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I'm trying to figure out a way to optimize the power consumption of a group of two high-side switches. They are not supposed to be switched more than once every minute.

I don't know if my way to connect them is right. I have two loads (both constituted of step-down and step-up voltage regulators, electronic speed controllers, etc). The goal is to command them by manual switches. There is a pull-down resistor on each MOSFET's gate (if one of the switch has a malfunction, the loads are still working).

If the load 2 is turned off, load 1 should be automatically turned off (thanks to the diode).

Loads are between 10 and 60 amps. (The load values are for illustrative purpose)

Schematic

I have several questions:

  • I chose P-Channel MOSFETs (IPD042P03L3), they probably increase the power loss compared to a N-Channel MOSFET. Is there another "easy" way to have the same functionnality without P-Channel MOSFETs ? In the datasheet, they say the Rds(on) is about 4.2 mOhms for Vgs = 10 V (Vgs should be higher that 10 V in this circuit), that seems a decent value to me, am I right ? If my calculation is right, the power loss would be about 0.25 W for a load of 60 Amps.

  • I decided to use the same MOSFET for both loads, the reason is that I didn't see any other P-Channel MOSFET with a lower Rds(on) on Digikey. Should I use a MOSFET rated for a smaller current ? My guess is that the gate charge could be smaller with a smaller MOSFET, but does that really matter if I only switch once in a while. Is there any parameters that does matter in my case ?

  • The pull-down resistor is about 1k. If I increase it, that should reduce the power consumption but how do I know how far I can go ?

Thank you in advance,

Please feel free to comment the design !


EDIT 1

I updated the design with the P-Channel MOSFETs that Russell was talking about.

The voltage source is a 4-cell LiPo battery (voltage goes from 12.8 V to 16.8 V), I use a Zener diode to keep the minimum value of Vgs at around -10 V regardless of the voltage source.

The Zener diode I chose is the MMBZ5240BLT1G. When switch is open, the current through the Zener diode is between 0.9 mA (Vsource 12.8 V) and 2.3 mA (Vsource 16.8 V). Power dissipated by the zener diode: between 9 mW and 23 mW (far below 225 mW).

With this design, Rds(on) should stay below 2mOhms... Power loss would be about 7 W for a maximum load of 60 Amps (load values on the schematic are for illustrative purpose only !).

New design v2

Keep posting your comments on the design ! Thank you very much !


EDIT 2

I'm thinking about another aspect of this design.

In the case the MOSFETs are turned off during a long period of time (over several months).

Since the system is powered by a battery, it may be a good idea to reduce the quiescent current to a negligeable value.

For example, in the previous schematic, the current needed by the resistors when the switch is closed is about 2x (16.8 V / 3000 Ohms) = 11.2 mA. For a 4-cell battery of 5000 mAh, that means the battery will be dead after 19 days (and probably much less than that).

I don't think I will implement it in the first design of this board but how is it possible to reduce the quiescent current ?

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I recently received 4 of these :-) !!!
I haven't tried them yet.
IPB180P04P4L-02 :-) :-) :-) :-)
IPB180P04P4L-02 datasheet here P Channel MOSFET, 40V, 180A,
1.8 milliOhm typical at 100A (!) with Vgs = 10V at 25 C.
2.4 milliOhm max same conditions.
At Vgs = -10V Rdson is about ruler straight through 180- Amps.
It will be higher at higher temperatures. Pricing here $6.38/1 at E14 / Farnell and less in volume
$2.75/1 Mouser - 6000 in stock.
$1.31/1 Avnet !!!, - no stock.


Drive resistors are non critical for occasional on/off. With 1K you'll get maybe 10's of uS switch time worst case but minimal loss overall.

BUT Vgs_max matters. For the IPB180P04P4L-02 it's 16V max, so a 10V clamp (zener) woul;d be in order if using more than 10V drive. (You usually get to exceed Vgsmax only once per MOSFET lifetime.

Check data sheet for your FETs.


Your loss calculation has gang aglae.

4.2 mOhms for Vgs = 10 V ... If my calculation is right, the power loss would be about 0.25 W for a load of 60 Amps.

Power = I^2 x R = 3600 x 0.0042 =~ 15 Watts.

For the IPB180P04P4L-02 at 60A at say 2 mR that's ~= 7 Watts.
My package is the TO263-7-3.
3 n+ 1 cut off + 1 in TO263 pkg with bent pins for SMD.
Tjc = 1 K/W = 7 K rise at 7 Watts. You want some sort of semi real heatsink for this at say 10 Watts to be safe.
At 5 C/W (easily enough achieved that's about 60C rise for 10W.
Lower C/W heatsink = cooler.


