Continue to Site

# Estimating the maximum current for a microcontroller board's regulator

Status
Not open for further replies.

#### floppy32

##### Member level 1
Hi all,

I will use this STM32 microcontroller board and want to power it from an 12V supply. And I measured the current sinked by the regulator(node 3 below) will not exceed 100mA.

The regulator of the board is given as follows from its manual (VIN=12V in my case):

The board's voltage regulator(LD1117S50TR) datasheet is given here. And in my case VIN will be 12V.

The manual also has the following information(at Table 7. Power supply capabilities):

So from the user manual, what I understand is that VIN can be 12V but the current should not exceed 300mA.

Is that correct? Can I power this board by 12V external supply with 100mA input current safely? In my case the ambient temperature will not exceed 50°C.

The board requires "less than 300 mA" at 12V. Just because you measured 100mA for a particular situation, doesn't mean it's ALWAYS going to be 100mA. If your supply can only proviide 100mA, it's probably not going to work.

Further, those specifications make no sense. U6 is a linear regulator. Regardless of what you have for an input voltage, the output current (at 5V) will be the same, and, thus, so will the input current. Those specs look like it's for a switching regulator, which is not what you're showing.

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.

Dissipation = Iload x ( Vin - Vout ), in your case, (12 - 5) * 0.1 = 700mW

for 200mA it will be 1.4 watts

Unless you have designed for sufficient heatsinking to remove this heat without undue temp rise - the reg will shut down due to over temp

N.B. efficiency = Power out / power in, = Iload. Vout / Vin. Iload = Vout / Vin = 5/12 = 41.67 % always

So 58.34 % of the power into the reg has to be got rid of as heat ...

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

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 ---

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.
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.

Last edited:

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.

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.

Hello and thank you for your help. But I asked the same question at board vendor's forum. And someone replied please see the answer here: https://community.st.com/s/question...boards-regulator-when-powered-with-12v-supply

I just wanted to have your opinion of using a series resistor to ease the voltage regulator. Best regards,

Hi,

I just wanted to have your opinion of using a series resistor to ease the voltage regulator.
???
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

Hi
Using Linear Regulator is good choice for low noise application, but it not mean it can reject all noise from input and adapt quicky to load respond. That why some LDO's price is high and higher so much than others.
LDO normally have lower efficiency than switching power, but it simple.
Do you belive LDO have efficency higher than SMPS ?
Yes, this case exist.
When you design for low power application you will meet. SMPS will drop efficiency so much when low load. Because of that, some new SMPS support dual mode : LInear & switching.
Don't trust all and 100% to info in datasheet as if you are experiment engineer. All parameters is tested in some design condition, it is not your operating conditions.
I used serial resistor before LDO to share drop voltage and reduce heat on LDO.
To select right resistor value, have to ensure peak/max load current.
If temperature of LDO rising too high it will be spoil or drift output. If again, using this fo VDDA, the results of ADC also drift (without self-calibration process, with ... will make speed lower in some case).
Example:
Good drop voltage for load respond and all kind of LDO: V_drop = 2V
Voltage drop on serial resistor expect: V_res = Vin - V_load - V_drop = 12 - 3.3 - 2 = 6.7V
Dissipate power on LDO: P_ldo = 100mA x 2V = 200mW must be smaller of dissipate power of IC package
on serial resistor: P_res = 100mA x 6.7V = 670mW, must select size of resistor can withstand this peak power. normally power is 50mA, you only need select 50/100 = 1/2 power rating of 670mW instead of full power range.
Ensure input capacitor of LDO is high enough to store energy when load transient.
Serial input serial also act as LPF - Low pass filter, help to improve output ripple/noise.
You can use pre-power by BUCK to 5V then through LDO to 3v3 to improve efficiency.

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
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.

If OP doesn’t trust the manufacturer’s power supply data then why should they trust anything in the data sheet? This whole thread is about the OP’s failure to accept what the manufacturer is explicitly saying.

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.

Last edited:

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.
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,

In my application the current can be between 85mA up to 100mA. And if you are more conservative max 150mA.

How about using this zener diode in series?:

https://uk.rs-online.com/web/p/zener-diodes/8624887

The zener drops 5.6V and can handle 3W. It will drop VIN to 6.4V.

The 5V LDO(LD1117S50TR) in question needs 1.1V dropout voltage. So for VIN anything more than 6.1V should be fine I guess.

You mentioned the zener so I wonder whether this would suffice? (I would prefer 4.7V zener at 3W but I dont have time to wait that much) So I could only find this
https://uk.rs-online.com/web/p/zener-diodes/8624887

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

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
Hi again,

One last question. I already ordered 4.7V 3W zener 1N5917BRLG. I was a bit late to see your 3.3V suggestion Would 4.7V 3w zener also be fine? Thank you for your help

(Yes my 12V supply has +-%5 regulation)

EDIT: I have now ordered 3.3V 5W zener anyways.

Last edited:

Hi,

Would 4.7V 3w zener also be fine?
I think it´s better than the 5.6V one. It makes the circuit more error tolerant.

In the end it depends on your used devices (like the 12V supply specification), wiring, temperatures.
And on your style of designing electronics.

Klaus

Status
Not open for further replies.