Upgrading the FET's on the 18m2 High Power Board (CHI035)

jensmith25

Senior Member
I contacted Picaxe about upgrading the FET's on the 18m2 high power project board and they said it should be fine but to ask on here in case anyone has already done it and can suggest the right FET's to use. They pointed out they may need heatsinks if going a lot bigger.

Basically the PCB has 4 outputs at 1.5amp and 4 outputs at 1 amp and I just want to upgrade the 1amp ones to 1.5amp ideally.

Has anyone done this and can recommend a suitable FET? It needs the same pin layout as the existing ones.

For background info it's for a project where my customer wants to use 40 x 50mA grain of wheat bulbs spread over two outputs and so the current draw is much higher than with the LEDs I normally use. This links to the other post I've made about using a potentiometer to control flickering of the bulbs.

Thanks.
 

premelec

Senior Member
Just heat sinking the existing FETs could work - but depending on how they are configured they might need to be insulated from the sink... if they are just on or off heat dissipation will not be very high... what is the number on the FETs [type]? [I don't have schematic]
 

premelec

Senior Member
It says IRL520 on the schematic - 10 amp 100v RdsOn about .2 ohms - power to dissipate = IxIxR so at 2 amps about .8 watts - i didn't look at the data sheet free air dissipation value but likely will handle the power you are needing to - and adding a small metal 'flag' would keep the FET even cooler...
 

jensmith25

Senior Member
It says IRL520 on the schematic - 10 amp 100v RdsOn about .2 ohms - power to dissipate = IxIxR so at 2 amps about .8 watts - i didn't look at the data sheet free air dissipation value but likely will handle the power you are needing to - and adding a small metal 'flag' would keep the FET even cooler...
Thanks premelec. Sorry, I missed that on the schematic. To make sure I'm understanding you correctly - Picaxe are saying they are 1.5amp FETs but like you say, the datasheet says 10amp so you're saying that they are higher capacity and at 2amps they would still be fine?

Does that mean all the 4 FET's on the board are the same?
 

premelec

Senior Member
Yes the schematic indicated replicated circuits of the one that is shown... when you deal with power you have to take into account a number of issues - often heat dissipation is a determinant... Look carefully at the data sheet... there are specs for maximum temperatures and thermal resistance which indicates how hot the internal transistor chip will get if it has no heat sink, or infinite heat sink for the TO220 package. Also the transistor capability changes with voltage on it. That's why on the data sheet you have many graphs... The MOSFET will handle 10 amperes under appropriate heat sinking and voltage. In no heat sink conditions the chip heats up a lot more with the same current. 1.5 amp is a conservative spec for case to ambient air... IF the transistor is biased on well - if it isn't turned on well then heating can be a problem [Watts = Volts x Amps]...
 

cravenhaven

Senior Member
I meant to upload this yesterday,
I have had this CHI035 board for a few years now, but these are the original FETs.

2016-08-28 08.35.31.jpg
 

cravenhaven

Senior Member
I was just reviewing the original question and realised that you are not asking about the FETs that are onboard, but the output of the L293D optional motor driver. The Chi035 has 4 FET outputs and 4 L293D outputs. The CHI035 FET outputs are rated at 1.5A *( although the FETs themselves have a considerably higher capability) and L293D outputs are stated as 1 amp each.
As I showed in the above post the FETs are NOT IRL520 types on my board (That would be much better), but are STP14NF10 and would struggle to operate efficiently as directly driven from the PICAXE (according to the datasheet as I read it).
A quick read of the data sheet for the L293D suggests that the L293 might provide the output current that you require (2A vs 1A).
 

jensmith25

Senior Member
I was just reviewing the original question and realised that you are not asking about the FETs that are onboard, but the output of the L293D optional motor driver. The Chi035 has 4 FET outputs and 4 L293D outputs. The CHI035 FET outputs are rated at 1.5A *( although the FETs themselves have a considerably higher capability) and L293D outputs are stated as 1 amp each.
As I showed in the above post the FETs are NOT IRL520 types on my board (That would be much better), but are STP14NF10 and would struggle to operate efficiently as directly driven from the PICAXE (according to the datasheet as I read it).
A quick read of the data sheet for the L293D suggests that the L293 might provide the output current that you require (2A vs 1A).
Thanks cravenhaven. Looking at the datasheet they have updated the board so it is possible that they have also upgraded the FETs as these are clearly listed on the new datasheet as IRL520 but not listed as such on the old datasheet.

