Controlling regulator heat via sinking/shielding

[hollowman]

The image below shows a dual 5-VDC "mini" PSU circuit that uses LM340T-5 regs. The PSU was "designed" for a minimal-footprint layout (compactness) and, unfortunately, I did not take into account heat dissipation. One of the regs is used to power a 5V / 200mA digital decoder IC -- that's 1W of power draw. Additionally, my transformer outputs about 11.7VAC with load (I don't have anything smaller; the post-rectifier voltage, i.e., at the input of the reg., is 9.2VDC), so regs have to dissipate the "extra" voltage. Any hints on transformer sizing for this reg. is appreciated. My input AC is 116-117VAC.

Not sure if you can discern in the image, but I'm using two "stacked" heat sinks per regulator. Even then, the reg that powers the 200ma IC gets too hot to touch after about 15min. I don't have much room to stack more sinks (mostly because the sinks are pre-sized). I could rig some ad hoc sinks out of Al foil, Cu foil or other material -- any suggestions?

The main worry -- thx to thermodynamics! -- is that the nearby electro caps are "absorbing" some of this heat. I.e., they feel warm after the PSU has been running for about 20-30min, especially the aluminum part on top. Is there some way to "blanket" or insulate these caps?

This is just a proto-board ckt and the final version will be on PCB with better layout. But that won't be a reality for another two mos., so I need an interim solution. I don't want to re-layout on the proto-board because it does not have much room as is.

Thx!

[00940]

Originally posted by hollowman
Additionally, my transformer outputs about 11.7VAC with load (I don't have anything smaller; the post-rectifier voltage, i.e., at the input of the reg., is 9.2VDC), so regs have to dissipate the "extra" voltage.

(11.7 X 1.414) - 0.7 = 15.8VDC

You're certain you only have 9.2VDC at the input of the reg ? Because you cannot have at the same time 11.7VAC loaded at the output of the xformer and 9.2VDC at the input of the reg.

Or are you using a CRC filter ?

[hollowman]

Originally posted by 00940


(11.7 X 1.414) - 0.7 = 15.8VDC

You're certain you only have 9.2VDC at the input of the reg ? Because you cannot have at the same time 11.7VAC loaded at the output of the xformer and 9.2VDC at the input of the reg.

Or are you using a CRC filter ?

I'm using an Amveco 62034 xformer. It's wired for 230VAC config. -- for a native 115VAC line voltage -- so I can use the whole xformer (parallel) for 9VAC output (I do this 'cause I need all this xformer's current capacity).

Yes: 11.7VAC is the loaded voltage I measure on the secondary xformer outputs. The input of the reg is bypassed w/a 0.1uF film cap. I also have a std. 20mA LED at the input of reg. (parallel to the bypass cap), so I assume that drops 2-3VDC.

[DigitalJunkie]

I'm using an Amveco 62034 xformer. It's wired for 230VAC config. -- for a native 115VAC line voltage -- so I can use the whole xformer (parallel) for 9VAC output (I do this 'cause I need all this xformer's current capacity).

Got a schematic?

[jackinnj]

If you have a good idea of the amount of heat you want to burn off use a resistor ahead of the regulator -- less work (and less heat) for the regulator to deal with.

You can also use a smaller regulator with an outboard pass transistor -- if you are drawing a lot of current you might want to use one of the Darlington's.

[hollowman]

Originally posted by DigitalJunkie
Got a schematic?

No schematic -- nothing special here, folks
Pretty std. 4-diode rect. w/1000uF main cap + 0.1uF bypass. (The trans leads have 100R+0.1uF snubber, too.)

After the rect., see schema below (but disregard the 12VDC input).

[hollowman]

Originally posted by jackinnj
If you have a good idea of the amount of heat you want to burn off use a resistor ahead of the regulator -- less work (and less heat) for the regulator to deal with.

I don't have a good idea Given that my application requires 5VDC and 200mA, what do you suggest?
Thx!

[jackinnj]

I believe that the LM340T has dropout voltage of 2V, if you assume a line voltage swing of +/-15% the minimum you should be getting from the 9V winding is about 11.7 * 85% = 9.9V (probably during a brownout or something like that). So you need to drop a bit less than 2.9 V. Let's use 2V and allow the reg to do some work. 2V/0.2A = 10 ohms, at least 1 Watt (and preferably 2).

The regulator itself is a little piggy, drawing 6mA quiescent current.

You don't have to be overly fussy with the capacitors on the output side of these early regulators, it's the low dropout variety where the capacitor is an important part of the compensation.

[AndrewT]

5V regulated supply needs 6Vac to 7Vac transformer.

For regulated 9Vdc to 18Vdc use a tranformer with the same AC voltage as the final regulated DC voltage
From 20Vdc and upwards the AC voltage can be slightly lower than the required regulated DC voltage. 25Vac for a 26Vdc to 27Vdc supply.

[hollowman]

Originally posted by jackinnj
I believe that the LM340T has dropout voltage of 2V, if you assume a line voltage swing of +/-15% the minimum you should be getting from the 9V winding is about 11.7 * 85% = 9.9V (probably during a brownout or something like that). So you need to drop a bit less than 2.9 V. Let's use 2V and allow the reg to do some work. 2V/0.2A = 10 ohms, at least 1 Watt (and preferably 2).

Thx for that info. Earlier you noted: "If you have a good idea of the amount of heat you want to burn off use a resistor ahead of the regulator ..." By "ahead" do you mean: before the reg and in series with the input, or after the reg?

FYI, the datasheet for the LM340 notes that Vin has to be between 7.5 and 20vdc. I can get away with a 7VAC out xformer w/no issues, but a 5VAC xformer is not kosher.