Power relay from rectifier winding?

[Chuck H]

If i have a 12v relay and the coil is 150 ohms am i correct that it would pull 12.5 mA?

I think you buggered your math. I get 80mA (12/150=.08)

[daz]

I see. What'd i tell ya about my math "skills"? Thanks chuck. So that would be in amps, so .08=800mA?

[Chuck H]

I see. What'd i tell ya about my math "skills"? Thanks chuck. So that would be in amps, so .08=800mA?

Nope. .08 is the same as .080, so, that would be 80 thousandths of an amp (80mA).

Incidentally I too suck at math. My math conceptual skills are supposedly high according to SAT's but I never went far enough in school to learn the language or characters beyond the old +, -, x, ÷ and =

Of course some stuff has rubbed off on me along the way. And I've had good intentions to learn more but have never made it a priority.

[daz]

Nope. .08 is the same as .080, so, that would be 80 thousandths of an amp (80mA).

Thank god for that ! I just looked at a bunch of small relays and thinking .08 would be 800 mA none would trigger with a wall wart id that math were correct ! At least not the typical 100-200 mA warts. This is much better. I figured i must be wrong since i have done this many times in the past with many relays triggered by 100-200 mA warts with good results.

[daz]

Another Q about current draw....if i use a 9v power source for a 6v relay with a 100 ohm coil and i use a voltage divider to bring it the power supply down to 6v, will that cause the coil to draw less ?

If i calc 6v/100ohm=60mA. If i calc 9v/100ohm=90mA. So i guess the Q i'm asking is whether you should calculate it using the actual voltage the PSU provides or the lower voltage after using a voltage divider?

[Chuck H]

You don't want to use a "simple" voltage divider for a circuit that has considerable power demands (that is, voltage@current). Simple voltage dividers are generally for dividing something like signal voltages.

You really DO want the voltage from your source to be close to correct to avoid regulating a large difference because that regulation will generate a lot of heat and undue stress on the winding. With a low voltage, like 5V, you really would need to know specifics like the winding resistance and the unloaded voltage from the winding to know what your actual, unregulated voltage would be with a given load. As it is the winding you propose to use is designed to deliver 5V at it's rated current. So it will actually deliver MORE than that to a lower current. You also get roughly 1.4X the winding voltage and a lower current capacity when using a bridge rectifier. That lowered current capacity isn't going to be a problem for your relay circuit since the winding is enormously over rated for the task. But the voltages are adding up now! EXCEPT... There's also going to be a forward voltage drop imposed by the diodes. For a high voltage circuit this is utterly insignificant, but for a low voltage circuit it can matter. In the end you're likely to be close enough for a five or six volt relay without additional regulation, but I can't swear to that. If you MUST drop a little voltage you can just add a diode to either side of the forward portion of the rectifier to drop another .5V or you can use a simple shunt zener circuit, which is easier to implement than a resistive voltage divider.

[daz]

I think you may be looking at what i was doing before i decided to remove it all from the amp and do it externally. I'm making a project box with the relay and using a 9v/200mA adapter on a 6v relay. Is it really critical/ because i've done it this way in the past tho usually not dropping voltage, tho this is only 9 down to 6 and i'd be running LEDs off it. I drove about 20 miles to get a friggin relay only to find all they has at this huge electronics surplus warehouse were 6v and under or 24 and up. Nothing in between ! Can't win.

[g1]

The allowable range would be on the spec sheet. 9V would probably be ok for most 6V relays.
Here's an example spec sheet:


You should measure it in actual operation to see what the supply is actually putting on the relay.

[daz]

The allowable range would be on the spec sheet. 9V would probably be ok for most 6V relays.
Here's an example spec sheet:
[ATTACH=CONFIG]40458[/ATTACH]

You should measure it in actual operation to see what the supply is actually putting on the relay.

9.65vdc under load. I know chuck said it's not a good idea to drop the voltage but what if i just drop 1v, maybe 1.5 with a voltage divider since it's so close?

[Steve A.]

