60v power regulator for MCU required

[johnny78]

hi Guys
i need to power an MCU on 60v supply
unfortunately the HV version on lm317 ot 2596 is not available
a friend suggested to connect 2 lm2596 on series for 60v input & use one output
But it didnt work
as a voltage devider to read the 60v dc on analog pin i have used 100k as R1 & 9,09k as R2
using voltage devider calculator & it works But do you have any ideas if i need to use another values?

any suggestion for 60v regulator?

thanks in advance
Johnny

 

[johnny78]

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Order today, ships today. SCWN06C-05 – Isolated Module DC DC Converter 1 Output 5V 1A 36V - 72V Input from MEAN WELL USA Inc.. Pricing and Availability on millions of electronic components from Digi-Key Electronics.

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Will 5V 1A work?

more than enough But ordering parts is not a solution for me
here cant get anything actually

i need to fo this with common parts old regulator circuits can help
actually lm317 was a perfect solution for all tests but the 60v was not in mind when i started to work on this
& the HV version of it isn't available here

thanks

 

[KlausST]

Hi,

i have posted here what happened exactly

No, you did not. Sorry for my honest words.

You miss to answer questions, you missed to tell how you tested, you missed to show an almost complete schematic..and so on.
This is far from "exactly".

It seems every single poster spent more time in schematics, maths, simulations, and a lot of guessing ... than you did.
At least you don't show any effort ... even in communication.

This is a rather simple task and should be solved within 6 posts or so. Now we are over 40.
Every single poster showed an appropriate solution - with the given informations.

Drawing a sketch, using a calculator, taking some photos has nothing to do with language barrier.

Klaus

 

[danadakk]

What is your max load current needed for the design ?

What is goal for regulated MCU V, 5V +/- 5% ?


Regards, Dana.

 

[BradtheRad]

hi Brad
i have tested this circuit on 12v & used 4.7v Zener diode with 470ohm so i get 4.9v which its what i need
but when i have supplied 60v for test the resistor & transistor got burnt
im using bd139 npn transistor which in datasheet the emmiter collector voltage is 80v

Evidently your load is greater than 10mA. BD139 is TO220 package able to carry a couple Amperes.

So let's say your load is 100mA. To provide 5V the NPN operates as a resistance, dropping 55 V (or 5.5 Watts of heat). Consider sharing partial burden through a plain resistor placed inline with the supply. Ohm value is not crucial. The NPN regulator is okay dissipating 1 or 2W. Hence the plain resistor can have a 5W rating, dissipating 2 to 4W.

It's easier than cascading two voltage regulators, the first down to 30v, the second down to 5V.

 

[danadakk]

Brads design, power for key elements (2 values Rload to produce 50 & 100 mA load) :

You will need a heatsink on Q1 and appropriate wattage ratings on R's (with some margin).

https://www.ti.com/lit/an/slva462/slva462.pdf?ts=1681787349470&ref_url=https%253A%252F%252Fwww.google.com%252F

https://www.infineon.com/dgdl/smdpack.pdf?fileId=db3a304330f6860601311905ea1d4599

Regards, Dana.

 

[johnny78]

What is your max load current needed for the design ?

What is goal for regulated MCU V, 5V +/- 5% ?


Regards, Dana.

hi
i need 4.9v 0.5A max not more & +/- 5% is ok

thanks

--- Updated Apr 18, 2023 ---

Hi,

No, you did not. Sorry for my honest words.

You miss to answer questions, you missed to tell how you tested, you missed to show an almost complete schematic..and so on.
This is far from "exactly".

It seems every single poster spent more time in schematics, maths, simulations, and a lot of guessing ... than you did.
At least you don't show any effort ... even in communication.

This is a rather simple task and should be solved within 6 posts or so. Now we are over 40.
Every single poster showed an appropriate solution - with the given informations.

Drawing a sketch, using a calculator, taking some photos has nothing to do with language barrier.

Klaus

hi
in post 29 i have explained how did i supplied the npn regulator circuit on 12v & it works but on 60v everything got burnt
& what schematic should i show here ?
i was talking about 3 zener diodes connected on series & the NPN regulator circuit is taken from post 23
then i had problems when i supplied 60v to the NPN regulator

in post 32 FVM has gave me some explanation on why the lm2596 module got burnt when testing on zener diodes

anyway all the information i got here is useful
thank you all

 

[D.A.(Tony)Stewart]

hi
i need 4.9v 0.5A max not more & +/- 5% is ok

thanks

This rules out all linear suggestions.

