Saturating a MOSFET with a constant current sink driver - Page 1
Forum section: electronics / projects

author: Oleenick
author_rank: Contributor
timestamp: December 29, 2019, 11:30:27 am
content: Hello EEVBlog forum,I'm designing an LED cube and for its LED drivers I'm using TLC5927 16 Channel constant current sink drivers (https://datasheet.lcsc.com/szlcsc/1804162351_Texas-Instruments-TLC5927IDBQR_C181418.pdf).I was wondering if I could use the constant current sink outputs of the TLC5927 to drive a P-channel MOSFET (for the ~7amp max layer current).As per the diagrams bellow I just want to use some outputs from the TLC5927 to drive the logic level of the gate of the MOSFET.Intuitively it seems that it would work fine in my head, but I'm not experienced enough to know if there will be any problems saturating a MOSFET with a constant current sink driver.Thanks in advance!

author: Wimberleytech
author_rank: Super Contributor
timestamp: December 30, 2019, 07:20:49 pm
content: I think you want to exchange the source and drain connections.

author: moffy
author_rank: Super Contributor
timestamp: December 31, 2019, 04:28:50 am
content: If you are looking at 7A current, then you would tend to use a switchmode constant current source to limit power dissipation in the switching device.

author: Oleenick
author_rank: Contributor
timestamp: December 31, 2019, 08:58:26 am
content: Quote from: moffy on December 31, 2019, 04:28:50 amThat makes sense, but in the design I'm already using these constant current sink drivers for 512 LEDs (30mA sink per channel).I don't have much experience yet, so would you say it's worth it to implement switchmode constant current source ICs for the 8 anode control point (the ~7A points)? This is just for a one off project.Quote from: Wimberleytech on December 30, 2019, 07:20:49 pmAt the moment the top pin the the drain and the bottom is the source on the MOSFET in that diagram. When it is saturated I intend for the LVL0 net to be placed at a +5V potential. I think I see what you mean, that the drain pin should be where the current is "drawn" from through source. But then that diode on the diagram shows that current flows from drain to source, or am I wrong assuming that? Could you elaborate on this to help me out of my confusion.For reference I based it on this design from Kevin Darrah with the same component. Maybe the symbols have the D and S in different places?Thanks.

author: mikerj
author_rank: Super Contributor
timestamp: December 31, 2019, 12:36:26 pm
content: Quote from: Oleenick on December 31, 2019, 08:58:26 amThe direction of the body diode on your MOSFET should give you a big clue something is not correctYou have a P channel device, so the gate must be at a lower potential then the source to turn it on. This means you need the source pin connected to the supply rail, and the gate pin then gets pulled toward 0v to switch the transistor on.

author: Wimberleytech
author_rank: Super Contributor
timestamp: December 31, 2019, 01:28:26 pm
content: Quote from: Oleenick on December 31, 2019, 08:58:26 amFirst, see mikerj's comment.I will elaborate.A MOSFET is fundamentally a four-terminal device: Drain, Source, Gate, Body.The Body terminal is also called: substrate, bulk, well, tub(rare...but I can think of one company that used this term).A MOSFET is a "majority carrier" device. For an nch MOSFET, the majority carriers are electrons, for pch, they are holes.In order for current to flow from source to drain (or drain to source), a channel must be created...a path for majority carrier current to flow.For a pch device, the Body (substrate) is lightly-doped n-type silicon and the drain/source are heavily-doped p-type silicon.In order to create a channel, the lightly-doped substrate must be "inverted" (in simple terms, converted to p-type) so that majority carriers (holes) can flow.An appropriate potential "relative to the substrate" must be applied to the gate in order to form the inversion layer. For a pch MOSFET, this voltage is negative with respect to the substrate terminal. Therefore, the gate voltage must be lower in potential than the substrate terminal. Moreover, you must also consider the potential of the drain/source terminals. At the device level for conventional MOSFETS (no DMOS or asymetric MOSFETS), there is no physical distinction between drain and source. They are interchangeable. But, in order for the first carrier (electron or hole) to leave the source terminal, the region at the edge of the source must be inverted. Thus, for a pch device, the source must be at a higher potential than the gate. Either physical drain/source terminal can serve as the source depending solely on terminal conditions (voltage relative to the other terminals).OK this may be a lot to absorb. Here is a simple rule for a four-terminal pch MOSFET.1) the substrate voltage must be equal or greater than the other to conducting terminals (not the gate terminal)2) the source terminal is the terminal with the highest voltage of the two conducting terminals3) the drain terminal is the terminal with the lowest voltage of the two conducting terminals.In the case of the transistor you used, the substrate is tied internally to one of the other terminals. By applying the rule above, you will see that the terminal attached to the substrate must be the source terminal.

