Embedded project needs devices which can work continuously at +125 degree C

[andre_luis]

It is for measuring furnace temperature

As you are giving information in small portions, one would presume that it is forge for metal molten, and in that case the environment 125 degrees above mentioned now would make sense.

By the way, an idea for cooling I saw in a local TV documentary long time ago, it was presented a very creative solution for decrease the temperature on the vivarium shed of a chicken farm in an extremely hot region: A ceramic clay pipe was buried 10m deep down in the ground, 10m outward, rising there 10m, that is, an "U" shape, pulling hot air from the ouside environment, exchanging heat in the subsoil, and injecting it inside; no freon gas compressor needed, no peltier cells needed, just a a fan. It was interesting how much the air was cooled in the path; I can bet this would reduce some tens of degrees the temperature in your system, and once the air flow necessary to cool a board is quite small, even a smaller scale apparatus would do the job; you will probably have restrictions to implement this, but I leave the suggestion here just for the sake of sharing the information.

 

[baileychic]

@ andre_teprom

Okay, I will ask client (Client's client) to see if any cooling can be provided.

Here is the latest almost finalised schematic. The 7 Segment display and delay setting buttons were removed as client could not find any suitable 7 Segment display which can operate at 125 deg C. The setting of delay will be done through WiFi communication.

Only requirement now is to read the 8x thermocouples and send data to PC using ESP8266.

@betwixt

Which headers can I use for connecting thermocouples to PCB ? Please provide datasheet of the headers.

 

[betwixt]

That's fine.
A better strategy would be to place the thermocouples inside the furnace and keep everything else away from it. The thermocouple wires should be long enough to extend out of the furnace and reach to the MAX31855 devices which can all be on a PCB where the temperature is cooler. The length of thermocouple wires is irrelevant (within reason) as it is only the welded tip that produces any voltage.

You must understand one of the basics of thermocouples though - almost anywhere dissimilar metals contact each other will produce a voltage, the thermocouples you can buy are just ready made devices with a very predictable temperature to voltage characteristics. To use them to measure temperature it is essential that an offset is applied at the amplifier and that is what cold compensation (also called ice-point) is all about. It has nothing to do with the thermocouple tip, the compensation is to remove the error created when the dissimilar wire materials are attached to the MAX31855, that is why they can measure their own temperature, it is to calculate the offset needed at the connections. Ideally, the wires are jointed as close to the IC as possible to minimize any difference between the IC and the joints.

If follows that if you need to extend the thermocouple wires, you must use the correct kind of connector so it doesn't introduce any further errors. The connectors are usually made of the same metals as the wires so no voltage is created and obviously, you have to wire the connector the correct way around because the metals are different.

Incidentally, have you realized the shutdown control in the power supply cannot possibly work. The emitter of Q1 cannot possibly rise more than VDD - Vbe or about 2.6V. You need an additional transistor to turn it on or off and to ensure it can be given enough base current.


Brian.

 

[baileychic]

@ experts

I was checking these for thermocouple to PCB connectors. They suit my requirements but in the below datasheet what is it that is mentioned as TO-92 transistor for cold junction temperature measurement ? Do I still need any TO-92 package style transistor. If yes, which one ?

https://in.omega.com/pptst/PCC-OST-SMP.html

https://br.omega.com/omegaFiles/temperature/pdf/PCC-OST-SMP.pdf

 

[betwixt]

You don't need the version with a TO-92 clip. TO-92 is a package size, not a particular device so it could include the many thermometer types with that construction (LM35, MCP7901, DS1802 etc). You are doing the compensation and amplification in the MAX device so it isn't needed. As I mentioned before, the reason for measuring the temperature at the connector is to remove the error voltage it creates by itself, hence the need to know the temperature at that point. What you must do it mount the connector as close to the MAX as possible so they are at the same temperature as each other.