Why did it take 46 posts to find out you need 0.5A? :{

 

[johnny78]

Evidently your load is greater than 10mA. BD139 is TO220 package able to carry a couple Amperes.

is it right as i've checked the datasheet the dc139 can handle up to 80v?

So let's say your load is 100mA. To provide 5V the NPN operates as a resistance, dropping 55 V (or 5.5 Watts of heat). Consider sharing partial burden through a plain resistor placed inline with the supply. Ohm value is not crucial. The NPN regulator is okay dissipating 1 or 2W. Hence the plain resistor can have a 5W rating, dissipating 2 to 4W.

It's easier than cascading two voltage regulators, the first down to 30v, the second down to 5V.

what voltage regulator i can use to step down to 30v ?

View attachment 182335

 

[D.A.(Tony)Stewart]

is it right as i've checked the datasheet the dc139 can handle up to 80v?

what voltage regulator i can use to step down to 30v ?

No linear solution will work for you.

 

[KlausST]

hi
in post 29 i have explained how did i supplied the npn regulator circuit on 12v & it works but on 60v everything got burnt
& what schematic should i show here ?
i was talking about 3 zener diodes connected on series & the NPN regulator circuit is taken from post 23
then i had problems when i supplied 60v to the NPN regulator

Either I am confused or you. So didn´t you test the LM2595 with my zeners? Now you talk about NPN..

Until Post29 we did not know that you used a module at all, and we had no schematic about the module. How should we know about capacitors and all the schematic around LM2595? You even didn´t show how you connected the module, the supply, the zeners... showing 3 zeners in a row was all your effort.

Why did it take 46 posts to find out you need 0.5A? :{

That´s exactly the point.Since post#5 we ask for it!

from post#11:

Additionally I'd not use a linear supply to step down 60V to 5V with load currents above 10mA or so.

But you still tried, although you expect it to work up to 50 times the current!

Now at post#50 I guess it´s time to leave.

Good luck.

 

[danadakk]

Juggling this I see this as one half decent approach :

Two curves shown, load of 500 mA and 100 mA.

Zener needs to be heatsunk and ~ 10W. Note device shown is inadequate for current, pick a bigger one with
margin.
R2 should be 2W for margin.
Q1 needs to be heatsunk. Earlier post shows you how to figure out that one, its needed thermals.
Same process applies for heatsinking the Zener.

Solution is a terrible waste of energy/power, this design a strong candidate for a switcher.
Also it has no short circuit or thermal protection like a Zener plus a LM317 solution would
have, or better yet a switcher. And T performance uncertain, that you will have to investigate
from Vz as a f(Temperature) and the inherent Vbe drift of a bipolar.

Regards, Dana.

--- Updated Apr 18, 2023 ---

Below is the LTspice sim of a circuit that uses two LM317 to give 5V output from a 60V supply.
The first is set to give 30V out so the voltage is never more than 30V across each LM317 even if the output is shorted.
The LM317's will need to be on heat-sinks, their size depending upon the output load current.
The nominal dissipation will be 30V * load-current for U2 and 25V * load-current for U1.

View attachment 181899

This approach has a potential hazard at starup of exceeding the first LM317 input V.
Due to C1 at turn on with no charge on it.

That should be addressed.


Regards, Dana.

 

[johnny78]

Either I am confused or you. So didn´t you test the LM2595 with my zeners? Now you talk about NPN..

Until Post29 we did not know that you used a module at all, and we had no schematic about the module. How should we know about capacitors and all the schematic around LM2595? You even didn´t show how you connected the module, the supply, the zeners... showing 3 zeners in a row was all your effort.


That´s exactly the point.Since post#5 we ask for it!

from post#11:

But you still tried, although you expect it to work up to 50 times the current!

Now at post#50 I guess it´s time to leave.

Good luck.

hi
i was trying to test all the solutions i have but i've failed
me also so confused & maybe i was'nt clear to explain exactly what's going on
my best solution was to reduce the 60v to supply the lm2596 module

so thats what i need to do which i've failed & have burnt some parts

any idea to reduce the 60v to supply the lm2596 module will help

the mddule have 100uF 50v cap at the input 470uH inductor & 220uF 35v cap at the output

i have tried any idea to step down the 60v to supply this module

so would you please for the last time help me with this?

--- Updated Apr 19, 2023 ---

No linear solution will work for you.

unfortunately its a bad news

thanks anyway

 

[KlausST]

Hi,

---

I can only repeat: it's still unclear how you connected the zeners with the module and the power supply.

