mic29502wt

Question

Hi

i don't really know much about this device ,but i would like to use it in a regulated vacuum tube filament supply , output voltage should be able to move from min voltage to about 10vdc ,6.3v @ 3A max load.

I have played with the lm317 and the 5A version to that device and i could never get them to regulated properly and they very easy to damage ,

My luck with SS parts , are a joke maybe it is ,the lack of knowledge in this area , which is stopping me from understanding how these reg work and @ £ 7 a pop i need to be pretty sure

about what resistors and capacitors are needed for the project before i buy a single Ldo

cheers

shabaz · Answer

Hi Denis,

3A is very high, this would be > 18W at 6.3V (P=Voltage x Current). The reason this becomes important is that

regulators dissipate power as heat in order to perform their function.

So if (say) your source was a 12V supply and you're regulating to 6.3V with a load taking 3A, then the power dissipated

as heat by the regulator is approx (12-6.3)*3 = approx 18W again, i.e. the little regulator needs to dissipate 18W which needs

a very large heatsink (without that, it will heat up rapidly and then possibly get damaged. This is regardless of if it is rated for 5A, it will need an external heatsink to achieve that.

You didn't mention if you have a heatsink, or the input voltage, but these are the areas to focus on if they are getting damaged.

The 317 is a very common part, there will be circuit diagrams to show how to add a variable resistor.

The Micrel part has a similar method, where you can replace one of the resistors with a variable one, and the output voltage will be based on the formula you found. But 3A is approaching the limit for what is practical with a linear supply (which is what the LM317 and the Micrel part are) due to heatsink size eventually becoming impractical as you go beyond a few amps (depends on your input voltage as mentioned)

and then other (more complex) circuit topologies are preferred.

jc2048 · Answer

If you are managing to blow up LM137s you need to be very cautious before using the MIC29502 If it were me I'd try and understand why the LM317s were dying before doing anything else(Are you sure you are killing them They have an internal temperature monitor and will shut down and refuse to operate if they get too hot Most LM317s are rated at 1.5A for normal use the current limit will be higher than that but you don't know exactly where it will cut in and you don't want to be operating close under it Given how cheap the LM317s are compared to the MIC29502 couldn't you use a pair of them with each one running 4 filaments(4 x 300mA 1.2A[For the cost of the MIC29502 you could give each filament its own LM317 If you do want to try a MIC29502, here's how it would look The calculation of the resistors works differently to the LM317. The LM317 is a floating regulator and its control loop compares the voltage between the ADJ pin and the output pin with the internal reference to control the pass element (the internal transistor that adjusts the current to the output). The MIC29502 is ground referenced and is comparing the voltage at the ADJ pin relative to GND with the reference voltage. Here's how I got the values on the drawing. I started with the potentiometer. They come in far fewer different values than fixed resistors, typically in a 1,2,5 sequence. You wanted an adjustment range of about 1.5V. A 200 ohm pot seemed like a good start. With 1.5V across it, that would mean a current of 1.5/200 = 7.5mA (the current flows from the output down through R1, VR1, and R2 to ground) which is reasonable. The regulator tries to hold the ADJ pin at 1.24V, so resistor R2 would be 1.24V/7.5mA = 165 ohms. Here we have a choice - 150 and 180 are standard values - and I chose 180 ohms; that compromises your adjustment range a little but I reckoned it would still be acceptable (you can rework the values if it's not). Now the current isn't 7.5mA any longer, instead the regulator will maintain a current of 1.24V / 180ohms = 6.88mA which I'll use for the rest of the calculations. We have two of the values, we just need a value for R1. Let's say we have the pot at the halfway point. The resistance will be 100 ohms. We want that to correspond to 6.3V. The total resistance (R1+VR1+R2) is 6.3V / 6.88mA = 916 ohms. So, R1 = 916 - (180 + 100) = 636 ohms. Nearest value is 620. With the pot set to zero, the output voltage will be 5.5V. With the pot set to 200ohms, the output will be 6.88V. If you want to check that I've got it right, take the output voltages above and the R2 value and plug them into the formula on the datasheet and see what value it comes up with for R1 in each case. If it matches R1 + VR1 then we're there. One additional thing to note is that the part has a minimum current of 10mA and the current I've chosen is less than that. It won't matter once you get your filaments connected, but if you want to test it first before you connect them put a small load on the output (1K will do) otherwise it may not regulate properly. Alternatively you could scale all the resistor values so that the current through the chain is more than the 10mA. Couple of other observations 1/ don't forget the diode drops in the bridge when calculating the transformer voltage you'll need (2 x 0.9V = 1.8V at 3A) 2/ the ripple on 20000uF at 50Hz is about 1.5V at 3A and 1.2V at 2.4A

michaelkellett · Answer

You might do better using a ready built bridge rectifier rather than 4 diodes - I like this style because they are easy to heat sink and attach thick wires - this was the cheapest I found quickly at Farnell CM1508 - MULTICOMP - Bridge Rectifier Diode, Single, 800 V, 15 A, 1.2 V, 4 | Farnell element14 if you want you can get a pcb mounting part even more cheaply. In fact they are so cheap you can easily afford to over specify on current and voltage which will make you design more robust. There is no point in putting a huge capacitor after the regulator - the values in the data sheet will be sufficient - and it cautions you against using a very low ESR part so don't stick a big ceramic in ! 47uF should be OK. The output impedance of the regulator will be of the order of 0.01ohms up to a few hundred Hz (check the graphs on the data sheet) so putting big cap there won't help. (At 100Hz you need 159,000 uF to get down to 0.01R !!). The type of capacitors you've chosen should be OK (did you mean them to be pcb mounted ?). MK

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