the datasheets. (Thanks R.G.).

Again, thanks to everyone.

[merlinb]

My idea is to have many regulators, one for each output, and what I wanted to accomplish is to have an LED that indicates when power is being drawn from that particular output. Is this possible without adding anything else than an LED + Resistor?

No, you need a bit more than that. You would need to add a small resistor (perhaps 10 ohms) before the regulator, then use a comparator or opamp to measure the voltage drop across that resistor and light an LED accordingly.

The 'simplest' solution would be to use an LM317L (because it has a built-in 200mA limit) instead of an ordinary LM317, and put a high-power LED directly in series with it. The LED should then light according to how much current is drawn. Bare in mind that high-power LEDs are usually surface mount, though.

[juan_felt]

Thanks again for the replies, I have an idea of what I'll do with the LED. I'll probably just have a "power on" one.

On the other hand, I have a doubt abut the transformer's rating.

I'm planning on using one transformer with different secondaries to obtain isolated outputs. On one of these regulated outputs I want to be able to switch between 9v/18v. My question is how will this affect the Amperes, so I can properly choose the transformer.
Are the secondaries of the transformer rated in VA or in Amperes?

Thanks

[PRR]

> Are the secondaries of the transformer rated in VA or in Amperes?

Either way. Show us the transformer.

[antonis]

I'm planning on using one transformer with different secondaries to obtain isolated outputs. On one of these regulated outputs I want to be able to switch between 9v/18v. My question is how will this affect the Amperes, so I can properly choose the transformer.

Current (Amperes) will be the same because you switch between winding turns and not between wire diameter..

Your transformer should be rated (as Paul said..) either at VA (i.e. 2 X 13.5VA whitch gives 1.5A PER output (9V) or 27VA whitch also gives 1.5A at 18V..) or at Amperes (i.e. 2 X 9V, 1.5A).

P.S.
If your transformer is "trully" isolated (2 indepedent outputs - not center tapped..) you'll have to make double switching (two outer wires to rectifier and two inner wires short-circuit..)

[merlinb]

I'm planning on using one transformer with different secondaries to obtain isolated outputs. On one of these regulated outputs I want to be able to switch between 9v/18v. My question is how will this affect the Amperes, so I can properly choose the transformer.
Are the secondaries of the transformer rated in VA or in Amperes?

Don't forget that however much AC current the transformer is rated for (VA divided by AC volts) you can only draw about half that in DC, owing to power factor losses. In other words, if you want a 1A DC power supply, you need a transformer rated for at least 2A AC.

[juan_felt]

My idea to change the voltage was to change the resistor in the ADJ pin of the regulator. I want to know if I'll still get the same Amperes from the secondary when I set it for 18V.

Thanks for all the kind replies, and sorry for all the questions!!

[juan_felt]

Also, lets say I want to have 4 outputs in one secondary. I can have one diode rectifier and 4 separate regulators (with its filters) or 4 separate rectifiers, each with its regulator. Does this have any electrical benefit, or it makes no difference?

Thanks!

[PRR]

Since all the outputs go to the same audio system ground(?), separate rectifiers is very bad.

Separate regulators is probably pointless unless you expect to exceed the limit of one regulator.

Plans exist on the Web. Have you looked at some/many for ideas and trends?

[juan_felt]

Since all the outputs go to the same audio system ground(?), separate rectifiers is very bad.

Separate regulators is probably pointless unless you expect to exceed the limit of one regulator.

Plans exist on the Web. Have you looked at some/many for ideas and trends?

PRR many thanks for this (and all other) replies.
Why is this very bad?

About the regulators, the reason is if I want to use many 78L09 regulators, so yes, that is the reason.

I have seen some ideas online, I'd like to use the TonePad schematic, with a switch to select different voltages, plus another secondary with the 78L09 regulators.

[merlinb]

I can have one diode rectifier and 4 separate regulators (with its filters) or 4 separate rectifiers, each with its regulator. Does this have any electrical benefit, or it makes no difference?

You can do both, but it would be a waste of rectifiers.

My idea to change the voltage was to change the resistor in the ADJ pin of the regulator. I want to know if I'll still get the same Amperes from the secondary when I set it for 18V.

Let's say you use an 18V transformer rated for 18VA, which is 1 amp AC. You can therefore demand about 0.5 amp DC. It doesn't matter what you set the regulator output voltage to, whether it is 2Vdc or 18Vdc, you can only draw up to 0.5 amp DC. There may be other limitations beyond that, such as regulator overheating, but 0.5 amp DC is your upper limit.

The situation can be different if you actually switch to a different winding or tapping point on the transformer; it becomes application specific.

[antonis]

My idea to change the voltage was to change the resistor in the ADJ pin of the regulator. I want to know if I'll still get the same Amperes from the secondary when I set it for 18V.

My bad..
(I thought that you intended to switch between secondary windings..)

As Merlin said, you can take as much current as transformer in capable to provide, despite of your final voltage...!!
(you allways get the same secondary voltage before your regulator - as long as you don't overcome it's current rating..)

Another matter is possible regulator overheating because at 9Vdc mode it will have to "absorb" about 12V X whatever is your current..
(18V x 1.41 - 1.5V(rect.) - 3V(reg. In-Out))

[juan_felt]

Let's say you use an 18V transformer rated for 18VA, which is 1 amp AC. You can therefore demand about 0.5 amp DC.

Merlinb, thanks for your reply. I didn't know this happened, why is this? Is it always half the current?

[PRR]

> Is it always half the current?

You can always get more DC current if you accept that the voltage will go down and the heat will go up.

A careful analysis of a rectifier capacitor supply gives 1/1.6 to 1/1.8 depending on whether heat or regulation is the limiting factor. For work where you can not custom-wind and "exact optimum" transformer, "half" is a convenient and reasonably conservative estimate.

In small transformers like this, heat is not likely to be the first problem. Instead the voltage will sag. In that case maybe the 9V setting "can" support more current than the 18V.

Stepping back further: why use "one size" AC winding to support either 18V or 9V DC? Isn't that like using 24 inch timber for every stick in a house? We usually save the big lumber for the main beams and use 12" or 8" lumber for smaller loads like floor joists.