lm1117impx-3.3 get hot when a load is connected

[abdullahseba]

Hi
I have circuit that has an lm1117impx-3.3 to power the omega2 plus (U4). But when I plug the omega into the socket the regulator gets really hot and after a bit the omega loses power.
I tried connecting just the power pins only in case there was a problem with the rest of the circuit but the same happened. The omega only drew around a 100 mili amps well under the regulators 800.
It provides 3.3v fine and powers the on board led. Total circuit resistance was a 8-9 mega ohms, so no shorts.
Any one have an idea why this is?
I wont mind if it got hot but kept the thing powered but its soooo puzzling 
Oh and this is on a pcb board.
Also the atmega 2560 (U1) is not on the board yet.
U5 is the pcm2901 and the other two is the txs0102.
Thanks in advance.

[MK14]

The omega only drew around a 100 mili amps well under the regulators 800.

Although the voltage regulator is rated for up to 800 milliamps. That is ONLY when it is sufficiently cooled down. E.g. Heatsinks.

I wont mind if it got hot

The voltage regulator, will automatically shut down, if its internal temperature, reaches around 125 deg C (maybe a bit higher).

I'm not clear what type of package your regulator is in. How much current (maximum) is the voltage regulator seeing ?
100mA is mentioned, but is that the maximum you are giving it (i.e. using on the output terminals) ?
Any heatsinking on it ?

You seem to be running it at 12 Volts so it has to dissipate any voltage drop across it as heat (12V - 3.3V = 8.7V). The higher the current, the hotter it will get, until it automatically shuts down when it gets too hot.

(Voltage drop across regulator) x (output current) = power dissipation of voltage regulator

[newbrain]

According to the datasheet, IMPX is SOT223, with Tja = 61.6K/W.

The power to be dissipated is (12V -3.3V)*0.1A = 0.87W.

The temperature raise will then be in the ballpark of: 61.6K/W * 0.87W ~=  54K.

The case will reach (assuming 25C ambient temperature) about 80 degree Celsius.

So yes, it will get very hot to the touch.
Still some room to Tjmax, but higher load current or ambient temperatures might get it very close.

[Zero999]

Where did you people get 12V from? The schematic seems to show a 5V supply.

There are two lm1117impx-3.3s on the schematic with their outputs both connected to the 3.3V, implying they're connected together in parallel which is never a good thing.

Some of the 0V pins on the other ICs aren't connected and there a lots of pins not connected to anything so you should check they have internal pull-up/down resistors enabled, if applicable.

[MK14]

Where did you people get 12V from? The schematic seems to show a 5V supply.

There are two lm1117impx-3.3s on the schematic with their outputs both connected to the 3.3V, implying they're connected together in parallel which is never a good thing.

Some of the 0V pins on the other ICs aren't connected and there a lots of pins not connected to anything so you should check they have internal pull-up/down resistors enabled, if applicable.

If you look in the bottom right hand corner (then an inch or so to the left) of the circuit diagram. The regulators (3.3V and 5V) seem to say they are being fed 12 Volts.

[wraper]

Where did you people get 12V from? The schematic seems to show a 5V supply.

Unless we live in parallel dimensions, schematic shows 12V

[MK14]

The power to be dissipated is (12V -3.3V)*0.1A = 0.87W.

I don't want to disagree, with your nicely constructed post. But I think the current is a lot higher than 100 milliamps, which is why I suspect the overheating (thermal shutdown). I think the 100 mA's he was referring to, was for just one part of the circuit.

But I could easily be mistaken. It really needs the OP to clarify it.

[Zero999]

Where did you people get 12V from? The schematic seems to show a 5V supply.

There are two lm1117impx-3.3s on the schematic with their outputs both connected to the 3.3V, implying they're connected together in parallel which is never a good thing.

Some of the 0V pins on the other ICs aren't connected and there a lots of pins not connected to anything so you should check they have internal pull-up/down resistors enabled, if applicable.