I will have to wait for the board to arrive to know for sure.

I think you slightly misunderstood. It's not the motor driver itself that I need but the two PWM pins. One is on B.3 and the other is on B.6. B.3 is served by the 1.5amp FETs but B.6 is associated with the L293D and 1 amp FETs. As I understood it, I thought I could use the two 1 amp FETs (upgraded if not IRL520 already) WITHOUT the L293D to drive the other set of lights.

Is this correct?
 

jensmith25

Senior Member
Yes the schematic indicated replicated circuits of the one that is shown... when you deal with power you have to take into account a number of issues - often heat dissipation is a determinant... Look carefully at the data sheet... there are specs for maximum temperatures and thermal resistance which indicates how hot the internal transistor chip will get if it has no heat sink, or infinite heat sink for the TO220 package. Also the transistor capability changes with voltage on it. That's why on the data sheet you have many graphs... The MOSFET will handle 10 amperes under appropriate heat sinking and voltage. In no heat sink conditions the chip heats up a lot more with the same current. 1.5 amp is a conservative spec for case to ambient air... IF the transistor is biased on well - if it isn't turned on well then heating can be a problem [Watts = Volts x Amps]...
Thanks premelec. It looks like you can get heatsinks for them easily enough if I need them. I'll have a look at the data sheet. It's 12v input voltage x 1A so should be 12Watts. I just didn't want to risk running the FET at max hence wanting to make sure the 1amp ones would be ok at a higher amperage.
 

premelec

Senior Member
Please note that you may be switching 12 watts the power the FET must dissipate is current squared times Rds ON - much lower than the power you are trying to switch - e.g. if Rds ON = 1 ohm that's going to be 1 watt in the FET at 1 amp...
 

premelec

Senior Member
Probably ok BUT you must work with actual numbers... What is the actual current being controlled? What is the maximum ambient temperature the unit will be in? Actual test that the voltage across the FET when ON times the current looks ok for temperature rise [according to FET data sheet]. The data sheet values are often 'typical' and 'maximum' - and good practice suggests a good safety margin for long life of the unit... e.g. it's best to not cycle to very hot and back a lot of times as the thermal stresses reduce the FET life - so operate the FET cooler rather than up to near it's limits. For instance if you are actually controlling 2 amps the dissipated power in the same resistance will be 4 times as much. My assumption is that you are operating the power transistor on or off and not partially on... It's good to check actual values with the numbers as well as putting you finger on the transistor to see if it's very hot... ;-0
 

cravenhaven

Senior Member
Thanks cravenhaven. Looking at the datasheet they have updated the board so it is possible that they have also upgraded the FETs as these are clearly listed on the new datasheet as IRL520 but not listed as such on the old datasheet.

I will have to wait for the board to arrive to know for sure.

I think you slightly misunderstood. It's not the motor driver itself that I need but the two PWM pins. One is on B.3 and the other is on B.6. B.3 is served by the 1.5amp FETs but B.6 is associated with the L293D and 1 amp FETs. As I understood it, I thought I could use the two 1 amp FETs (upgraded if not IRL520 already) WITHOUT the L293D to drive the other set of lights.

Is this correct?
Ah! I didnt understand that you wanted to use the PWM outputs.
In any case there are NO FETs provided by the Chi035 on port B6, this only connects to the L293D socket. The L293/D has bipolar output transistors not FETs so the voltage drop when in the ON state is at around 1.2-1.8V. As I understand from the L293(D) layout, it could be used as a driver for the string of lights as you require though the heat dissipation would need to be tested.
Assuming you had 20 bulbs in parallel @ 50mA each that would be a total current of 1A and with the ON state voltage drop on the chip of 1.2-1.8V that would be a power dissipation in the chip of 1.2-1.8 watts, which is probably OK up to around 20 degrees ambient.
As I said before, the L293 would be a better chip than the L293D because it has a higher current handling capability, but is otherwise very similar.
 