author: Oleenick
author_rank: Contributor
timestamp: January 01, 2020, 11:29:55 am
content: Wow thank you all for taking the time to reply. Your explanations are great, though it may take a few reads to fully understand. For now atleast I know that I need to flip my source and drain connections.If I understand correctly: CLVL0, when pulled to GND by the constant current sink, should saturate the MOSFET and allow current to flow from source to drain, source should be connected to VCC and drain should be connected to the anode of each LED layer, which is then connected to ground through the LEDs.I realised what tripped me up was that I made this part of the circuit based on Kevin Darrah's design, which I attached before. In his diagram the MOSFET drain and source are flipped from the schematic that I used in EasyEDA. Trap for young players indeed.Would you have any idea if using a constant current sink would be an issue on the gate?

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 01, 2020, 09:00:51 pm
content: Quote from: Oleenick on January 01, 2020, 11:29:55 amSome additional theory and terminology for you.MOSFETs can operate in two regions when conducting: saturation, and linearFor a pch device, If the gate to drain voltage is negative by at least the threshold voltage, the device is operating in the linear region. Otherwise, it is in saturation.If you pull CLVL0 to ground, the transistor will turn on likely be in the linear region of operation. That is fine for this application. You just want it to be fully on.QuoteNot an issue.Note: I ignore the subthreshold region of operation in my comments above. It is yet another region but not of concern for this application.

author: Oleenick
author_rank: Contributor
timestamp: January 03, 2020, 10:38:10 am
content: Quote from: Wimberleytech on January 01, 2020, 09:00:51 pmSorry to keep dragging this on but this is important for me to understand.Would it be worth using another discrete transistor to switch the P-channel MOSFET from the microcontroller?The P-channel that I was going to use has 100mΩ RDSon which is capable of -10V -10A, but the power dissipation is a bit lacking since I'd be running max 7amps:P = R * I^2= 0.1 * 7^2= 4.9 wattsThe Ta (ambiently cooled) @ 25C rating for the IRF9Z34NSTRLPBF is only 3.8 W. Even if I give it a good copper plane to cool, there's going to be 8 of them in opperation at once on the PCB.I know Kevin Darrah used the THT version of the MOSFET with no problems, maybe because they're being switched at a 1/8 duty cycle (which is also what I'm doing)?Or should I use an N channel MOSFET for its lower Rds on, it may simplify things and reduce the power wasted as heat.Any advice would be greatly appreciated.

author: mikerj
author_rank: Super Contributor
timestamp: January 03, 2020, 12:38:37 pm
content: Quote from: Oleenick on January 01, 2020, 11:29:55 amIt will be fine, provided:1) R15 is a high enough value that the current sink output saturates. No problem with a 30mA current set point.2) The saturated voltage is sufficient to fully turn on the MOSFET. No problem, datasheet characteristics suggest you will be within ~100mV of 0V at 5mA.3) The zero current gate voltage is close enough to Vcc to fully turn off the MOSFET. Leakage is only 1uA max so again no problem.

