Maybe I misarticulated. My power supply is 12V 6A. The board is sinking 100mA through its regulator. Thtas what I meant. You wrote "If your board draws no more than 300mA you are safe" But this type of regulator cannot be heatsinked. I'm wondering how they dissipate such power on the Nucleo board. Also they dont mention the max ambient temperature in the table.The dissipation is the voltage DROPPED by the regulator (Vin - Vout) multiplied by the current through it. That is why higher input voltages restrict the current you can take from them before overheating.
If your board draws no more than 300mA you are safe but your statement "12V external supply with 100mA input current safely" implies the supply itself can only provide 100mA so you may overload it if you draw more the the 100mA rating. It is the source that would be overloaded not the regulator or board.
Brian.
Why are you “confused about how they dissipate such heat”? You don’t give us any information about the physical aspects of the board, so we sure can’t tell you. But if the manufacturer tells you a specific value, why do you not trust that? Why would you, instead, trust a bunch of strangers on the internet?Hello again,
I think some of you misunderstood. My 12V power supply can source current up to 6A.
That table is current limitation for the board if one uses VIN. Please see this updated manual for my board here:
https://www.st.com/resource/en/user...ucleo-32-boards-mb1180-stmicroelectronics.pdf
View attachment 179122
It says when Vin<=12 the current limitation is 250mA so that the regulator wouldn't heat too much.
But the regulator they use on the board is not a type to be heatsinked and I'm wondering whether I can rely on what they claim. Also no ambient temperature range is mentioned.
Again in my case my power supply is 12V 6A and the voltage regulator of the board sinks 100mA in my case. The table says 250mA limit but I'm a bit confused how they dissipate such heat.
I'm using the Nucleo board and power it by VIN 12V and the current is 100mA in my case. How can I know whether the regulator on this board can withstand the heating? Can I rely on the table?
--- Updated ---
Maybe I misarticulated. My power supply is 12V 6A. The board is sinking 100mA through its regulator. Thtas what I meant. You wrote "If your board draws no more than 300mA you are safe" But this type of regulator cannot be heatsinked. I'm wondering how they dissipate such power on the Nucleo board. Also they dont mention the max ambient temperature in the table.
Why are you “confused about how they dissipate such heat”? You don’t give us any information about the physical aspects of the board, so we sure can’t tell you. But if the manufacturer tells you a specific value, why do you not trust that? Why would you, instead, trust a bunch of strangers on the internet?
Apparently the regulator can dissipate about 1.6W. (7V-5V)x.8. Or, (9-5)x.45. Or, (12-5)x.25. Maybe they have liquid cooling. Maybe they’ve got a lot of copper area on the PCB.
And, if you don’t trust the manufacturer’s power supply spec, why would you trust any of their other specs? ST is a pretty reputable company.
???I just wanted to have your opinion of using a series resistor to ease the voltage regulator.
I don’t see a problem with the previously-unmentioned series resistor, as long as the input to the regulator is high enough. This is done all the time. But I don’t see the necessity, either.Hi,
???
it´s the first time you mention a series resistor.
Resistor to drop the voltage?
I can not recommend it.
I´d use a buck converter.
From 12 V to the lowest useful voltage.
Klaus
Of course. you have to do your due-dilligence; that’s engineering. I didn’t say throw ANY resistor in there. But using an inexpensive large-wattage resistor is a common practice as opposed to dealing with heat sinks, etc.Hi,
the big problem with a series resistor is that the voltage drop depends on load current.
You may have measured 100mA ... but did you really do a "peak" current measurement during all expectable conditions:
* different temperatures
* during start up
* during programming
* and so on
the booting process with intentially floating inputs at the microcontroller or any periferals may draw a lot more current than during normal condition. And due to the voltage on floating pins is not determined, also the current is non determined. And "floating pins" is just one parameter that infuences the current.
In my eyes its more likely thet the application becomes upset because of the added series resistor, than the datasheet is wrong.
In the many years of electronics development, I found a couple of datasheet mistakes. But surely the majority of application problems were else where.
If my word contradicts a datasheet: better trust the datasheet. Sad but true.
A lot of unknown conditions.
Some time ago I had a problem with an application (not my design) that started well on all my computers, but refused to start at the customer. I found out that this was a bad situation of an unusually weak USB port at the customer´s PC and a high current drawn form the device. The result was: wild display flicker, but communication via USB did work well. So the high current made the USB voltage to drop and (maybe due to slowly rising supply voltage) the microcontroller did start well, but the display refused to work properly.