If you look in the bottom right hand corner (then an inch or so to the left) of the circuit diagram. The regulators (3.3V and 5V) seem to say they are being fed 12 Volts.

Oh I see now. I was looking at the wrong IC! No parallel dimensions, I just looked at the wrong dimension.

[newbrain]

The power to be dissipated is (12V -3.3V)*0.1A = 0.87W.

I don't want to disagree, with your nicely constructed post. But I think the current is a lot higher than 100 milliamps, which is why I suspect the overheating (thermal shutdown). I think the 100 mA's he was referring to, was for just one part of the circuit.

But I could easily be mistaken. It really needs the OP to clarify it.

Feel free to disagree! I did not know what an Omega2 was until 30" ago, so I went with OP provided info.
The post was intended to show the very basic of thermal calculations, which I suspect were overlooked by the OP.

[abdullahseba]

Where did you people get 12V from? The schematic seems to show a 5V supply.

It is 12 volts the 5V is just for other circuits.

[abdullahseba]

Thanks to every one for helping me!  It was the 12v been dissipated into heat. I used a 5v supply and it works fine even the USB part through the 5v reg. 
The omega draws a 100 mA but not sure what the rest draws.
I'll have to use a voltage divider or something on the outside for the 12 volts I guess

[MK14]

Thanks to every one for helping me!  It was the 12v been dissipated into heat. I used a 5v supply and it works fine even the USB part through the 5v reg. 
The omega draws a 100 mA but not sure what the rest draws.
I'll have to use a voltage divider or something on the outside for the 12 volts I guess

It would be useful to find out the total current (which you seem to be confirming was above the 100mA's). Then we need to estimate what is the most it realistically may use.

There are a number of solutions, but knowing what the current really is, would help.

Also when the Omega is doing stuff (running programs, as idle current may be considerably lower), it may use a lot more current. Presumably up to the recommended 500mA's for it (from a quick look at the spec/datasheet on my part).

[Zero999]

Thanks to every one for helping me!  It was the 12v been dissipated into heat. I used a 5v supply and it works fine even the USB part through the 5v reg. 
The omega draws a 100 mA but not sure what the rest draws.
I'll have to use a voltage divider or something on the outside for the 12 volts I guess

That's interesting. What's the voltage on the 5V line with 5V in to the 5V low drop-out regulator? I suspect the voltage will drop considerably and although it may work, it's not good practise.

If you don't want to modify the board, you could power it from a 6.5V power supply (preferably switched mode but a linear regulator, such as the LM317, will do if you have room for a large heat sink) which might reduce the power dissipation of the lm1117impx-3.3 to an acceptable level. Another option is to remove the 5V regulator, bypass it with bodge wires and power the entire board of 5V. Reasonably priced, efficient 12V to 5V buck regulator modules are widely available so this might be your best bet and it will result in lower over all power dissipation too.

[abdullahseba]

Thanks to every one for helping me!  It was the 12v been dissipated into heat. I used a 5v supply and it works fine even the USB part through the 5v reg. 
The omega draws a 100 mA but not sure what the rest draws.
I'll have to use a voltage divider or something on the outside for the 12 volts I guess

It would be useful to find out the total current (which you seem to be confirming was above the 100mA's). Then we need to estimate what is the most it realistically may use.

There are a number of solutions, but knowing what the current really is, would help.

Also when the Omega is doing stuff (running programs, as idle current may be considerably lower), it may use a lot more current. Presumably up to the recommended 500mA's for it (from a quick look at the spec/datasheet on my part).

well the whole circuit doesn't seem to draw more than 250mA total with wifi.

[abdullahseba]

Thanks to every one for helping me!  It was the 12v been dissipated into heat. I used a 5v supply and it works fine even the USB part through the 5v reg. 
The omega draws a 100 mA but not sure what the rest draws.
I'll have to use a voltage divider or something on the outside for the 12 volts I guess

That's interesting. What's the voltage on the 5V line with 5V in to the 5V low drop-out regulator? I suspect the voltage will drop considerably and although it may work, it's not good practise.