author: Oleenick
author_rank: Contributor
timestamp: January 06, 2020, 10:40:18 am
content: Quote from: mikerj on January 03, 2020, 12:38:37 pmI'd just like to clarify my understanding of your help.1) Do you mean that the constant current from CLVL0 will flow through R15 (30mA), which will make the gate see a low enough voltage to saturate the MOSFET?2) Do you mean that the datasheet suggests that for the transistor to turn on it must be within less than + or - 0.100mV?3) So when CLVL0 is high, the current through R15 to VCC will be low enough to turn off the MOSFET?I've revised the circuit but I'm considering switching to an N-channel MOSFET which has better thermal characteristics.Should I be using a resistor on the gate (between CLVL0 and gate) even though I'm using a constant current source?I'm not sure if thats how it works, my understanding is that it should reduce oscillation on the edges of the switching singal to the gate.Because of my current P-channel MOSFET RDSon value of 0.1ohm I'd like to find something less than 20mOhm RDSon.I don't really need it to be P-channel.I'm struggling to source a nice MOSFET from LCSC. Something close to the CSD18542KTT from TI would be good but I can't find anything like it on LCSC.EDIT: I think I found a good one from LCSC, herehttps://datasheet.lcsc.com/szlcsc/1808281537_Infineon-Technologies-IRLS3036TRLPBF_C73577.pdfIt's an N-channel logic level device, which has a very low RDSon max at 2.4mOhm and a PD max of 3.75W while my load will dissipate only 2.4W

author: mikerj
author_rank: Super Contributor
timestamp: January 06, 2020, 11:47:02 am
content: Quote from: Oleenick on January 06, 2020, 10:40:18 amCorrect. However the actual current flowing through R15 may be limited by the supply voltage rather than the TLC5927 target current, in which case it's output will saturate somewhere close to 0v. For some reason I thought that Vcc was 5v, but checking again I'm not sure you specified this value and it is important. Not only does Vgs need to be high enough to ensure the MOSFET switches on, it also has to be below the maximum Vgs permitted by the MOSFET or you will kill it almost instantly.Quote from: Oleenick on January 06, 2020, 10:40:18 amNo, I simply mean that Vgs must be high enough to fully saturate the transistor at the drain current you require. Vgs in this case is the difference between Vcc and the output of the TLC5927. The 100mV was based on my (faulty) assumption that Vcc was 5v giving a maximum current into the TLC5927 of 5mA (5v/1k), again you need to specify this value if it's something else.Quote from: Oleenick on January 06, 2020, 10:40:18 amCorrect, with the output of the TLC5927 switched off, the voltage applied to the gate must be (significantly) lower than Vgs(thr) to ensure the MOSFET is fully turned off.Quote from: Oleenick on January 06, 2020, 10:40:18 amPlease turn the transistor around rather than the voltage rails! Most people expect the higher potential rail to be at the top of the schematic and this makes reading the schematic easier i.e. conventional current flow from top to bottom.[ Attachment Invalid Or Does Not Exist ]Quote from: Oleenick on January 06, 2020, 10:40:18 amShouldn't make any difference in this case as the impedance of the output during transition is determined by R15. If Vcc is high enough that the Vgs is close to or exceeding the maximum for the device chosen then you will have to add additional components to limit it. This could be as simple another resistor to form a potential divider with R15.

author: Oleenick
author_rank: Contributor
timestamp: January 06, 2020, 11:54:02 am
content: Quote from: mikerj on January 06, 2020, 11:47:02 amThank you thats very valuable help.I am using +5V as VCC, I should have stated that earlier.Also I think I will be going with the IRLS3036TRLPBF N-channel MOSFET now anyways because of its very low RDSon.That will fix my thermal issues as I'd like it to be capable of a continuous current of each layer just incase the program hangs during switching between the 8 levels.There's the schematic that I would follow with the new MOSFET. The resistor to ground would be 620ohms and the capacitor 100uF.In this case should I use a resistor on the gate?