So if you really want to uselessly dissipate about 2.5 times the power of your application into heat (and this is what any linear power supply does - independent of used series resistors) then better use a series zener.
And switch mode converters aren´t as bad as you might think. I used two independent swtichers in a handheld measurement device, and with ggod PCB design and minimal filters, the 16 bit ADC still worked down to less than 1LSB RMS noise (datsheet specification) @ 10kSamples/s.
Also: not every linear regulator is an LDO. For dropping 12V to 5V you surely don´t need an LDO.
An LDO is a linear regulator with - like the name says - low dropout voltage.
Using (noisy, unstable) supply voltage as the REFerence for an ADC is i most cases the killer for ADC performance.
Klaus
If you use a 12V power supply with a +/-5% accuracy, then the circuit of post#9 is limited to currents of about 115mA. It may work, but for my taste this calls for problems.
Hi again,Hi,
the big problem with a series resistor is that the voltage drop depends on load current.
You may have measured 100mA ... but did you really do a "peak" current measurement during all expectable conditions:
* different temperatures
* during start up
* during programming
* and so on
the booting process with intentially floating inputs at the microcontroller or any periferals may draw a lot more current than during normal condition. And due to the voltage on floating pins is not determined, also the current is non determined. And "floating pins" is just one parameter that infuences the current.
In my eyes its more likely thet the application becomes upset because of the added series resistor, than the datasheet is wrong.
In the many years of electronics development, I found a couple of datasheet mistakes. But surely the majority of application problems were else where.
If my word contradicts a datasheet: better trust the datasheet. Sad but true.
A lot of unknown conditions.
Some time ago I had a problem with an application (not my design) that started well on all my computers, but refused to start at the customer. I found out that this was a bad situation of an unusually weak USB port at the customer´s PC and a high current drawn form the device. The result was: wild display flicker, but communication via USB did work well. So the high current made the USB voltage to drop and (maybe due to slowly rising supply voltage) the microcontroller did start well, but the display refused to work properly.
So if you really want to uselessly dissipate about 2.5 times the power of your application into heat (and this is what any linear power supply does - independent of used series resistors) then better use a series zener.
And switch mode converters aren´t as bad as you might think. I used two independent swtichers in a handheld measurement device, and with ggod PCB design and minimal filters, the 16 bit ADC still worked down to less than 1LSB RMS noise (datsheet specification) @ 10kSamples/s.
Also: not every linear regulator is an LDO. For dropping 12V to 5V you surely don´t need an LDO.
An LDO is a linear regulator with - like the name says - low dropout voltage.
Using (noisy, unstable) supply voltage as the REFerence for an ADC is i most cases the killer for ADC performance.
Klaus
If you use a 12V power supply with a +/-5% accuracy, then the circuit of post#9 is limited to currents of about 115mA. It may work, but for my taste this calls for problems.
Hi again,Hi,
The benefit of the zener is the almost constant voltage drop (regarding current).
So if the application needs more current for a short time there is no problem.
The source impedance seen from the regulator is more low ohmic, thus easier to handle.
The voltage at the regulator input is more stable, tus the output is more stable, too.
But I don´t like to go that close to the limits as you do.
The dropout is 1.1V typ, but it can be higher. You say the application current is 125mA max. As already written I´ve made bad experience where applications dar a lot more current for some 10ms .. 100ms.
Also a 12V suuply does not perfectly output 12.00V. You may expect at least a drop of 5%.
And why at all now going to extremes ragarding disspated power,
12V to 5V with 125mA is 0.875W. Why no make it difficult for the reegulator and give him the absolute minimum to operate properly?
A 5.6V zener @ 125mA means 700mW
the regulator with 1.4V @ 125mA means 175mW
I´d relax everything. Use a 3.3V zener (410mW) leaving 3.7V for the regulator (470mW). delta-T of maybe 55°C
No stress for short time higher current, no stress for short time lower input voltage, no thermal stress ofr the regulator nor the diode. Stable, relaxed, reliable.
In my eyes going to the extremes always results in reduced reliability.
(All calculations are just "over the thumb" and not meant to be perfect)
Klaus
I think it´s better than the 5.6V one. It makes the circuit more error tolerant.Would 4.7V 3w zener also be fine?
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