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  • 1
    \$\begingroup\$ Power = I^2 x R = 3600 x 0.0042 =~ 15 Watts not 0.042 \$\endgroup\$ – PlasmaHH Feb 19 '15 at 9:27
  • \$\begingroup\$ @PlasmaHH Indeed. Thanks. Calculation was right, what I wrote was entypo'd. You are welcome to correct such manifest errors when seen if you want to. Unlike some things which may be contentious and annoy people if corrected, something like this is very clearly an error. \$\endgroup\$ – Russell McMahon Feb 19 '15 at 13:19
  • \$\begingroup\$ I am still lacking the reputation to do these kind of changes. \$\endgroup\$ – PlasmaHH Feb 19 '15 at 13:41
  • \$\begingroup\$ Thank you for correcting my loss calculation, that was a mistake. What do you actually call a "semi real heatsink" ? If I understand correctly, the best way to reduce the C/W is to solder a heatsink on the PCB (link). I don't see any good reason not to use the MOSFET you are talking about. I will redo a schematic with your MOSFET and zener diode to limit Vgs. Thank you ! \$\endgroup\$ – Marmoz Feb 20 '15 at 3:48
  • \$\begingroup\$ @PlasmaHH - +40 on rep :-) (+1 score on 4 out of 5 of your answers) \$\endgroup\$ – Russell McMahon Feb 20 '15 at 8:47
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"They are supposed to be switched more than once every minute"

If they are only being switched at such low rates, then the only thing you should worry about is Rds_on. This is always inherently higher in P-channel FETs. Try to design your circuit with low-side N-FETs which you will find have lower Rds_on values for similar package/technology/capabilities.

The main losses from using FETs as switches is the Rds_on for DC style currents, and the switching loss when they are being switched very fast, usually with PWM methods.

You are not switching often enough to worry about switching loss, and the 1K resistor is okay, but could be much lower if you wanted.

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  • \$\begingroup\$ I thought about low-side MOSFET but I really need to have all my load connected to ground at all time. About the 1K resistor, is there any way to know what is the highest value of resistor I can use ? \$\endgroup\$ – Marmoz Feb 19 '15 at 2:45
  • \$\begingroup\$ @Marmoz in that case, then you are stuck with P-fets. For the resistor, it's a simple capacitor (at the FET's gate) discharge equation through a resistor, where the speed of turning on/off the FET is determined by how fast the gate is charged/discharged. Use a SPICE simulator and have a capacitor at full starting voltage (whatever that is) and then connect it to ground through a resistor of whatever value, and you shall see the time taken. Otherwise, find some fancy equations and make a spreadsheet for resistor values and pick a nice one. Power lost is the highest during switching! less time! \$\endgroup\$ – KyranF Feb 19 '15 at 3:01
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First off, your choice of p-type will not make a great deal of difference in the power dissipated. The reason? The unit you've chosen has a nominal Rds(on) of .0042 ohms, and you will have a great deal of trouble making a PC board with trace resistances less than this. Handling these currents will be a challenge if you don't have any experience.

Next, your power calculation is off. Your calculation (.0042 x 60) gives you the voltage across the MOSFET when it is on. Power is this voltage times the load current, or about 15 watts. Actually, assuming your load resistances are 3:1, your worst-case current for a single MOSFET is 45 amps, and the power will be ~ 8.5 watts.

Assuming your load impedances are correct, your supply voltage is in excess of 20 volts, and your gate drive will exceed the maximum Vgs of 20 volts. This also implies that your load power is in excess of 1.2 kW. You haven't told us why you need to minimize MOSFET power. If you're worried about overall efficiency, you're really wasting your time. A MOSFET power dissipation of ~ 10 watts is less than 0.1% of your system total power.

Likewise, the power dissipated in your gate pulldown resistors is insignificant (~ .4 watts per resistor at 1 k). Increasing the resistance to 10k will drop this to .04 watts, and this is so much less than the MOSFET power dissipation that it makes no sense to worry about it.

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  • \$\begingroup\$ I understand that the pcb layout will be a challenge. That may take some time but I will probably have questions when I will get there. OK for the power calculation, that's a mistake. I agree with your result. Unfortunately, I don't understand when you say "your supply voltage is in excess of 20 volts, and your gate drive will exceed the maximum Vgs of 20 V". It seems to me that the supply voltage is about 17 volts maximum, it shouldn't go above the maximum Vgs. I understand for the pulldown resistor, it doesn't cost me anything to pick a higher resistor. It is a system powered by a battery. \$\endgroup\$ – Marmoz Feb 20 '15 at 3:39
  • \$\begingroup\$ I said that I was assuming your load values were correct. With a .5 and a 1.5 ohm load, and 60 amps draw, you need 22 volts. You don't mention duty cycle. Let's say 5 seconds on once per minute. 17 volts x 60 amps x (5/60) gives an average power of 85 watts. Then a 1k resistor will dissipate 25 mW, and have essentially no effect on battery duration. \$\endgroup\$ – WhatRoughBeast Feb 20 '15 at 4:25
  • \$\begingroup\$ The load values are not correct. The source is a 4-cell battery. The loads are between 10 and 60 amps. The resistors load 1 and load 2 were for illustrative purpose, sorry for the confusion. I use this MOSFET as a way to connect and disconnect a load (from the power supply). Is it appropriate to talk about duty cycle in this case ? \$\endgroup\$ – Marmoz Feb 20 '15 at 5:58
  • \$\begingroup\$ I use this MOSFET as a way to connect and disconnect a load (from the power supply). So, I will connect it for a minute, then disconnect it for another minute or two, connect it again, etc. \$\endgroup\$ – Marmoz Feb 20 '15 at 6:13
  • \$\begingroup\$ Oh yes, it's not only appropriate to talk about duty cycle, it's critical. What is the amp-hour rating of your batteries? Let's say it's 100 H-hr. Then running 60 amps continuously will discharge the batteries in 1.6 hours. If you run for 1 minute on, 1 minute off, your duty cycle is 50%, and your batteries will last 3.2 hours. This assumes that a complete discharge will not harm the batteries, and for regular lead-acid cells this is not true. \$\endgroup\$ – WhatRoughBeast Feb 20 '15 at 16:14

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