Inductance

Thread Starter

Chillum

Joined Nov 13, 2014
546
the people at Mantech hates me for changing the order the whole time. I decided to build 2 seperate models a 12V 100uH and a 36V 150uH using 680uF/63V for C2 on both models. I have 63V package of the 100uF for both models. R1 and R2 remains the same R1 1K 2W and R2 0.1W but I'm probably gonna use 47k Min Lin Pot @ 0.125W instead of the trimmer I currently have, any comments welcome

I misplaced calculations, could somebody please show me how to calculate power through each resistor R1 1K and VR2 47K
 

takao21203

Joined Apr 28, 2012
3,682
you'd see one inductor value wont work without limits for all voltages.

It depends on the difference between input and output, and the current.
Generally, for larger difference, you need a larger inductor.
Which costs more for the same Amperage.

A larger (uH value) inductor has a larger DCR too, which pretty much limits the current, besides the possibility of saturation, or partial saturation (which produces heat and brings down the efficiency).

You'd find for some desired currents, and a given voltage, a certain inductor just wont do it anymore.

If you only need low currents for a higher voltage it can be better to boost up from somewhere with a smaller circuit. Most higher voltages are small current for ordinary electronics.
 

Thread Starter

Chillum

Joined Nov 13, 2014
546
you'd see one inductor value wont work without limits for all voltages.

It depends on the difference between input and output, and the current.
Generally, for larger difference, you need a larger inductor.
Which costs more for the same Amperage.

A larger (uH value) inductor has a larger DCR too, which pretty much limits the current, besides the possibility of saturation, or partial saturation (which produces heat and brings down the efficiency).

You'd find for some desired currents, and a given voltage, a certain inductor just wont do it anymore.

If you only need low currents for a higher voltage it can be better to boost up from somewhere with a smaller circuit. Most higher voltages are small current for ordinary electronics.
reason why the 2 models, I just need to know if mine resistors will handle the wattage, so If you know where to find those calculations (it was on some thread of mine somewhere), it be great help
 

takao21203

Joined Apr 28, 2012
3,682
reason why the 2 models, I just need to know if mine resistors will handle the wattage, so If you know where to find those calculations (it was on some thread of mine somewhere), it be great help
Resistors can handle the wattage for which they are rated, if it is more, they slowly blacken and at some point the laquer coating burns off. If you overload them too much they start to smoke and to glow and thats bad.

Larger ones need cooling / a heatsink. Also larger resistors are often used for high voltage, 1/2W or 1W instead of 1/4 Watt.
 

Thread Starter

Chillum

Joined Nov 13, 2014
546
Resistors can handle the wattage for which they are rated, if it is more, they slowly blacken and at some point the laquer coating burns off. If you overload them too much they start to smoke and to glow and thats bad.

Larger ones need cooling / a heatsink. Also larger resistors are often used for high voltage, 1/2W or 1W instead of 1/4 Watt.
good to know, how do I calculate what the power needs to be, it had something todo with the feedback voltage of 1.23V
EDIT: found it on the last page of my first thread, the one that spawned this step down psu
R1=0.15mW
R2=0.12W
if I'm not mistaken
 
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takao21203

Joined Apr 28, 2012
3,682
they are almost always 1/4 or 1/6w for this IC or small SMD, in the 10K range. At higher voltages you must be careful if you use a trimmer, the track burns out easily. If you see smoke its too late.
 

ronv

Joined Nov 12, 2008
3,768
The regulator will increase or decrease the output voltage to keep the voltage at the feedback pin at 1.25 volts. So knowing this you can use ohms law to see the current in the 1k will be .00125 amps. then IXE=.00156 watts. You can then do the same thing for the adjustable resistor.
 

Thread Starter

Chillum

Joined Nov 13, 2014
546
The regulator will increase or decrease the output voltage to keep the voltage at the feedback pin at 1.25 volts. So knowing this you can use ohms law to see the current in the 1k will be .00125 amps. then IXE=.00156 watts. You can then do the same thing for the adjustable resistor.
R1=0.15mW
R2=0.12W
if I'm not mistaken
Will you please verify my calculation of R2? (show steps if you might) (what is IXE)

P=V^2/R=(36-1.25)^2/47000= 25mw
P=V^2/R=(36-1.25)^2/3000=402mw

Is that correct, cause it's different than my first calculation.
 
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ronv

Joined Nov 12, 2008
3,768
Ahh, I see the confusion. :)
Remember back in post 27 for the regulator to be happy it wants to see 1.25 volts at the FB pin. So the 1K resistor will always have .00125 amps ( I = E/R or 1.25 /1000)
So if you want the output to be 2.5 volts, 2.5 volts will be across R1 and R2. Since we know .00125 amps must flow to get 1.25 volts across R1 we can calculate the total resistance usin R = E/I or 2.5/.00125 or 2000 ohms. So R2 (the pot) will need to be set to 1K so R1 + R2 = 2000 ohms. In this case the power in both is P = I X E (Current times voltage) or as you have it V^2/R or .00156 watts - 1.56 mw.
Using the same formulas if you want 36 volts out the total resistance would need to be 28800 ohms. So 27,800 ohms for the pot. So now you can use P = I^2 X R or either of the other 2 to get the power. Lets use I^2 X R just so we use them all. P (R2) = .00125^2 X 27800 or .0434 watts - 43 mw.
Did I make it better or worse? :D
 

Thread Starter

Chillum

Joined Nov 13, 2014
546
better!

So the other way to calculate it would be:
P=V^2/R=(36-1.25)^2/27800=0.0434375 or 43mW
for finding max power necessary to handle we would make R2 1K?
P=V^2/R=AND HERE LIES THE QUESTION (bold and caps to announce not shout) is V still 36-1.25 or is V the Voltage at 1K then squared then divided by 1K ?

And please, what would I gain using a fast recovery Schottky vs normal Schottky?
 

ronv

Joined Nov 12, 2008
3,768
Since you always need .00125 amps thru the 2 resistor the maximum power is where R2 is highest. so 27,800. P = I^2 X R.
Power for the 1k is the voltage at the 1k. Maximum power for the pot is 36 -1.25.
Any schottky diode will be fine as long as it has a rating of 3 amps or more and over 40 volts or so.
 

Thread Starter

Chillum

Joined Nov 13, 2014
546
SR360 BP
DIODE SKY AXL 60V 3A

but why the option for fast recovery, surely it would have some effect on the operation of the circuit, what is it?

PS for the 12V 100uH model I'll be using the SR360 cause the BAT 85 that I initially used isn't rated for 3A
 

takao21203

Joined Apr 28, 2012
3,682
SR360 BP
DIODE SKY AXL 60V 3A

but why the option for fast recovery, surely it would have some effect on the operation of the circuit, what is it?

PS for the 12V 100uH model I'll be using the SR360 cause the BAT 85 that I initially used isn't rated for 3A
not at low frequency
 

Thread Starter

Chillum

Joined Nov 13, 2014
546
not at low frequency
what? not at low frequency what?

if the load changes at a low frequency it doesn't make a difference, at high frequency (which is what exactly, what frequency?) it does make a difference? did I guess right?
 
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Thread Starter

Chillum

Joined Nov 13, 2014
546
then what should I look for in frequency wise? not above x hz?

and fast recovery diode can do high frequency I guess?
 

takao21203

Joined Apr 28, 2012
3,682
then what should I look for in frequency wise? not above x hz?

and fast recovery diode can do high frequency I guess?
yup. 1n4148 has 4ns recovery, pretty good. Before you wonder again- you can parallel more than one. It could make sense for just 200mA, if you dont have good diodes available.
 
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