could "TREAD" LED pts drive a relay?
Jun 2, 2005 at 3:12 PM Thread Starter Post #1 of 15

Secret Squirrel

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I am currently preparing to build the “Revised Millet-Hybrid” and intend to use the “TREAD” power supply at 24 Volts. I would like to use a relay triggered by a momentary contact switch for the “power on/off” switching. My question:

The “TREAD” board includes a provision for a LED to indicate “TREAD” power on. Could these connection points be used to drive the relay switching circuit? I am looking at using a 6 Volt relay and adjust the LED resistor to supply the voltage. The relay circuit would be controlled by a MC4069 IC chip and a 2N4401. Another option would be to tap into the AC voltage at test points 1, 3 on the “TREAD” board.

Thank you
 
Jun 2, 2005 at 5:41 PM Post #2 of 15
Quote:

Originally Posted by Secret Squirrel
Could these connection points be used to drive the relay switching circuit?


I'd have to see your proposed circuit to know for sure, but provisionally I don't see why it wouldn't work. A relay only needs a few tens of mA, so there's no worry of burning up traces or anything like that.

Is there anything specific that leads you to be concerned? If so, what?
 
Jun 2, 2005 at 8:30 PM Post #5 of 15
I'd be tempted to just leave the TREAD powered up all the time, and use the relay to make/break the V+ line to the amp. Much simpler than trying to hack the TREAD circuit to incorporate that one.

EDIT: Oh, and to power the pushbutton activation circuit, I'd just hang a 78L09 off the TREAD's output.
 
Jun 2, 2005 at 9:05 PM Post #6 of 15
I was looking at Make/breaking the V+ of the amp with the relay and powering the soft start circuit by connecting it to the TREAD LED connection points. I would not use the "power on" LED of the TREAD.

I like your idea using the 78L09.
 
Jun 12, 2005 at 5:52 PM Post #7 of 15
I wired up a 78L09 to the output of the TREAD. Without the load of the amp, it powers the relay circuit at 9 volts. The tread output was 18 volts as a test case. The maximum imput of the 78L09 is 30 Volts and I intend to use the TREAD at 24 to 28 volts. When I used the 18 volt input to the 78L09, the chip was hot to touch.
My Qustion: Would an input to the 78L09 at 28 volts be a problem? (Shutdown or stability problems)
 
Jun 12, 2005 at 9:17 PM Post #8 of 15
Quote:

Originally Posted by Secret Squirrel
I wired up a 78L09 to the output of the TREAD. Without the load of the amp, it powers the relay circuit at 9 volts. The tread output was 18 volts as a test case. The maximum imput of the 78L09 is 30 Volts and I intend to use the TREAD at 24 to 28 volts. When I used the 18 volt input to the 78L09, the chip was hot to touch.
My Qustion: Would an input to the 78L09 at 28 volts be a problem? (Shutdown or stability problems)



How much current is going through the relay coil?
You might be able to limit that a bit with a low-Ohm resistor.

How hot is "hot to touch"?
It's typical for a linear regulator to be hot... the remaining question is "how hot". Hot enough to make a sizzling sound or take off finger skin?

If it's really hot, substitute the TO220 casing LM7809 for the 78L09. If the LM7809 is still too hot (which I doubt, since you don't report the 78L09 shutting off (yet), you could even put a minimal heatsink on it.
 
Jun 12, 2005 at 9:29 PM Post #9 of 15
I measured the current used by the relay circuit at 100 mA when it was connected to a 9volt battery. I guess I could try to limit the current. I am not sure to the lower limit of the circuit, but I could try to similate it with my power supply. As far as being hot- It is just hot to touch. It would not burn me at 18 volts. I'm not so sure at 24-28 volts. I will try my power supply and measure with a thermocouple. The data sheet indicates that it is thermally protected, so I should be safe from frying the chip. I'm just looking for a stable solution.
 
Jun 12, 2005 at 9:49 PM Post #10 of 15
calculate the power dissipation: P = (Vin-Vout)*I = (24-9) * 0,1 = 1,5W
thermal resistance is then: Rth = (Tmax - Tambient)/P = (125° - 50°)/1,5W = 50 K/W
most likely that's too much for that little ic (check the datasheet), and things will get worse with increasing output voltage of your TREAD. furthermore the 78L09 is at its limits at 100mA. there are to92-heatsinks, that could help you. but i'd recommend the bigger to220-7809, just as mono. pin it to your metal-case (you'd have to isolate it, if there's ac-ground already) to cool it. it's cheap enough to make further tweaks unnecessary.
 
Jun 13, 2005 at 12:13 AM Post #13 of 15
I am interested in using the circuit Secret Squirrel posted above with a Bulgin switch for a Gilmore Dynamic w/ Dynahi psu. I would like to switch the secondary side of my transformer and use one of the secondaries to power the momentary circuit. Can someone suggest a DPST relay that would work well (something from Mouser would be great)? I don't know anything about relays yet...
 
Jun 14, 2005 at 2:02 AM Post #15 of 15
FWIW, if a power supply is implemented such that it has only a limited adjustment, ie- voltage adjustments stay within a pre-planned boundary, one could simply use a resistor to drive the relay rather than a linear regulator. I'm not proficient in magnetics and changing resistance of coils but a guess at calculations would be to measure the coil resistance and factor for the higher voltage of the power supply.

In other words, if you have a (wild guess) 90 ohm relay coil spec'd for 9V, putting a 90 ohm resistor in series between it and the power source should allow using an 18V supply. Since I haven't done exactly this, I would start out with a conservatively higher resistance and tweak it downward in resistance till an effective Ohm value is determined, such that the relay simply isn't activated rather than burning it out from excess current.

So if one had a relay that worked from 40mA minimum to 90mA nominal current, using resistance that allows >~40mA at lowest anticipated user-set voltage range (in the case of adjustable power supplies), then just a ramping of current up to the ~ 90mA as user adjusts power supply higher, would suffice within the limits of the coil's current tolerance.
 

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