The connectors are fine but they accept standard thermocouple plugs with round or rectangular pins. If your product is a single assembly, you can save considerable costs by soldering the thermocouple wires directly to the PCB. If you do that though, be careful to use suitable solder as some wires do not easily bond to standard Sn/Cu solder. If my memory serves me well, back in the 1980s I had to use a 100% Pb solder with flux but please double check.

Brian.

 

[baileychic]

@ betwixt

Okay, I will ask client if they can directly solder the thermocouple probe to PCB else we will use the Omega connectors.

We are now thinking about omitting the Peltier and use wireless power to charge the battery. We need 5V @ 1A max to charge the 1100mAh battery. I want to know,

At what distance the wireless charging device can work and where to find good wireless charging (Tx and Rx) circuits.

 

[baileychic]

If we use 4x 3.7V batteries in series to get fixed 12V 350mA for fan and 3.3V 150mA for other parts of the circuit then can we charge the 3x 3.7V batteries in series by applying 4 * 4.2V = 16.8V ?

We need 12V 350mA initially from battery to cool the one side of Peltier to get Peltier output voltage of 16 to 20V. After Peltier provides output it is used for charging battery and also power the device. If Peltier voltage drops then battery is used to run the device.

Fan is 12V 350mA 3500 RPM Brushless DC type.

 

[betwixt]

Firstly, wireless is very short range, ideally with the transmit and receive coils almost touching. All you need to do is drive a suitably powerful signal into the transmit coil and rectify the resulting voltage from the receive coil.

If I understand - you want to use battery power to create a temperature difference across the Peltier then 'reverse' the process to recharge the battery. Given the low efficiency of these devices and the need to maintain the temperature difference I can't see that working.

Brian.

 

[baileychic]

@betwixt

Okay, we will see if Peltier cooling can be increased by some other means but assuming Peltier can give 18 to 20V then can we charge 4x 3.7V batteries in series by applying Peltier voltage regulated to 4x 4.2V = 16.8V ? Or is this charging voltage high ?

 

[betwixt]

To charge the batteries there are two main considerations:
1. You start with more voltage than the fully charged battery voltage,
2. You use a suitable charging current/voltage regulator to tailor the charge rate for the battery.

Point 2 is why you need the excess voltage, some will be lost in the charge regulator. 18V to 20V should be adequate but you have to be sure the Peltier can produce that voltage at full charging current. This is my biggest concern, voltage off-load and under full load will be different and the voltage itself will depend on the temperature difference between hot and old sides. You are going to need quite a large temperature step over a short distance and keeping the cold side temperature low enough will be difficult. A heat sink alone will not work, it will just heat up to nearly the hot side temperature. If you force cool it, the ambient temperature still needs to be low and you waste more power in the cooling process. If you want to follow the Peltier route, I would suggest looking at water cooled systems, it isn't as difficult as it sounds because hundreds of types of liquid cooling mechanisms are available for 'overclocked' PC applications. They allow you to shift the heat outlet away from the heat source and are less bulky than a massive metal heatsink.

Brian.

 

[baileychic]

I decided to use LT1963A LDOs. My ground is 0V.

I am using both LT1963A-ADJ (to get 12V) and LT1963A-3.3V (to get fixed 3.3V).

I want to know the R1 and R2 resistors values for each LDO to get 12V and 3.3V.

It is mentioned in the datasheet that IAdj = 3uA at 25 deg C.

Datasheet gives equation

Vout = 1.21V * (1 + (R2/R1)) + (Iadj) * (R2)

Datasheet also mentioned that Vadj = 1.21V

So, how to choose resistors R1 and R2 to get 12V and 3.3V from the above regulators ?

It is mentioned in datasheet page no. 14 that R1 should not exceed 4.17k and so if I choose R1 = 3.3k fixed then how to find out R2 ? Should I assume Iadj as 3uA only at 125 deg C ?

- - - Updated - - -

Can I choose these resistor values to get 12V and 20V ?