jensmith25

Senior Member
Ah! I didnt understand that you wanted to use the PWM outputs.
In any case there are NO FETs provided by the Chi035 on port B6, this only connects to the L293D socket. The L293/D has bipolar output transistors not FETs so the voltage drop when in the ON state is at around 1.2-1.8V. As I understand from the L293(D) layout, it could be used as a driver for the string of lights as you require though the heat dissipation would need to be tested.
Assuming you had 20 bulbs in parallel @ 50mA each that would be a total current of 1A and with the ON state voltage drop on the chip of 1.2-1.8V that would be a power dissipation in the chip of 1.2-1.8 watts, which is probably OK up to around 20 degrees ambient.
As I said before, the L293 would be a better chip than the L293D because it has a higher current handling capability, but is otherwise very similar.
Thank you cravenhaven. Ah, that complicates things a bit. Ok, I'll have a look at the L239 chip.
 

jensmith25

Senior Member
Following on from this it seems the L239 chip is impossible to get hold of so I'm going to have to stick with the L239D.

I have recieved the high power board this morning and I can confirm that the FETs are IRL520N.

I was considering other options ie using a different PIC chip with more PWM outputs and making up 2-3 FET circuits but the IRL520 FETs are really hard to get hold of. CPC/Farnell only have 2 in stock, Rapid don't stock them at all...

The Picaxe shop only sells the IRF520 which I've researched in not a good option. It needs to be a Logic FET so which would be best? I've seen the IRL530 and IRL540 which appear to be higher spec but there's also the IRL510 which is 5.1amp which would be plenty for what I need. I read that bigger isn't better due to power dissipation? so would the IRL510 actually be the best option?

I have bought some heatsinks so that should help whichever one is best.
 

Jeremy Harris

Senior Member
When choosing power FETs there are four main considerations, generally.

- Is the gate turn-on voltage low enough for the drive voltage you have available?

- is the Vds rating high enough for the maximum voltage you're ever likely to see across the FET (allowing for spikes)?

- is the Rdson low enough so that the power (and hence heat generated) is adequate for the heat sink or device package, in terms of keeping Tjunc well below the maximum limit?

- is the gate capacitance low enough for the gate to be charged by the available gate drive current within the desired turn-on time?

Generally there's some trade-offs here, as low Rdson often goes together with high gate capacitance, so you reduce I²R heating at the expense of switching loss. The same goes for max Vds and Rdson, generally the higher the max Vds the higher the Rdson.

In your case I'd choose a FET with the lowest gate turn-on threshold voltage you can find, together with the lowest Rdson, and then make sure that the max Vds and gate capacitance are within limits. If you're not switching these FETs on and off fast (as in a few kHz or more) then I'd not worry about gate capacitance too much, as switching losses will be a lot lower than the I²R losses due to Rdson.
 

jensmith25

Senior Member
Ok, thanks. The IRL510/520/530 are all 2v threshold voltage and 5v Rds on so those are all ok. The difference is in the Rds on ohm rating and power dissipation.

IRL510 is 0.54ohm Rds On, 43w pd, Vgs test volatge 5v, continuous drain current 5.6amp
IRL520 is 0.18 ohm Rds on, 48w pd, Vgs test volatge 10v, continuous drain current 10amp
IRL530 is 0.1 ohm Rds ON, 63 w pd, Vgs test volatge 10v, continuous drain current 15amp
 

Jeremy Harris

Senior Member
You can pretty much ignore the FET continuous current rating, as it rarely, if ever, forms a limit. 99% of the time I²R losses from Rdson and the current in the application are the limiting factor, by causing Tjunc to get too high (and heats inking is often limited by the package thermal resistance, so doesn't always provide a complete solution). I don't know what current you need per FET, but assuming it's 5A then the numbers work out as:

IRL530:
Vdson from Rdson of 0.1 ohm x 5 A = 0.5 V
I²R loss from Rdson would be 5² x 0.1 = 2.5 W

IRL520:
Vdson from Rdson of 0.18 ohm x 5 A = 0.9 V
I²R loss from Rdson would be 5² x 0.18 = 4.5 W

IRL510:
Vdson from Rdson of 0.54 ohm x 5 A = 2.7 V
I²R loss from Rdson would be 5² x 0.54 = 13.5 W
 

jensmith25

Senior Member
Its L293 NOT L239.
Element14 in the UK have 156 in stock.
Sorry, that was a typo on my part. I was looking for the right part number.

