Follow up to a blown, failed attempt at a buck, I've rethought the actual features I want and the requirements for a power supply have gone down. As a result, I think I can use a linear regulator. The supply is a doorbell transformer (8/16/24VAC selectable)
Can someone confirm my math on the right? As for rectification capacitors I have two 470uF. The regular is 500mA @ 3.3VDC.
(An aside, it looks like I could probably run it at 16VAC and get away with it, generating double the heat)
The problem is that you measured the dc with no load. If you plan to draw 1/2 Ampere you will find the source has dropped considerably. As long as it's above the threshold required by the regulator, that will be an advantage because the regulator dissipation will be a lot less.
Right. The regulator dropout is 2V, so as long as I have 5.3VDC I'm good.
No, you’re not good. At 500 mA output, better hope the thermal shutdown works
The power dissipated by the regulator is (Vin - Vout) * Iload which means you’ll have squat for available output current. Ideally, you want about 5.5-6 volts DC input for 3.3 volts out. Even at 8 vac input, for 500 mA out, you’ll be dumping close to 4 watts into the regulator. That’s a lot heat for a to-220. How big is the heat sink?
Edit: sure that transformer is good for an amp output? Most doorbell transformers are rated for very limited duty cycles. It might not like constant current at your desired output. That cap at 470uf is pretty marginal as well, you’ll have about 8-9 volts of ripple at 500 mA.
Right, 5.85 watts at 10VAC.
Can I not use the bottom pad and create a heatsink out of the PCB to get a rise of 3 degrees C per watt? (It will not be sustained 500mA, in fact 50mA will be typical)
If you use a TO-220 package then it's easy to put a heat sink on it. I have a giant box filled with various heat sinks salvaged from old equipment. Some big, some small, some with fins, some without. I also have another box filled with fans from similar sources.
There's two of those 470uF capacitors, so the ripple at full
load should be around 4V: It varies from 11 to 15V with a 13V
average, i.e. a 1W lower dissipation.
How can I size these input capacitors to reduce the ripple?
The regular is indeed TO-220 so I can use a heatsink.
The regulator will eliminate most of the ripple because it senses any ripple as a change in voltage which it then corrects for so the ripple on the capacitors probably won't be a problem. One cheap way to decrease the ripple and lower the dissipation of your regulator at the same time is to put a 5 ohm 2 watt resistor between the bridge terminal 1 and C1 and another between C1 and C2. The resistors will still dissipate some heat but they lower what voltage the regulator has to waste. This forms 2 low pass filters and while these RC circuits aren't as effective as LC circuits, they should work just fine. Just make sure that your input AC voltage give you enough headroom for the regulator. Here is a link with more detailed information. Scan down in the article to the 'Low Pass Filters' section.
https://learnabout-electronics.org/PSU/psu12.php
Right, 5.85 watts at 10VAC.
Can I not use the bottom pad and create a heatsink out of the PCB to get a rise of 3 degrees C per watt? (It will not be sustained 500mA, in fact 50mA will be typical)
3C/Watt is only the thermal impedance between the junction and the case. Thermal resistances add up like resistors in series, so add on the impedance of the thermal interface (should be low if bolted to a flat heatsink with grease or thin silpad) and most importantly the thermal impedance of the heatsink itself. In general PCBs make fairly poor heatsinks, a small 50x50mm double sided FR4 board will be in the 30+ C/W region. You should be able to see that keeping this cool may be a challenge if you have do it in a small space.
I've found a To-220 substitute which is a buck converter itself - CUI VXO7803-500, 3v3 500mA max. C1/C4 from datasheet, C2/C3
Thoughts?
- Trying to provide overvoltage protection with zener D2
- Load resistor R14 is to provide minimum load (datasheet said 0.2mA, went with 2mA)
I realize that your schematic capture and simulation software thinks that the ground symbol is a perfect equipotential node, but in practice you must wire the negative side of the bridge directly to the negative side of the capacitor bank, and from thence wire it to the common side of the regulator, and another from the regulator to the output. There is a lot of 100 or 120 Hz pulsating current through the first of these wires. The Zener D2 should mount next to the regulator as well. If a PCB layout, treat the traces between components as wires, and run the output trace to the ground plane. The negative route should look very like the positive route on your drawing.
I was wondering that actually. Since I do want C2/C3 to be located in a such a manner on the PCB that minimizes heat to them (eg place them near the bottom of the board as it is to be mounted vertically). So in that case, I will run large traces first to these.
Perhaps I can check-in again in a day or two once I've done the PCB layout. I'm also wondering about ground pours and such with 60Hz if it's really needed or not for those first few wires as you mentioned. Typically I just pour ground on top and bottom for simple designs. This one the AC section might be best isolated... or.. maybe not?
That zener diode isn't a good thing. The rated voltage is nominal and with a tolerance. If the voltage rises enough to make it conduct, there is no impedance to limit the current so it could fail. If it fails it usually will short, making the fault current even higher and causing smoke or a fire. Or a burned out source.
There is a fuse on the input side which should blow.
Perhaps but the point of fault conduction isn't well defined. Most zener diodes have a soft knee and so might end up stuck in partial conduction, getting hot.
There are better ways to do this.