If you don't want to modify the board, you could power it from a 6.5V power supply (preferably switched mode but a linear regulator, such as the LM317, will do if you have room for a large heat sink) which might reduce the power dissipation of the lm1117impx-3.3 to an acceptable level. Another option is to remove the 5V regulator, bypass it with bodge wires and power the entire board of 5V. Reasonably priced, efficient 12V to 5V buck regulator modules are widely available so this might be your best bet and it will result in lower over all power dissipation too.

well I could desolder the 5v reg and supply 5v direct to the header.

[Zero999]

Thanks to every one for helping me!  It was the 12v been dissipated into heat. I used a 5v supply and it works fine even the USB part through the 5v reg. 
The omega draws a 100 mA but not sure what the rest draws.
I'll have to use a voltage divider or something on the outside for the 12 volts I guess

That's interesting. What's the voltage on the 5V line with 5V in to the 5V low drop-out regulator? I suspect the voltage will drop considerably and although it may work, it's not good practise.

If you don't want to modify the board, you could power it from a 6.5V power supply (preferably switched mode but a linear regulator, such as the LM317, will do if you have room for a large heat sink) which might reduce the power dissipation of the lm1117impx-3.3 to an acceptable level. Another option is to remove the 5V regulator, bypass it with bodge wires and power the entire board of 5V. Reasonably priced, efficient 12V to 5V buck regulator modules are widely available so this might be your best bet and it will result in lower over all power dissipation too.

well I could desolder the 5v reg and supply 5v direct to the header.

That's sensible but you'd also need to apply 5V to the input of the lm1117impx-3.3 too.

[abdullahseba]

That's sensible but you'd also need to apply 5V to the input of the lm1117impx-3.3 too.

Yeah that would be thorough the 12v header. So just connect the 12 v and 5v header together.

[MK14]

well the whole circuit doesn't seem to draw more than 250mA total with wifi.

Shamelessly stealing Newbrain's calculations, and using the 250mA total.

We get:

The temperature raise will then be in the ballpark of: 61.6K/W * 0.87W ~=  54K.

Then being lazy, and saying 250mA's is 2.5x 100mA's.

54 x 2.5 + 25 (room temperature) = 160 deg C

Which would explain why you seem to be getting thermal shutdown, soon after power up. Because the regulators internal (junction) temperature should be a lot lower than 160 deg C. Max is something like 125 deg C at which point it thermally shuts down automatically (with 150 deg C as the absolute maximum datasheet value).

The solutions being offered by me and others, including reducing the incoming voltage, or using switching voltage converters, or heatsinking (e.g. using the TO-220 version + heatsink), would all work.

[Zero999]

That's sensible but you'd also need to apply 5V to the input of the lm1117impx-3.3 too.

Yeah that would be thorough the 12v header. So just connect the 12 v and 5v header together.

That's fine but beware that only connecting one of the headers, whist the power is applied, might cause trouble. Consider what would happen if there was a 3.3V supply and no 5V or vice versa?

[abdullahseba]

That's sensible but you'd also need to apply 5V to the input of the lm1117impx-3.3 too.

Yeah that would be thorough the 12v header. So just connect the 12 v and 5v header together.

That's fine but beware that only connecting one of the headers, whist the power is applied, might cause trouble. Consider what would happen if there was a 3.3V supply and no 5V or vice versa?

yeah I just noticed that I need to apply 3.3v first than 5v, not sure why something to do with the USB. BAD CIRCUIT DESIGN.

[Monkeh]

yeah I just noticed that I need to apply 3.3v first than 5v, not sure why something to do with the USB. BAD CIRCUIT DESIGN.

Requiring power staging is not bad circuit design..

[Zero999]

yeah I just noticed that I need to apply 3.3v first than 5v, not sure why something to do with the USB. BAD CIRCUIT DESIGN.

Requiring power staging is not bad circuit design..

However it can be a nuisance if you've not planned it from the start. Unfortunately the regulators don't have enable pins so you need an external transistor before the input if you want to delay the turn on.