Thanks. I couldn't find any of them in a google search, only the L293D

Did you mean the L293NE? That seems to be 2A and there is 156 in stock...
 

jensmith25

Senior Member
You can pretty much ignore the FET continuous current rating, as it rarely, if ever, forms a limit. 99% of the time I²R losses from Rdson and the current in the application are the limiting factor, by causing Tjunc to get too high (and heats inking is often limited by the package thermal resistance, so doesn't always provide a complete solution). I don't know what current you need per FET, but assuming it's 5A then the numbers work out as:

IRL530:
Vdson from Rdson of 0.1 ohm x 5 A = 0.5 V
I²R loss from Rdson would be 5² x 0.1 = 2.5 W

IRL520:
Vdson from Rdson of 0.18 ohm x 5 A = 0.9 V
I²R loss from Rdson would be 5² x 0.18 = 4.5 W

IRL510:
Vdson from Rdson of 0.54 ohm x 5 A = 2.7 V
I²R loss from Rdson would be 5² x 0.54 = 13.5 W
Thanks Jeremy. I am trying very hard to understand all this so please bare with me. It's for powering filament lamps.

In reality I only need 1A, max 1.5A per FET so:
IRL530:
Vdson from Rdson of 0.1 ohm x 1.5 A = 0.15 V
I²R loss from Rdson would be 1.5² x 0.1 = 0.23 W

IRL520:
Vdson from Rdson of 0.18 ohm x 1.5 A = 0.27 V
I²R loss from Rdson would be 1.5² x 0.18 = 0.4 W

IRL510:
Vdson from Rdson of 0.54 ohm x 1.5 A = 0.81 V
I²R loss from Rdson would be 1.5² x 0.54 = 1.22 W

so which is the better option? Higher power dissipation or lower voltage Vdson from Rdson? Sorry if I'm asking the obvious!
 

Jeremy Harris

Senior Member
The best is the IRL530, as it wastes a LOT less power than the worst FET, the IRL510.

Rdson almost always gives a good indication of efficiency and overall power loss/heating of the device for any switching application where the FET isn't switched on and off rapidly. For repetitive high speed switching, then switching losses tend to dominate over the resistive loss due to Rdson, but for your application the lower the Rdson the better.

I should add that 0.1 ohm Rdson on is still very high, but acceptable for a logic-level FET at a pinch. The non-logic-level FETs I used in my electric boat brushless motor controller have an Rdson of 0.003 ohms, for example, so don't even need heat sinks in reality, even when powering the motor that's driving a 16ft long motor launch.
 

jensmith25

Senior Member
Thanks Jeremy,

I you have a better suggestion please let me know. I understood than none logic level FETs required high voltages 10-20v so logic level were best for picaxe control at 5v.
 

Jeremy Harris

Senior Member
Vgson tends to be around 7 to 8 V for non-logic-level FETs, although many have a Vgsthreshold of around 3.5 V, so you're right, you do really need a logic-level FET for direct Picaxe gate control (at low speeds).

For fast switching a Picaxe pin can't supply enough current to charge the gate capacitance of even a modest power FET, so a gate driver is needed, and usually when you have to use a gate driver the need for having a logic-level FET goes away, as you can feed the driver with the Picaxe pin output and supply it with a higher voltage, to allow the use of FETs with a much lower Rdson.

What you're looking for here is the lowest power dissipation in the FET when turned on by the voltage available from a Picaxe pin, at your maximum load current of 1.5A, and the IRL530 looks reasonable, as it only wastes 0.23 W when turned on, barely enough to get it lukewarm, I suspect.
 

Jeremy Harris

Senior Member
If you want to really reduce the losses in the FET, so reducing the heat generated when it's turned on and conducting, then the STP36NF06L is a reasonable bet. Around 50p each from Rapid Electronics, logic level gate, Rdson of 0.04 ohm, max Ids of 30 A, max Vds of 60 V.

