RELAY POWER SUPPLY

[studiodunn]

In the vein of "why wouldn’t I" do these perform the same task at a lower cost and form factor?

https://www.ebay.com/itm/AC-DC-110V-220 ... 2213639211

[bmx]

It should work. I have used various Chinese AC-DC boards to do relays and even heaters. Some of the boards have been junk, and some have been convenient and high quality.

[ViperDoc]

From my experience, relays are pretty tolerant of voltage. A 5 volt relay will work just fine off 9 volts or 12 volts. I'd just get a small cheap 5V supply like this one:

https://www.jameco.com/z/RS-15-5-MEAN-W ... 19043.html

AC to regulated DC in a small package.

I see the maximum input current is listed as 350 mA. Is that too low for a 7A heater winding? How would you reduce the current before going into a board like this? Thanks.

[pompeiisneaks]

From my experience, relays are pretty tolerant of voltage. A 5 volt relay will work just fine off 9 volts or 12 volts. I'd just get a small cheap 5V supply like this one:

https://www.jameco.com/z/RS-15-5-MEAN-W ... 19043.html

AC to regulated DC in a small package.

I see the maximum input current is listed as 350 mA. Is that too low for a 7A heater winding? How would you reduce the current before going into a board like this? Thanks.

With transformers, as you lower voltage, current can go up. I.e. a 240V input at say 350mA can handle dropping to 6.3v about a 38fold increase in current. i.e. about 13A. That's of course in a perfect world, there are losses etc, and they choose how thick to wind the 6.3 volt side to handle the exact current they want.

Hope that makes sense?

If you go UP the current rating goes down, so say 240vAC input to a 480VAC output means you'd only be able to handle 175mA on that side of things.

~Phil

[ViperDoc]

Thanks, Phil. I am interested in that kind of 5V power supply for a relay setup. The spec sheet says it can handle a max input of 350 mA. If the PT heater winding is set to 7A, will that be too much, or do I need an auxillary transformer with a lower current rating?

[pompeiisneaks]

Thanks, Phil. I am interested in that kind of 5V power supply for a relay setup. The spec sheet says it can handle a max input of 350 mA. If the PT heater winding is set to 7A, will that be too much, or do I need an auxillary transformer with a lower current rating?

If you're just runnign relays off it, 7A is massive overkill, BUT i'm guessing that 6.3V rail is for tubes also. You'd need to do the math.

Add up the current rating for all power tubes and preamp tubes and see if that comes up under 7A. You can also look up what current the voltage regulator you'd use needs as well as the per relay current requirements and add them in.

It's simple math, but I don't know exactly what your amp is.

here's an example. Say you have 4xEL34's. they need 1.5A per tube, that's 6A for the power tubes. 3 12AX7's they want 300mA per tube, that's 900mA... you're now at 6.9A and have 100mA to spare. That's probably enough for just adding relays, but you'd want to double check. On top of that, though, some older transformers would easily go well over their current ratings, as everything tended to be built overspec. Nowadays it seems all transformers are wound to be at almost 'fail if 1% over rating' so you'd not want to exceed the current rating in those cases.

In many cases just buying a separate 6.3V transformer, like dumble uses, is a lot easier and super cheap. A 1A 6.3v is often like 20$ if that much.

~Phil

[ViperDoc]

Thanks again, Phil! The auxillary trafo route was my first choice, I've just never done that. So I'll make sure once my design is complete, that the current on the aux is low enough.

[bal704]

I guess I don't understand the question. Anymore I run my relays off a dedicated DC power supply like the one I linked to. I don't use the heater windings for anything related to the relays. As an example, on my last build, I used a 12V LED indicator light on the front panel for power, and 12V LED's for the channel indicator LED's. I just powered the LED"s and relays off one dedicated DC supply. I have my own little PCB's that I use for my channel switching circuit.

The AC input powers this transformer, and you get 5VDC out in the case of this one.

Hope that clears it up.

From my experience, relays are pretty tolerant of voltage. A 5 volt relay will work just fine off 9 volts or 12 volts. I'd just get a small cheap 5V supply like this one:

https://www.jameco.com/z/RS-15-5-MEAN-W ... 19043.html

AC to regulated DC in a small package.

I see the maximum input current is listed as 350 mA. Is that too low for a 7A heater winding? How would you reduce the current before going into a board like this? Thanks.

[ViperDoc]

Thank you, bal704. Simplest answer: I have no idea how to power a relay properly. I'm working off of an SDM Jubilee layout with an antiquated point-to-point power supply using heater current. I'd rather use a dedicated relay transformer and updated power supply, I'm just unsure how to connect it up and spec it so it doesn't cook.