****
What I'd do:
* Use the zeners (and zeners also need some kind of heatsink for fully specified power dissipation: datasheet says short wiring and wire ands kept at ambient temperature)
* for the use with capacitors I'd add a series resistor to limit inrush current
* use the ON feature of the LM2596 for delayed enable
* measure voltage, current with a scope
on error I'd :
* show the complete setup in a photo (so that users can detect any wiring mistake, missing heatsinks)
* show the complete (complete!) schematic
* show all exact part names (like diodes, power supply ...give links to their datasheets)
* explain the test setup and the steps
* explain the error: like did the diodes get hot abd hotter, smoke, fail, or did they explode in the very first moment...
* especially newbies should consider to take a video ... so we can analyze things the newbie even isn't aware of. (ESD, wrong test setup, wrong expectations ...)

Klaus

 

[johnny78]

Hi,

---

I can only repeat: it's still unclear how you connected the zeners with the module and the power supply.

****
What I'd do:
* Use the zeners (and zeners also need some kind of heatsink for fully specified power dissipation: datasheet says short wiring and wire ands kept at ambient temperature)
* for the use with capacitors I'd add a series resistor to limit inrush current
* use the ON feature of the LM2596 for delayed enable
* measure voltage, current with a scope
on error I'd :
* show the complete setup in a photo (so that users can detect any wiring mistake, missing heatsinks)
* show the complete (complete!) schematic
* show all exact part names (like diodes, power supply ...give links to their datasheets)
* explain the test setup and the steps
* explain the error: like did the diodes get hot abd hotter, smoke, fail, or did they explode in the very first moment...
* especially newbies should consider to take a video ... so we can analyze things the newbie even isn't aware of. (ESD, wrong test setup, wrong expectations ...)

Klaus

here is my connection first of all

i have used 3 power supplies to get 60v for test
& the zener diode output measurements was ok without connecting the lm2596 module
but when its connected everything got burnt at the moment

here is the zener diode datasheet
https://www.alldatasheet.com/view.jsp?
Searchword=1n4742%20datasheet&gclid=CjwKCAjwov6hBhBsEiwAvrvN6ILZVBSp_Hj51vxQXTuAR9QMxwvDAiHGu5V7H4DZLswQr4i4sSLJdRoCp-0QAvD_BwE

c1 is 100uf 50v & c2 is 220uF 35v & the inductor 470

 

[danadakk]

The Zeners in post #54 may be inadequate power rating (they are 1W), ratings should be higher
with margin :

Power Designer

webench.ti.com

Do a steady state sim as well to gain further insight. The above is startup transient.

Also adjust component values to meet your module values. Note Webench allows
you to change BOM, optimization goal, input and output V.......


Regards, Dana.

 

[johnny78]

Hi guys

this is the circuit i've used & it was on test since yesterday & the lm317 heat is in normal range with 3x2cm heatsink

any idea to enhance this will be great
thank you all

Johnny

 

[danadakk]

You want to drop 20V by the LM317, to meet LM2596 max Vin = 40V, now you have ~ 16V, so
recalc the R2

VOUT= 1.25V (1 + R2/R1) + Iadj x R2

Its wattage is = (Vout - 1.25)^2 / R2

Use worst case numbers.

Then with a scope look at noise at LM2596 input and output. With full
load on LM2596 output. To make sure your load meets Voutmin and
the LM317 output is stable. Especially when load change creates a
transient.

I would also look at startup transient. Datasheet for LM317 states
(regarding load C) :

Regards, Dana.

--- Updated Apr 24, 2023 ---

The point about load transient is to make sure the compound circuit
connection, LM317 + LM2596, doe not produce over V on its output
frying load components. And vice versa, too low a V causes loads to
stop functioning.

--- Updated Apr 24, 2023 ---

The point about load transient is to make sure the compound circuit
connection, LM317 + LM2596, doe not produce over V on its output
frying load components. And vice versa, too low a V causes loads to
stop functioning.

--- Updated Apr 24, 2023 ---

Dont forget you have a .5A Iloadmax target, so Plm317 = 20V x .5A = 10W.

Design for max Tjunction + some margin your heatsink, its thermal R, from earlier design
references posted.

Regards, Dana.

 

[KlausST]

Hi,

Dont forget you have a .5A Iloadmax target, so Plm317 = 20V x .5A = 10W.

0.5A is the load curent on 5V, at the output of the LM2596.
But the LM317 is on the input of the LM2596. Since the LM2596 is a buck converter the input current is (much) less then the output current.

****

The OP is now a 6 years member in this forum. I expect him to be able to use Ohm´s law and some really basic calculations on V, R, P and I. Maybe I´m expecting too much.