As a comparison with the IRL530, at 1.5 A Ids, the power loss from Rdson would be just 0.09 W and the voltage across the FET when turned on would be only 0.06 V. That's just 39% of the power loss of the IRL530 at that current.
 

AllyCat

Senior Member
Hi Jen,

In #16 of a related thread you wrote:

".. I have tested the grain of wheat bulbs I have and the current doesn't spike on power up."

I think if you measure the "cold" resistance of a lamp with a multimeter it will be much less than the "expected" 60 - 150 ohms indicated by ohms law for normal operation (I'm not sure if they're 5v or 12v bulbs). I didn't pick that up at the time because if the current is limited (e.g. by the FET or power supply, etc.) then the lamp(s) will just switch on slightly more slowly (probably imperceptably so). But if you're now putting in FETs which have a much higher current capability than needed for "normal" operation then you might get some "unexpected" effects. Probably not catastrophic, but if you suddenly find the PICaxe mysteriously resetting, or a power supply "tripping out" for no obvious reason, then that might be the cause.

Cheers, Alan.
 

jensmith25

Senior Member
Hi Jen,

In #16 of a related thread you wrote:

".. I have tested the grain of wheat bulbs I have and the current doesn't spike on power up."

I think if you measure the "cold" resistance of a lamp with a multimeter it will be much less than the "expected" 60 - 150 ohms indicated by ohms law for normal operation (I'm not sure if they're 5v or 12v bulbs). I didn't pick that up at the time because if the current is limited (e.g. by the FET or power supply, etc.) then the lamp(s) will just switch on slightly more slowly (probably imperceptably so). But if you're now putting in FETs which have a much higher current capability than needed for "normal" operation then you might get some "unexpected" effects. Probably not catastrophic, but if you suddenly find the PICaxe mysteriously resetting, or a power supply "tripping out" for no obvious reason, then that might be the cause.

Cheers, Alan.
Hi Alan,

Now you've got me a bit worried as I can't afford anything like that happening if I'm selling the device. They are 12v bulbs. I've not had any issues during testing so far. What's the "cold" resistance? I powered the bulb with my 12v DC transformer and used the multimeter to measure the current flowing throught he bulb which was ~80mA as expected. No FET was involved in that case.

I'm not sure I understand why the FET will cause issues.
 
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AllyCat

Senior Member
Hi Jen,

If the bulbs are running from a separate (12 volt) supply then you probably won't have any problems (and if you did then they could be fixed).

The potential issue is that the filiament in an incandescent bulb runs literally red or white hot, where its resistance is much higher than when the bulb is not lit, i.e. cold (at room tempearture). A multimeter will read the un-lit (cold) resistance and Ohm's law indicates what "inrush" current would occur if the bulb is applied directly to a "solid" power supply, such as a rechargeable battery or a PSU with a large reservoir capacitor. With all this extra power the filiament warms up very rapidly and the current soon falls back to the normal running current.

FETs or Darlingtons, etc., rated for the "normal" current, probably will simply limit the current by dropping quite a lot of volts (until the filiament warms up), but not for long enough that the power dissipation becomes excessive. However, strictly one should check that the Absolute Maximum ratings are not being exceeded. By using highly rated FETs, you should be avoiding exceeding any device ratings, but do be aware that there might be some very large, but brief, current spikes in the system.

Cheers, Alan.
 
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fernando_g

Senior Member
Hi Alan,

Now you've got me a bit worried as I can't afford anything like that happening if I'm selling the device.
.
If you are selling the device, and want peace of mind (and repeated service calls) I would suggest that you consider auto-protected Mosfets.
I've used very successfully devices from STMicro, although Infineon, OnSemi and others offer similar devices.

http://www.st.com/content/st_com/en/products/automotive-analog-power-ics/high-and-low-side-drivers-switches/low-side-switches/vnp35n07-e.html
 

jensmith25

Senior Member
Hi Alan,

Thanks. I'm using the 18m2 High power board with single 12v power supply so the outputs are powered by the transformer and there's a voltage regulator for the picaxe chip and supply to the FETs etc.

The FETs on the board are IRL520 which have a maximum voltage of 100v and 10 amps.

The unit I've just 'sold' is running 13 grain of wheat bulbs, likely 50mA (customer doesn't know) off of 1 FET (I needed a pwm output so they're running off B.3) with a 12v 2amp DC transformer.