Why LaTeX became electroda.pl ?
[Moderator's note: 'tex' tag was misidentified by system interface. Therefore your equation was edited so it is now visible in quote box above.]

 

[d123]

Hi,

This part you are using is extremely similar/"identical" to the TPS73801 (even the 4.17k resistor value requirement is identical, as far as I remember). Just saying in case reading that datasheet helps with any doubts. 73801 is good (simple to grasp) for calculating power dissipation at the end of the datasheet, but same comment applies to page 19 of LT1963A datasheet.

LT1963A: Adjust pin bias current vs. temperature graph on page 10 shows it dropping 1uA to 2uA at 125ºC, that's third of the value at 25ºC... Specifications on page 5 for adj pin current say 3uA typical to 10uA max. I'd calculate for worst case and maybe dither and doubt and go for ~7uA and hope that might be the worst case at 125ºC (obviously a hobbyist with that non-academic approach). It's rather hit and miss seeing disparity between electrical specifications values and the graph example.

Your calculations are ones I'm unfamiliar with at first glance. I'll be copying them as they look useful, thanks. The datasheet has the one I know (page 14). From rough calculations I did, the 29k looks right. Using your results for R2, I get 3273 for the 12V and 3075 for the 20V for R1 without including adj pin current. I guess you are not expecting a precise 12 or 20V and a hundred mV or so up and down is irrelevant to your application.

 

[baileychic]

Okay, I considered Iadj as 3uA only. Here are my calculations and final schematic. Any changes needed ?

12V is for motor. It can vary a little like say 11.6V or 12.4V because it is a 12V 350mA BLDC motor.

For battery charging (battery nominal voltage is 3.7V and charging voltage is 4.2V) we are using 4 batteries is series that gives 4 * 4.2 = 16.8V.

The two calculkation for R2 (20V and 16.8V) are shown in attached PDF and according to it I need 30k and 42.5k. So, I can use 50k preset (trimmer) to get the values ?

Should I calculate for +125 deg C with ~7uA (Iadj) ? I guess it will fit in the 50k trimmer value.

I use Maple software to do the calculations.

I had checked two devices.

LT1963A and TL1963A and found they are similar.

https://www.ti.com/lit/ds/symlink/tl1963a-q1.pdf

- - - Updated - - -

because it is a 12V 350mA BLDC motor.

correct:

because it is a 12V 350mA BLDC fan.

I will use either 4x 3.7V (nominal voltage and 4.2V charging voltage) batteries to get 16.8V and 12V (12V BLDC fan) or I will use 2x 3.7V batteries to get 8.4V and 5V (5V BLDC fan).

According to my calculations using Maple We can use 50k trimmer resistor to get all desired voltage values between 1.21V and 20V for Iadj between (3uA and 7uA) and I have used a 3.3k fixed resistor for R1 and 50k trimmer for R2.

 

[d123]

Hi,

Thanks for all the info. Super. Apology: I have a job I need to finish by 9AM my time, so need to hurry a little bit.

Just reviewing the schematic "quickly" - and without having done any voltage regulator PD calculations - my personal observations and opinions are:

Trimpots: Good practice is not to set trimpots anywhere near the high/low ends of the range. It's recommended to use a fixed value resistor in conjunction with a trimpot whose wiper ideally doesn't travel much away from approximately half the trimpot value, ~42k is a little far for my liking but should be fine.

The motor on the same line as the 3.3V regulator input: !!! OMG! . Don't do that, I think. That voltage line will presumably have transient dips on it when the motor turns on and when it turns off the output of the 12V will get the back emf kickback. If the PD of the 3.3V regulator when connected to the 16V regulator is within the permitted for the part, I'd put it in parallel with the 12V regulator. I'd personally have paralleled all three regulators from the main input rather than cascade them as they all have a job to do, but again, it depends on PD of each regulator.

