mic29502wt

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.

This is pre design

so if you included a pre resistor to drop the voltage at input of the reg to abut 7.3vdc, would that help with the dissipation of the  reg?

regards

yes, but the reg would only be dropping 1vdc , and that would help with dissipation ? or iam i missing something

regards

Does that mean the input VDC is 8.3? You can take the above assumption shabaz

explanation of 12 VDC and replace with yours to see what wattage needs to dissipate. Lower the better.

Clem

Yes,  input VDC is 8.3 (8.3-6.3)*3A = approx. 6W  ,  will this regulate  don't know my self ,and a heat sink help with dissipation?

how do would 1 determine the size of the heat sink for this reg

cheers

Hi Dennis,

The same power equation is used if you have a resistor, in other words the problem just moves to the resistor and then that will need heatsinking.

The calculated 6W will need a heatsink too.

Will you vary the input when you need 10V for the filament supply? If your input supply is adjustable, you don't need to use 8.3V as the input.

You should set it just slightly higher than the dropout voltage. For the Micrel part, the datasheet says 600mV (0.6V) so if you set your input to just a bit higher then

you'll dissipate less energy as heat. So if you have your input at 7.3V then you've halved it to 3W, which can be easily handled by bolting it to the chassis.

This is fairly high. I refer you tp TI's explanation with formulae to help: http://www.ti.com/lit/an/slva462/slva462.pdf.

They used 2.75 watts in their example which resulted in 11.25°C above ambient. If we assume linear (which it is not) you would expect ~ 33 ish to handle in the heat sink. Now the problem is the small area of contact to dissipate heat on IC. Over time this will continuously build up until the IC can't handle w/o heatsink. Blue smoke is the result.

Clem

Hi, I have not read the whole discussion but If you are really considering to blow up something(which is a fun thing to do) you might wanna add Peltier cooler on heatsink or directly on it. Who knows if Peltier fails, it will heat that thing up more and it can be really a fun thing to watch that piece giving beautiful color of smoke.

mudz

No i need 6.3vdc @ 3A shabaz J

Ok this is what i am thinking  6.3 Vac @10A TRANSFORMER  INPUT  TO BRIDGE=  8.883Vd.c and current=6.2Ad.c ok

if this is correct i need a 2.2vdc drop over a resistor at 6.2Ad.c (2.2vdc 6.2Ad.c)R=0.354  Watts=14 i could use 2 resistors to half the wattage less heat after rectification i will filter with 2* 10000uf  to reg .

ok now to program the output voltage

V_out = 1.24(1 + [R1 + VR1] / R2)

so this is the bit that  i can not grasp the above formula

regards

den

I don't understand why there is a resistor, it doesn't gain you anything - you now have two devices that are dissipating the same total amount of power.

Also, it now sounds like you are designing an entire AC power supply, not just connecting your Linear regulator to a filament. This would need a lot more thinking - for a start 2 x 10,000uF is too low since you can't afford to have too much ripple if you're running your linear regulator with a voltage close to the dropout voltage. You might need to double it. I've not checked any of your figures but I also didn't understand where the 6.2A came from.

Anyway if your 14W figure is correct, which it could be, then this needs a big heatsink. May I ask what this is all needed for?

If it is for a valve/tube, the filament might not need a DC supply. I know nothing about them (is this what it is required for?) but as I understand AC is quite common, since the heater/filament just needs to be hot.

If you want a regulated supply, it would be easier (and safer) to get an off-the-shelf one - they even come in chassis mount variants, and are efficient if you use a switched-mode one. 7V ones are available, which could just have a 2W resistor (and maybe a thermistor in series too) connected to the filament. This XP Power oneThis XP Power one probably costs less than the parts to build one (external ones are cheaper still).

Also, where was 3A derived from, is that an estimate or you know for sure this is what is needed? Because that would drive a lot of valve/tube filaments - my only book on the topic says that current requirement is 0.1 to 0.3A for a range of typical valve filaments.

(My apologies if you have already considered all these points. I have no idea if you have, hence the reason I'm asking to confirm that you have considered them).

Regarding the formula, you can choose to have either R1 or R2 fixed, and you choose any value for it. There may be restrictions in the datasheet, e.g. R2 to be between x ohms and y ohms. Once you have picked a value for it, you can see what Vout will be for various values or the other resistor. Pick a variable resistor with that range. For example if you need between 1kohm and 11kohm to adjust between 6V and 10V, then your resistor can be a 10k variable resistor in series with a 1k resistor. This is not dissimilar to the way you must have used the LM317, just the formula may be different but it is the same format.

I think i should clarify the the project at this moment.

1) regulated 6.3vdc

2) should be able to supply current of  2.4A-3A  (0.300*8)=2.4A

3) the lowest low ripple i can achieve from reg

where the 6.2A came fromshabaz Jan 28, 2016 4:28 AM

its the max current the tx can deliver after rectification at  8.883Vd.c i think, correct me if i am wrong

Is this for a valve/tube filament, yes

I think 5.5v to 7v would be a good place for the adjustment range

regards

So you're paralleling up valve filaments.. I'm guessing the adjustable nature is so that it is possible to control the precise current through the filament? If so then an adjustable supply is not much use for this since the same voltage will be applied to all of them and there is no granular control since there could be variations of filament resistance - you may as well use a fixed supply.

Rather that all this complication, better to go for the fixed 7V supply such as the XP Power one (off-the-shelf, safe, efficient, far less heat dissipation issues).

Then stick a 0.5W fixed resistor (and maybe thermistor - your decision) in series with each filament. You therefore get precise control per filament - use a wirewound potentiometer to do any trimming if this is what you really want (I have no idea), before you replace with a fixed resistor - and it is completely regulated, and it protects your filaments. All advantages.

So you're paralleling up valve filaments.. I'm guessing the adjustable nature is so that it is possible to control the precise current through the filament? If so then an adjustable supply is not much use for this since the same voltage will be applied to all of them and there is no granular control since there could be variations of filament resistance - you may as well use a fixed supply.

Rather that all this complication, better to go for the fixed 7V supply such as the XP Power one (off-the-shelf, safe, efficient, far less heat dissipation issues).

Then stick a 0.5W fixed resistor (and maybe thermistor - your decision) in series with each filament. You therefore get precise control per filament - use a wirewound potentiometer to do any trimming if this is what you really want (I have no idea), before you replace with a fixed resistor - and it is completely regulated, and it protects your filaments. All advantages.