How do I work out the maximum possible ratings? When I read the 80mA grain of wheat bulbs I have with the multimeter from 'cold' off to 'hot' they didn't spike. I just saw them warm up from 0 to ~80mA in a second. I then assumed that all was well.

TBH, I'm much happier dealing with LEDs which I know, but the customer already has the lights and filament bulbs are still 'the norm' in dolls houses.
 

premelec

Senior Member
I think fernando_g has your answer... it's certainly possible that doll house wiring can get shorted out and so forth - somewhere in a good design current should be limited by a fuse or PTC - the power you reported is not very high [ca 1 amp] so you have a large margin for normal operation...
 

Jeremy Harris

Senior Member
If you can measure the resistance of the lamp filament when cold (with a multimeter) then do a bit of Ohm's Law, you can work out the initial turn on current. The equation is I = V (supply voltage) / R (measured lamp filament cold resistance). You will probably find that these small lamps draw around three times their working current for a fraction of a second as they turn on, with the filament resistance very rapidly rising and reducing the current to the value you measured of around 80 mA per lamp.

I'd not worry about it if using a FET with a high current rating and low Rdson for two reasons. Firstly, you're very unlikely to exceed the FET peak current rating - I doubt that you'll get more than a few percent of it. Secondly, the current peak will be so short as to have virtually no impact on heating the FET, as the energy will be very low (energy being power x time, and time will be a very short period, probably around couple of hundred ms at most).

As an example of what decent power FETs will tolerate, I have an electronic power-on switch on my electric bike, that deliberately turns the power on slowly, to limit the inrush current to the speed controller capacitors. I do this by sticking a capacitor across the gate and adding a gate series resistor, so that the FET turns on very slowly. The peak current is pretty high (in the tens of amps region) over a period of around half a second or so, but the FETs (two IRF4110 FETs in parallel) don't even get the slightest bit warm.
 
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jensmith25

Senior Member
Thanks Jeremy, premelec & Fernando.

I need to be careful about changing the FETs because the 18m2 Picaxe High Power Board is obviously designed to work. I did ask REV Ed the question and they didn't have a problem with it but the pins need to be the same.

The PCB does come with 1n4001 diodes for all the FETs so I thought this would provide some backwards protection. I may be wrong.

I will try and do the calculation though I did keep my fingers on the FET and it never even hinted at getting warm at any stage during testing.
 

jensmith25

Senior Member
I managed to measure the 'cold' resistance in ohms and got peaks of around 170 - 200ohm, dropping to about 15 ohm.

If it's in ohms and I need it in amps should it not be I(A) = V / R (ohms) ie 12v / 220 ohm = 0.05 amp??

So with 13 bulbs the max should be 0.65amp?

Is that right?
 

Jeremy Harris

Senior Member
15 ohms sounds reasonable for the cold resistance of a small low voltage bulb, Yes it should be V/R to get I, a typo on my part above (I'll edit it). The cold inrush current is likely to be around 800 mA, dropping right back to your measured running current of 80 mA, but the high current pulse will be very short, just a couple of hundred ms at most and not worth worry about, as power MOS FETs have a massive surge current capability. The likelihood is that the supply voltage will drop during switch-on of several lamps at the same time, so reducing the very short peak current, and no harm will occur.
 

inglewoodpete

Senior Member
....The PCB does come with 1n4001 diodes for all the FETs so I thought this would provide some backwards protection. I may be wrong.
The "protection" diodes will only have an effect with reactive loads (motors, solenoids, relays).

Grain-of-wheat bulbs are unnecessarily heavy current users. I gather this is for a doll's house? I would suggest using warm-white SMD LEDs: MUCH less current.
 
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jensmith25

Senior Member
The "protection" diodes will only have an effect with reactive loads (motors, solenoids, relays).

Grain-of-wheat bulbs are unnecessarily heavy current users. I gather this is for a doll's house? I would suggest using warm-white SMD LEDs: MUCH less current.
I'd love to but unfortunately, the customer already has the bulbs as I did mention earlier. Unfortunately, grain of wheat or similar bulbs are still the most commonly used lighting in dolls houses.
 
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