The motor needs a flyback diode, and a 100nF across the motor terminals won't hurt. I'd possibly be OTT and consider a series diode at the output of the 12V regulator and factor in its voltage drop to get 12V + expected diode Vf as the regulator output voltage.

I'm far from being an expert like many members here but hope the comments are helpful. I'll be back, like the proverbial bad penny .

 

[baileychic]

Hi,

Thanks for all the info. Super. Apology: I have a job I need to finish by 9AM my time, so need to hurry a little bit.

Just reviewing the schematic "quickly" - and without having done any voltage regulator PD calculations - my personal observations and opinions are:

Trimpots: Good practice is not to set trimpots anywhere near the high/low ends of the range. It's recommended to use a fixed value resistor in conjunction with a trimpot whose wiper ideally doesn't travel much away from approximately half the trimpot value, ~42k is a little far for my liking but should be fine.

Okay, I finalised the R2 calculations. For Vout = 1.21V to 20V and Iadj = 3uA to 7uA the R2 value lies between 0 Ohms and 50k Ohms.

My requirement is to get 4 x 4.2V = 16.8V (battery charging) and 12V (12V 350mA BLDC fan) or 2 x 4.2V = 8.4V (battery charging) and 5V (5V 350mA BLDC fan)

So, I will not be using trimmer at extremes to get these values. Or should I use a say a 5k resistor in series with trimmer wiper ?

To get 3.3V fixed voltage I am using same LDO but 3.3V fixed output type and so need for R1 and R2 is eliminated for that.

The motor on the same line as the 3.3V regulator input: !!! OMG! . Don't do that, I think. That voltage line will presumably have transient dips on it when the motor turns on and when it turns off the output of the 12V will get the back emf kickback. If the PD of the 3.3V regulator when connected to the 16V regulator is within the permitted for the part, I'd put it in parallel with the 12V regulator. I'd personally have paralleled all three regulators from the main input rather than cascade them as they all have a job to do, but again, it depends on PD of each regulator.

The motor needs a flyback diode, and a 100nF across the motor terminals won't hurt. I'd possibly be OTT and consider a series diode at the output of the 12V regulator and factor in its voltage drop to get 12V + expected diode Vf as the regulator output voltage.

I'm far from being an expert like many members here but hope the comments are helpful. I'll be back, like the proverbial bad penny .

Added the flyback diode and 100nF 50V was already these. Attached latest circuit.

@d123 and experts

So, should I parallel the 3.3V regulator circuit with 12V regulator circuit or series as in the attached will do ? Which is better ? because the LDO has to drop 20V to 3.3V. If used in parallel mode. In series mode it is dropping 12V to 3.3V.

For final circuit all SMD components of proper ratings will be used.

Practically which circuit is better from the 2 attached ?


For U1,U2 and U3 sense pins I am using 100k pullup resistors and for U1 I have a connection betweeen SENSE pin and microcontroller 3.3V pin. Is that fine ? Or should I use a NPN transistor to make the SENSE pin low to turn Off battery charging retaining the 100k pull-up resistor ? because there will be max 20V at SENSE pin and microcontroller is 3.3V type (PIC18LF26K22).

 

[betwixt]

Why use a "Low Drop Out" regulator at all when you have such a large voltage to drop? A cheap switch mode regulator will be FAR more efficient and run almost cold instead of dissipating (and wasting) quite a lot of energy.

Question: what does the fan actually do?

Brian.

 

[baileychic]

@ betwixt

Mention one. They need external inductors ? I need 1.5A output current from primary regulator.

Fan is for cooling Peltier heat sink.