Klaus

 

[danadakk]

Hi,

0.5A is the load curent on 5V, at the output of the LM2596.
But the LM317 is on the input of the LM2596. Since the LM2596 is a buck converter the input current is (much) less then the output current.

****

The OP is now a 6 years member in this forum. I expect him to be able to use Ohm´s law and some really basic calculations on V, R, P and I. Maybe I´m expecting too much.

Klaus

Thanks for pointing out my error. The LM2596 is delivering 5V at .5 A = 2.5W.
Say its 82% efficient, so need 2.5/.82 = ~ 3W from the LM317 output. Thats at
40V Vin2596 so L M317 current = 3 W / 40 V = 75 mA (not accounting for ref
network power).

The efficiency of the reg is Pout/Pin = 3 / (60 x .075) = 66%. So LM317 must dissipate
Pin - Pout = 4.5 - 3 = 1.5W. Clipon heatsink should work.

But I am having a prob with simulation. Because the LM2596 module has a large Cin
when power first applied to circuit the differential across LM317 too high.....looking into this.


Regards, Dana.

--- Updated Apr 24, 2023 ---

Here is problem sim. Caused by fact output of LM317 is shorted to ground at
startup of power, by LM2596 module input C, causing, for a 0 - 60V turnon
of power supply, excessive transient input V to LM317.

If I slow down the 60V Trise of the power supply input it looks OK, allows the
Cin 2596 to "catch up" sos's to speak. So if you can confirm the supply does in fact
take some time, mS area or larger, to get to 60 V all might be fine.

What again was problem using Zener on input to drop V to LM2596 ?


Regards, Dana.

--- Updated Apr 24, 2023 ---

Final sim, corercted for fact I left out the LM2596 Cin estimate.

Still looks OK if trise 60V supply well into mS region for startup.

Regards, Dana.

 

[johnny78]

Hi,

0.5A is the load curent on 5V, at the output of the LM2596.
But the LM317 is on the input of the LM2596. Since the LM2596 is a buck converter the input current is (much) less then the output current.

****

The OP is now a 6 years member in this forum. I expect him to be able to use Ohm´s law and some really basic calculations on V, R, P and I. Maybe I´m expecting too much.

Klaus

hi Klaus
actually its rarely to work on project which needs new calculations using Ohm's law which im not good with it as i've mentioned before
& thats why i ask you for help
anyway i appreciate your help giving me ideas & helping with calculations
maybe even if im an expert I don't need to recalculate every step i work on because me or someone else has did this before

thanks

--- Updated Apr 25, 2023 ---

Thanks for pointing out my error. The LM2596 is delivering 5V at .5 A = 2.5W.
Say its 82% efficient, so need 2.5/.82 = ~ 3W from the LM317 output. Thats at
40V Vin2596 so L M317 current = 3 W / 40 V = 75 mA (not accounting for ref
network power).

The efficiency of the reg is Pout/Pin = 3 / (60 x .075) = 66%. So LM317 must dissipate
Pin - Pout = 4.5 - 3 = 1.5W. Clipon heatsink should work.

But I am having a prob with simulation. Because the LM2596 module has a large Cin
when power first applied to circuit the differential across LM317 too high.....looking into this.


Regards, Dana.

--- Updated Apr 24, 2023 ---

Here is problem sim. Caused by fact output of LM317 is shorted to ground at
startup of power, by LM2596 module input C, causing, for a 0 - 60V turnon
of power supply, excessive transient input V to LM317.

View attachment 182451

If I slow down the 60V Trise of the power supply input it looks OK, allows the
Cin 2596 to "catch up" sos's to speak. So if you can confirm the supply does in fact
take some time, mS area or larger, to get to 60 V all might be fine.

View attachment 182452


What again was problem using Zener on input to drop V to LM2596 ?


Regards, Dana.

--- Updated Apr 24, 2023 ---

Final sim, corercted for fact I left out the LM2596 Cin estimate.

Still looks OK if trise 60V supply well into mS region for startup.

View attachment 182453


Regards, Dana.

hi Dana

the resistors i've used with the LM317 was the available so i made the calulations on lm317 calculator & used them
& i will change the R1 to 240 & recalculate the R2 value

would you help with calculating the value of the 5w Resistor? & how to calculate it ?
its working since yesterday even without modifying the input capacitor inrush of the lm2596
& i was thinking of this when Mr Klaus have mentioned about it & was thinking of adding a resistance on series to charge the
input Cap slowly at first startup But its ok even without doing this its working

thanks for help & explaining alot
Johnny