[sluckey]

You have a misconception (or none at all) about a very simple principle. Just because you have a transformer capable of delivering 7A does not mean that it will deliver 7A to anything it is connected to. The load resistance determines the amount of current that will be drawn from the transformer. A big ole EL34 has a low resistance (big load) and will cause 1.5A to flow from the PT. And a little 12AX7 has a higher resistance (smaller load) and will only cause .3A to flow from that same PT. The PT is not gonna try to shove the entire 7A capacity into the load. Light load = high resistance = light current; heavy load = low resistance = heavy current. It's ohm's law. Look it up. Understand it. Never forget it.

Your little power supply will only pull .35A from the PT when it needs to supply it's maximum current to it's load. The 7A PT will happily provide .35A to that light load forever without even breaking a sweat. It ain't gonna shove the entire 7A into that light load. It can't! Why? Ohm's law.

So, all you need for that little power supply is a small/cheap transformer that can supply about .5A. Sure, a 7A transformer will work just as well but it's a waste of money and has a much bigger footprint.

[ViperDoc]

You have a misconception (or none at all) about a very simple principle. Just because you have a transformer capable of delivering 7A does not mean that it will deliver 7A to anything it is connected to. The load resistance determines the amount of current that will be drawn from the transformer. A big ole EL34 has a low resistance (big load) and will cause 1.5A to flow from the PT. And a little 12AX7 has a higher resistance (smaller load) and will only cause .3A to flow from that same PT. The PT is not gonna try to shove the entire 7A capacity into the load. Light load = high resistance = light current; heavy load = low resistance = heavy current. It's ohm's law. Look it up. Understand it. Never forget it.

Your little power supply will only pull .35A from the PT when it needs to supply it's maximum current to it's load. The 7A PT will happily provide .35A to that light load forever without even breaking a sweat. It ain't gonna shove the entire 7A into that light load. It can't! Why? Ohm's law.

So, all you need for that little power supply is a small/cheap transformer that can supply about .5A. Sure, a 7A transformer will work just as well but it's a waste of money and has a much bigger footprint.

Thanks for that clarification, Sluckey. I get V = IR mathematically, but I don't understand "what that means" in terms of electrical behavior/physics, what pushes and what pulls, etc. So a 7A trafo allows up to 7A of current draw, say, but won't blow anything up. That is wonderful news! I'll get it, I just need to keep studying and working through this. As always, many thanks!

[pompeiisneaks]

You have a misconception (or none at all) about a very simple principle. Just because you have a transformer capable of delivering 7A does not mean that it will deliver 7A to anything it is connected to. The load resistance determines the amount of current that will be drawn from the transformer. A big ole EL34 has a low resistance (big load) and will cause 1.5A to flow from the PT. And a little 12AX7 has a higher resistance (smaller load) and will only cause .3A to flow from that same PT. The PT is not gonna try to shove the entire 7A capacity into the load. Light load = high resistance = light current; heavy load = low resistance = heavy current. It's ohm's law. Look it up. Understand it. Never forget it.

Your little power supply will only pull .35A from the PT when it needs to supply it's maximum current to it's load. The 7A PT will happily provide .35A to that light load forever without even breaking a sweat. It ain't gonna shove the entire 7A into that light load. It can't! Why? Ohm's law.

So, all you need for that little power supply is a small/cheap transformer that can supply about .5A. Sure, a 7A transformer will work just as well but it's a waste of money and has a much bigger footprint.

Thanks for that clarification, Sluckey. I get V = IR mathematically, but I don't understand "what that means" in terms of electrical behavior/physics, what pushes and what pulls, etc. So a 7A trafo allows up to 7A of current draw, say, but won't blow anything up. That is wonderful news! I'll get it, I just need to keep studying and working through this. As always, many thanks!

Another way to think of it is that the transformer can handle up to 7A of current being pulled through it, but it will ONLY deliver what's being asked of it... (If 9A is being asked it will still try, but then you get melted transformer windings or blown fuses if you did it right)

It's like hooking a 1" input pipe up to a system with only 1/2" pipe system through a reduction fitting. The 1/2" pipes will only allow 1/2" worth of water to flow, but the 1" could let more, it just won't since the downstream is limited.

~Phil

[bal704]

Most of the circuitry in a heater powered relay setup is the AC/DC rectification and filtering. My PCB looks something like the hoffman board with plated holes rather than turrets.

https://hoffmanamps.com/MyStore/perlsho ... D=!ORDERID!

https://hoffmanamps.com/MyStore/perlsho ... D=!ORDERID!


It has 2 parts on it. The relay, and a diode. That's it. The relay itself is just a DPDT switch. The power and switching is accomplished by basically shorting the hot and ground of the power together across the diode, which causes the relay to open/close.

[martin manning]

Bad link. Try this: https://el34world.com/projects/relay_switch.htm

[ViperDoc]

That looks quite simple the more I look at it. I think I'll try that one first. Thanks!