I found these. Can one of these be used ?

https://www.analog.com/media/en/technical-documentation/data-sheets/3681f.pdf

https://datasheets.maximintegrated.com/en/ds/MAX17524.pdf

https://www.ti.com/lit/ds/symlink/lm2592hv.pdf

https://www.onsemi.com/pub/Collateral/ENA2104-D.PDFv

 

[betwixt]

All SMPS need inductors but they are small and inexpensive. I use four 2A 12V regulators at full load 24hrs a day on an Al heatsink inside a box and it runs warm but not hot. Their combined size is only about 40mm x 40mm x 6mm. All the ones you listed will work perfectly but there are some with pre-set output voltages of 12V, 5V and 3.3V that do not require as many components, basically the feedback resistors are inside the IC. The power you waste in a linear regulator is "(input V - Output V) * load current" so for example dropping 20V to 12V at 1.5A would dissipate 12W and need a substantial heatsink of it own. If you use SMPS you can get efficiencies of better than 90%, not only running cool but greatly reducing the overall temperature.

I still have concerns about the cold side of the Peltier. A fan can only remove some of the heat and it cannot drop it below ambient air temperature. In an enclosed building and beside a furnace the ambient temperature is likely to be quite high so you may not achieve enough power output from it. Certainly dropping the power loss in the regulators will help considerably.

Brian.

 

[baileychic]

What is the difference between SMPS and LM2595SX ?

I decided too use LM2595SX-ADJ to get either 16.8V or 8.4V from 20V max input and LM2595SX-12 and LM2595SX-3.3 to get 12V and 3.3V fixed voltages.
For adjustable version the R2 values are as in attached. I choose 330uF 63V high temperature electrolytic Capacitors for switcher inputs and outputs.
I used 1N5822 diodes for all.

For Adj version to get 16.8 or 8.4V what should be the output inductor value to get 1.2A max output current ?
For fixed voltage version of 12V and 3.3V what should be the output inductor value to get 750mA output current ?

 

[d123]

So, I will not be using trimmer at extremes to get these values. Or should I use a say a 5k resistor in series with trimmer wiper ?

Good question. I'd need to review e.g. Bourns trimpot free online books/booklets to see if that's a standard approach. Frankly, I wouldn't worry so long as neither side of the trimpot ends have a very low resistance value on them.

So, should I parallel the 3.3V regulator circuit with 12V regulator circuit or series as in the attached will do ? Which is better ? because the LDO has to drop 20V to 3.3V. If used in parallel mode. In series mode it is dropping 12V to 3.3V.

Only the PD calculations can answer that in a meaningful way... As you'll know, usually the answer found regarding power dissipation in linear regulators is not the one the designer hoped to see and it's often back to the drawing board or "rip it up and start again".

Practically which circuit is better from the 2 attached ?

Personally, TMS-PS-NCN-S-COLOR makes me wince in empathy for the LT1963A at seeing a motor at the input to the regulator. That may (or may not) affect reliable functioning of the SHUTDOWN pin at any voltage dips.

For U1,U2 and U3 sense pins I am using 100k pullup resistors and for U1 I have a connection betweeen SENSE pin and microcontroller 3.3V pin. Is that fine ? Or should I use a NPN transistor to make the SENSE pin low to turn Off battery charging retaining the 100k pull-up resistor ? because there will be max 20V at SENSE pin and microcontroller is 3.3V type (PIC18LF26K22).

Do you mean SENSE or SHUTDOWN pin? Not understanding this paragraph between words and schematics. I'd guess from <0.25V to 0.75V = off and >0.9V to 2V = on, perhaps the micro pin could be connected directly without the need for an additional NPN. Microcontroller I/O pins can be configured as open collector output, right? (I'm an analog person).

All this said, and seeing how challenging your circuit is re working environment temperature, I'm under the impression Brian is very spot on with the SMPS advice. 125ºC operating temperature is a PD killer, and linear regulators run out of PD room for manoeuvre very quickly as ((Vin - Vout) * I out) often reaches a non-extreme operating ambient temperature PD limit.

Where possible, simulation can answer a lot of questions before wasting money on parts and time on prototypes (I should know .)

One thing I can say with complete certainty is the mantra of: "derating, derating, derating", even down to the resistor wattages - if the project isn't too penny-pinching.