I want to tap the heater secondaries for a voltage doubler. (to power relays)
Problem is that the heater secondaries use a CT or a simulated CT with the two 100ohm resistors.
Since the doubler does not want a CT used, how to I deal with this?
Problem is that the heater secondaries use a CT or a simulated CT with the two 100ohm resistors.
Since the doubler does not want a CT used, how to I deal with this?
You could forego the hum reduction, and not use the CT connection.
Or just use an extra small transformer and power supply for the relay coils.
Or use relays with coils suitable for your local AC line voltage.
Or just use an extra small transformer and power supply for the relay coils.
Or use relays with coils suitable for your local AC line voltage.
If the voltage doubler and relay circuit can be isolated from ground, the existing heater circuit can be used as-is. If you need to interface with grounded circuitry, it would be better to use a separate winding or transformer.
Can you provide a schematic of the "CT" or "Simulated CT"? Including voltages and desired voltages. And the circuit you want to use to switch the relays. If you can keep the complete relay circuit separated from ground you might be able to achieve what you want. Alternatively you could replace the resistors by a wirewound potentiometer and adjust for minimum hum.
I am actually going to wind a couple of 6.3V power transformers for a project where I am going to replace the input transformers in my mid-range and tweeter amps with discrete active balanced inputs. Not sure how this is going to turn out but I only need 50mA at +/-15V.
I bought some cores (laminated) and bobbins on eBay, have no idea whether it will work until I try it.
Make sure the DC is well enough filtered otherwise the relays will buzz and that can be very distracting.
I bought some cores (laminated) and bobbins on eBay, have no idea whether it will work until I try it.
Make sure the DC is well enough filtered otherwise the relays will buzz and that can be very distracting.
I have customs made toroidal PTs with both 6.3 and 12.6 secondaries I use with a 5v regulator.
But I have seen items like the Mojotone relay regulator that is designed to use the 6.3 secondaries and want to figure out
how it works. (no comment from them yet)
I dont see how they are getting 5v without a doubler.
I guess the 5v relay can operate lower but barley.
But I have seen items like the Mojotone relay regulator that is designed to use the 6.3 secondaries and want to figure out
how it works. (no comment from them yet)
I dont see how they are getting 5v without a doubler.
I guess the 5v relay can operate lower but barley.
No way to tell without trying it.
Is there a reason you cannot use relays with coil voltage 120/240 VAC?
got this answer, can anyone explain it?
both sides of the 6.3 (3.15) are rectified separately and there are bridge rectifiers that combines both sides into ~6V (including loss) then it is into the first filter, into the regulator, then the last filter before the output.
both sides of the 6.3 (3.15) are rectified separately and there are bridge rectifiers that combines both sides into ~6V (including loss) then it is into the first filter, into the regulator, then the last filter before the output.
How is there regulator getting 5v without a doubler
https://www.mojotone.com/5V-Relay-P...V96-5VEGgQctAE8LG02BUv8TiaphOd4NiwTh0tXS98cTF
https://www.mojotone.com/5V-Relay-P...V96-5VEGgQctAE8LG02BUv8TiaphOd4NiwTh0tXS98cTF
6.3V rectified gives ~7.5V peak, ~8V with Schottky diodes. If ripple is low enough, a (low dropout - not 7808) regulator can deliver clean 5V from that. A doubler could supply a 12v regulator. Either could drop out at ripple troughs at turn-on, when tubes are drawing startup current, but doesn't sound like you'd be using the relays then. Not that a little ripple would be a problem for a relay coil. The Mojotone PCB may not be able to deliver regulated 5V (dropping out on ripple troughs) - again, not that it matters.
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but 6.3v is 315 0 315 so isn't it 3.15 turned into 4.4v?
What is the circuit that allows you to rectify the full 6.3?
What is the circuit that allows you to rectify the full 6.3?
Appears to be FW bridge rectifier (no connection to CT).
Note there are only 2 terminals on the input connector.
Followed by 3 pin regulator.
Note there are only 2 terminals on the input connector.
Followed by 3 pin regulator.
If the winding is center-tapped, the PCB delivers +4V to the positive side of the filter cap, -4v to the negative side (measured with respect to ground), so 8V across the cap. Measured (to ground) at the output terminal, there will be about +1V at the positive terminal, -4V at the negative (regulated 5V across the two). If you were planning to have your foot switch connect each relay to ground, it won't work - return has to be to the -4V negative of the PCB not ground. You can make this work if you isolate the relay circuit from ground - use a 1/4" plastic cased jack so that the sleeve terminal isn't grounded (instead it connects to the minus terminal of the 5V supply). Can use two foot switches with a 3-conductor plug/jack. Think of this supply like a battery with plus and minus terminals - a flashlight works fine without a ground, doesn't it?
True, something we've been dancing around without explicitly saying it - it could only be connected to ground if the 6.3V winding is disconnected from ground, but that will likely cause hum. Keep the output isolated as explained above, and all will be fine. if the 6.3V winding and the PCB output are both connected to ground, expect smoke. May happen occasionally, since output (-) is marked "ground" and someone may connect it there - dont!
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