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 06, 2020, 02:18:30 pm
content: Quote from: Oleenick on January 06, 2020, 11:54:02 amThe threshold voltage for that transistor is min of 1v and max of 2.5 volts. Therefore, your output voltage will be a maximum of about 4 volts and a minimum of about 2.5 volts.A gate resistor is not necessary.

author: Oleenick
author_rank: Contributor
timestamp: January 06, 2020, 11:16:34 pm
content: Quote from: Wimberleytech on January 06, 2020, 02:18:30 pmThat is potentially problematic. Do you mean that Vds (output voltage) can only be 2.5-4V?Where did you find that on the datasheet?Or by output voltage do you mean gate voltage?This figure from the datasheet shows that at a voltage of ~3V Id can be 10A, which would be enough for my application.Do you mean that the gate voltage needs to be between 2.5 and 4V to achieve the current I need?

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 07, 2020, 12:01:29 am
content: Quote from: Oleenick on January 06, 2020, 11:16:34 pmYour latest circuit is a source follower. Therefore, the source will be VDD-VGS and VGS will be what it needs to be for a given current. So, from the graph you posted, at 10A, VGS will be ~3V. The Gate voltage maximum is 5V, so the source voltage will be 5-3=2V. Are you happy with the output voltage being 2V? If so, you are fine.

author: Oleenick
author_rank: Contributor
timestamp: January 07, 2020, 12:08:28 am
content: Quote from: Wimberleytech on January 07, 2020, 12:01:29 amBy the way thank you so much for your help.Is there a way that I can make the circuit with this IRLS3036 MOSFET that will output 5V (max 10A)?Is that just not possible because this MOSFET is designed for fast switching operation?

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 07, 2020, 12:20:00 am
content: Quote from: Oleenick on January 07, 2020, 12:08:28 amYou need to provide more information. What are you driving that needs 10 amps at 5 volts? The TI part is driving all of the LEDs, right?

author: Oleenick
author_rank: Contributor
timestamp: January 07, 2020, 12:36:47 am
content: Quote from: Wimberleytech on January 07, 2020, 12:20:00 amI'll be using 8 of the MOSFETS to control each of the 8anodelayers of the LED Cube. So each MOSFET will have a duty cycle of 1/8.So you could call these the high side drivers for the LEDs. The TI TLC5927 are constant current sink drivers so they will be on the low side controlling 192 cathodes per layer.I'd be running the LEDs at an absolute max of 50mA (if 30mA is not bright enough): 192 * 0.05 = 9.6 or ~10A.If 30mA is bright enough, which I'm thinking it will be, the current required will be ~6A (I miss calculated and said 7A before).I'd like the MOSFET to be capable of a constant load of 10A at 5V so if the code breaks and the layer stays on for greater than 1/8th second, no overheating will occur.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 07, 2020, 01:28:56 am
content: Quote from: Oleenick on January 07, 2020, 12:36:47 amI think you need a PCH as in your first design. Here is one (Fuji 2SJ475-01) that can handle the power dissipation and the current. Looks like this is discontinued but seems to be still available.Here is another that may be just on the edge w/r to power dissipation:https://www.mouser.com/datasheet/2/427/siss05dn-1595673.pdf

author: Oleenick
author_rank: Contributor
timestamp: January 07, 2020, 02:42:47 am
content: So for a high side driver a PCH MOSFET should be used?Sorry to keep you explaning things, but before I jump to conclusions, why should this be a PCH?Also as noted before I should flip the MOSFET to have the +5V on top of the diagram

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 07, 2020, 03:11:51 am
content: Quote from: Oleenick on January 07, 2020, 02:42:47 amWell, you are driving the anode side of the LEDs toward 5V. If all you have is 5 volts, an NCH device cannot drive the LEDs to a high enough voltage. A PCH can.

