Strange power supply solution -- will this work?

eigenvictor

Joined Jul 16, 2014
28
Hello guys,

I am building an AD950 based signal generator, and I need to power a microcontroller (3.3v single rail) and AD9850 (5v) and also an op amp amplifier (5v dual rail). I also added a tube (for warmth) which requires 6.3v at 700mA (I only connect the heater, it is just as a joke, there was an unused hole in the case so I bought a suitable tube on ebay for $2 which will cover the hole and also radiate nice, warm gold-yellowish glow). So I came up with a simple linear power supply for which I rewound an existing transformer as a center tap which outputs about +12v -12v with no load (down to about 9 with full load). Unfortunately I ran out of room on the secondary bobbin so the lower part of the tapped winding on the schematic is a thinner wire only rated at about 200mA. So I cannot use the entire center tapped winding for both low current +-5v and high current 6.3v rails. So I added a separate second rectifier (the upper bridge on the schematic) which is only connected to the upper ("high current") half of the tapped winding. As a consequence, it doesn't have the real "ground" so it is kind of like a virtual ground, (about 9v difference between "+" and "-" nodes on the schematic regulated down to 6.3v). Now, this set-up generates heat unnecessarily so I added a thermistor to sense the temperature. So uC would automatically disconnect power from the tube after like 30 min or so. So I needed a way to interrupt power to the heater. There is no room for a relay but I have a few of those smallish high-current, low Rdson D-PAK mosfets that I am thinking to wire as shown on the schematic (shouldn't need to dissipate more than 50-100mW or so, or I could find an N channel mosfet for a lower Rdson, see second picture). Since I can't use the real ground in the heater circuit, I needed to figure out how to wire everything without causing interference between two different power rails. So I added an optocoupler. When it is not powered, R3 is floating so the mosfet is "pulled up" and not conducting. When 3.3v is fed to the OC from the uC, R3 pulls the gate down to the "virtual ground" and mosfet opens fully. Is there any issues with this set-up? (Never mind this being a little silly, I know, I am more curious from the theoretic perspective if it is possible to get this to work) BR-549 Joined Sep 22, 2013 4,938 I would install the proper transformer first. AlbertHall Joined Jun 4, 2014 11,538 If C4 is only providing current for the tube heater then it certainly doesn't need to be that big and indeed could be removed completely. mvas Joined Jun 19, 2017 538 Hello guys, I am building an AD950 based signal generator, and I need to power a microcontroller (3.3v single rail) and AD9850 (5v) and also an op amp amplifier (5v dual rail). I also added a tube (for warmth) which requires 6.3v at 700mA (I only connect the heater, it is just as a joke, there was an unused hole in the case so I bought a suitable tube on ebay for$2 which will cover the hole and also radiate nice, warm gold-yellowish glow).
So I came up with a simple linear power supply for which I rewound an existing transformer as a center tap which outputs about +12v -12v with no load (down to about 9 with full load). Unfortunately I ran out of room on the secondary bobbin so the lower part of the tapped winding on the schematic is a thinner wire only rated at about 200mA. So I cannot use the entire center tapped winding for both low current +-5v and high current 6.3v rails. So I added a separate second rectifier (the upper bridge on the schematic) which is only connected to the upper ("high current") half of the tapped winding. As a consequence, it doesn't have the real "ground" so it is kind of like a virtual ground, (about 9v difference between "+" and "-" nodes on the schematic regulated down to 6.3v). Now, this set-up generates heat unnecessarily so I added a thermistor to sense the temperature. So uC would automatically disconnect power from the tube after like 30 min or so. So I needed a way to interrupt power to the heater. There is no room for a relay but I have a few of those smallish high-current, low Rdson D-PAK mosfets that I am thinking to wire as shown on the schematic (shouldn't need to dissipate more than 50-100mW or so, or I could find an N channel mosfet for a lower Rdson, see second picture). Since I can't use the real ground in the heater circuit, I needed to figure out how to wire everything without causing interference between two different power rails. So I added an optocoupler. When it is not powered, R3 is floating so the mosfet is "pulled up" and not conducting. When 3.3v is fed to the OC from the uC, R3 pulls the gate down to the "virtual ground" and mosfet opens fully. Is there any issues with this set-up? (Never mind this being a little silly, I know, I am more curious from the theoretic perspective if it is possible to get this to work)
Your transformer with two secondary windings ( L6 & L7 ), but having different ratings, is acceptable.

AnalogKid

Joined Aug 1, 2013
9,408
C4 = 10,000 uF would yield approx. 0.8 Vpp ripple.

ak

eigenvictor

Joined Jul 16, 2014
28
If C4 is only providing current for the tube heater then it certainly doesn't need to be that big and indeed could be removed completely.
C4 = 10,000 uF would yield approx. 0.8 Vpp ripple.

ak
Yeah, I need a hefty capacitor there to avoid too much ripple and in fact 0.8v might be too much since the output is regulated down to about 6.9V by LM317. So I calculated the number of turns on the secondary so that unregulated output from the rectifier under full load would be just a tad above the headroom needed by LM317 (1.5 - 2v) to avoid dissipating too much heat. (I had to use a smallish 10C-15C/W heatsink and limit temperature to about 40C). If the ripple is kept low it can be neglected which makes the calculations easier.

AlbertHall

Joined Jun 4, 2014
11,538
Why does the tube heater current need to be ripple free?