Here is a discussion on the subject from the old AMPAGE forum (December 2001):

http://www.blueguitar.org/new/text/t...cool-stuff.pdf

Starting on page 10 I explain how I converted the preamp tube filament supply from 6VAC to DC in this 1998 article on the mods I did to my Pignose G40V amp (basically a 220X Marshall preamp driving a pair of 6L6's designed by Dennis Kager of Sundown Amps):

http://www.blueguitar.org/new/articl...s/g40v_mod.pdf

Here is a JPG of the relevant drawings from that article:





Steve Ahola

P.S. Reading the title of this thread I thought it was about using a relay to shut down and power up a guitar amp just like it was done with TV's in the 90's. It would be easy enough to wire up a relay from the filament supply which when de-energized would turn off the amp but powering the amp back up would be trickier...

BTW in the D-clone I built in 1999 based on #124 I used 3 relays powered off the 6.3VAC filament supply using the AC to DC circuit from a Mesa Boogie Mark IIIC as I recall. Without a voltage regulator it was really tricky getting the proper series resistance dropping the voltage to energize one, two or all three relays! I guess using a voltage doubler rectification circuit tapped off the 6.3VAC filament supply to drive a 6VDC voltage regulator would screw up the filament voltage going to the power tubes? I remember having to keep the jack and cable going to my 3 button footswitch isolated from the chassis to avoid a nasty hum or buzz. Ah, the good old days here...

[SoulFetish]

9.65vdc under load. I know chuck said it's not a good idea to drop the voltage but what if i just drop 1v, maybe 1.5 with a voltage divider since it's so close?

A voltage divider here will likely over complicate things. It's inefficient and unnecessary, but the good news is there is an easy trick which solves your problem. (I believe Chuck actually mentioned this a few posts back)...
Simply add two standard rectifier diodes in series back to back on the +9V line. The diodes will provide the 1.5V decrease (around 0.7ish volts for each diode) by their natural forward voltage drop. Just make sure to use diodes that can handle the full current load. It provides a cheap fix which won't add an unnecessary load, or limit the current on your power supply.

[Dave H]

Another Q about current draw....if i use a 9v power source for a 6v relay with a 100 ohm coil and i use a voltage divider to bring it the power supply down to 6v, will that cause the coil to draw less ?

If i calc 6v/100ohm=60mA. If i calc 9v/100ohm=90mA. So i guess the Q i'm asking is whether you should calculate it using the actual voltage the PSU provides or the lower voltage after using a voltage divider?

You don't need a voltage divider. A resistor in series with the relay coil will do it. The relay coil is then the 'bottom' resistor of the divider. Power supply is 9V, relay is 6V so we need to drop 3V across the resistor. You've already calculated the current is 60mA so we need a resistor that will drop 3V @ 60mA. R = V/I = 3/0.06 = 50ohms. Use 47R, 56R or 2 x 100R in parallel. Power is only I x V = 0.06 x 3 = 0.18W in the resistor.

[daz]

A voltage divider here will likely over complicate things. It's inefficient and unnecessary, but the good news is there is an easy trick which solves your problem. (I believe Chuck actually mentioned this a few posts back)...
Simply add two standard rectifier diodes in series back to back on the +9V line. The diodes will provide the 1.5V decrease (around 0.7ish volts for each diode) by their natural forward voltage drop. Just make sure to use diodes that can handle the full current load. It provides a cheap fix which won't add an unnecessary load, or limit the current on your power supply.

Will LED's do? Because one way or another i need a indicator and i'm thinking along the lines of several in series/parallel to drop it and so they will handle it w/o burning out. Is there a possible combo that will work?

[J M Fahey]

Plain Leds in series with the relay, either bypassed or not, are not a good idea because:
a) current that they handle is puny.
b) no matter what, bypassed or not, you lose 1.9V across them, which are substracted from what´s available at the relay.

I suggest you add a simple transistor per relay, chassis housed, not at the FSW floor case, and you can also add a case mounted LED per switch , if you want to get fancy another Led per relay on the chassis or front panel, how´s that?

FWIW this fancy new Transistor invention has been around for some 60 years now

[SoulFetish]

Will LED's do? Because one way or another i need a indicator and i'm thinking along the lines of several in series/parallel to drop it and so they will handle it w/o burning out. Is there a possible combo that will work?

No. In fact, it looks like Dave has been following along more closely. If in fact you need to drop 3V for the 6V relay when energized, just follow his instructions he posted. Its a better solution than the diodes in this application (I was being lazy giving you an "in general" solution)
If you need LED indicators then do what Juan said and use transistors. I'm sure he'd be more than happy to walk you through the entire design process.