author: Oleenick
author_rank: Contributor
timestamp: January 07, 2020, 04:39:34 am
content: Quote from: Wimberleytech on January 07, 2020, 03:11:51 amAhh i see. I've read over everything again and I understand why I'd need a PCH device.Because an NCH device is only useful for driving a load at a higher voltage than the gate (logic level) can supply.And because the Vgs is the voltage across gate and source.So if I have 5V on an NCH gate and the source is at 0V: Vgs = 5V-0V = 5V (NCH MOSFET on) ... 5V then goes from drain to source, Vgs = 5V-5V = 0V (NCH MOSFET off): So an NCH will never stay on.Whereas with a PCH MOSFET, Vgs = 0V-5V = -5V (PCH MOSFET on greater than -2.2V (SiSS05DN)) ... 5V then goes from source to drain, Vgs 0V-5V = -5V the PCH MOSFET is still on.Does my understanding make sense?Thanks again.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 07, 2020, 01:36:30 pm
content: Quote from: Oleenick on January 07, 2020, 04:39:34 amNot sure what you are trying to say here...I think you are still a little confused.QuoteCorrectQuoteNo. First of all voltage does not go anywhere...it is "across." If the NCH transistor is on, it will conduct as much current as it can based on its physical properties and external circuit elements.QuoteCorrectQuoteAgain voltage does not go anywhere. Current will flowQuoteYou are getting closer.

author: Oleenick
author_rank: Contributor
timestamp: January 07, 2020, 10:07:36 pm
content: Quote from: Wimberleytech on January 07, 2020, 01:36:30 pmI was trying to say that I an NCH device is only useful if another, higher voltage is availible. For example, switching 10V with 5V signal, would that have a Vgs (off) = 0V-0V = 0V and a Vgs (on) of 5V-10V = -5V?. Is it right that I have that -10V on the end there? I'm assuming that when the gate is at 5V the source is at 10V.QuoteSo apart from my irresponsible description of voltage (thanks I'll make sure to correct it in future), its just that I assumed that source will be at 5V after saturing the MOSFET given the load. In a previous case, does that mean if the MOSFET was on and 10A of current started flowing from drain to source, the voltage across drain and source would be ~3V?Thank you for continuing to help.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 07, 2020, 10:32:19 pm
content: Quote from: Oleenick on January 07, 2020, 10:07:36 pmDidn't mean to be pejorative.Think about this. Imagine an NPN BJT with the collector at 5 volts and the base at 5 volts. With some amount of current flowing collector to emitter, will the emitter voltage ever be five volts? MOSFETS are a different beast, but similar in this way.If the gate and drain of an NMOS transistor are connected to five volts, the source can never rise above 5V-Vth.If you know how to use LTSpice, I will generate a simulation for you to examine

author: Oleenick
author_rank: Contributor
timestamp: January 08, 2020, 07:33:07 am
content: Quote from: Wimberleytech on January 07, 2020, 10:32:19 pmNo no its fine I didn't take it that way. I'm glad you corrected me.Quote from: Wimberleytech on January 07, 2020, 10:32:19 pmThe emitter voltage would depend on the current flowing collector to emitter. So the voltage across base and emitter would be influenced by the current flowing collector to emitter, which is governed by the transistors RDSon. I think?Quote from: Wimberleytech on January 07, 2020, 10:32:19 pmI do know the basics of LTSpice, but I just "simulated" it on a breadboard with a PMOS and NMOS.Measuring voltage across source and ground (the load should be at 5V).NMOS (load is not at 5V):PMOS (load is 5V):

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 08, 2020, 02:51:19 pm
content: Please draw what you are measuring...like what I show here.

author: Oleenick
author_rank: Contributor
timestamp: January 09, 2020, 01:56:02 am
content: Quote from: Wimberleytech on January 08, 2020, 02:51:19 pmHere are the two circuits I was testing, minus the capacitor which I didn't add to the breadboard.My voltage measurements were across the blue and red nodes on each NCH and PCH diagram.

author: Oleenick
author_rank: Contributor
timestamp: January 12, 2020, 05:44:07 am
content: I think I've finally found a nice MOSFET. If you're curious (https://au.element14.com/infineon/ipb120p04p4l03atma1/mosfet-p-ch-40v-120a-to-263-3/dp/2480803)I've been looking more into MOSFETs, and I can see clearer now what's been described to me here.Thanks for all the support.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 12, 2020, 01:41:54 pm
content: Quote from: Oleenick on January 12, 2020, 05:44:07 amYes, that will work nicely.