# Multiple mosfet driving

#### Kevin Jones

Feb 2, 2017
5
Hello,

I have a PV charge controller which drives 4 50A mosfets to supply current to charge batteries in a small power station. I wish to modify it to handle larger current, by increasing the number of mosfets to 16.

Currently, the 4 mosfets are driven by a single emitter-follower circuit, through 10K resistors, one connected to each mosfet gate. eg, the emitter of the driver transistor is connected to 4 10k resistors, each resistor being connected to the gate of each mosfet.

My question is, can I simply connect the gates of the additional mosfets in the same way, to the same emitter-follower driver? So then I would have a single emitter-follow output connected to 16 10k resistors, each one connected to the gate of a mosfet. Or will I need to add additional emitter-follower driver circuits? It would seem to me that additional drivers should not be necessary, since the gate impedance of the mosfets are extremely high.

Also, I am curious what the purpose of the 10k resistors are.

#### duke37

Jan 9, 2011
5,364
I think that they can all be connected to the same emitter follower.
The gate resistor is said to damp radio frequency oscillations but I have not had this trouble without the resistor.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
10k gate resistors? Are you sure? Could they be 10Ω?

If they're really 10k then you should be able to drive more gates from the same source. This would also imply that the switching speed is very slow and this switching is infrequent.

#### Kevin Jones

Feb 2, 2017
5
~ 1KHz switching.

Thank you all very much.

Kevin

#### BobK

Jan 5, 2010
7,682
You really need a push pull circuit to ensure that the MOSFETs do not stay in the linear region long. If you are using 10K and a single sided driver, it is probably dissapating mich more power than it should currently.

Bob

#### Kevin Jones

Feb 2, 2017
5
Hmmmm, I would think at 1KHz time in the linear region would be a very small percentage.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Can you satisfy our curiosity by telling us the part number of the MOSFETs and if possible the supply voltage to the emitter follower and it's emitter resistor value.

It should be relatively straightforward to calculate the switching losses

Feb 2, 2017
5

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
I can do calculations based on 10k, but an emitter follower has a resistor connected between the emitter and a supply rail, with the collector connected to the other supply rail.

I'll go ahead with some very generous assumptions, but they may not be valid without knowing more about the driver circuit.

#### Chemelec

Jul 12, 2016
291
10K Seems VERY LARGE.
10 to 100 Ohms is Better.

Ideally each mosfet should have a Low Value (0.1 Ohms) SOURCE RESISTOR, to even out possible difference between the mosfets

#### Kevin Jones

Feb 2, 2017
5
10K Seems VERY LARGE.
10 to 100 Ohms is Better.

Ideally each mosfet should have a Low Value (0.1 Ohms) SOURCE RESISTOR, to even out possible difference between the mosfets

I am just stating what the current design is. This is a product I am modifying. I don't know the reason for using 10K in the design (I double-checked, and that is the value), but I suppose I could use smaller value resistors.

Typical current through these will be around 20 - 25A. With 0.1 Ohm source resistor, that's a 2 volt drop, which would be unacceptable.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
I realised that the original post doesn't contain details about the actual current being switched or the voltage across the mosfets when they are off. I'll make some assumptions and the OP can substitute real values into my equations.

Ids = 80A / 4 = 20A (assuming 80A total across 4 mosfets)
Vds = 3V
f = 1000Hz
duty = 100% (d = 1) This is the worst case for calculation
Rds = 0.018Ω
Q = 60nC
Vgs = 12V (the voltage available from the gate driver)
Vgs(th) = 50% Vgs (for ease of calculation)
Rg = 10kΩ

Firstly the power dissipated while the mosfet is ON (static power dissipation):
Pstatic = Ids * Ids * Rds * d = 20 * 20 * 0.018 * 1 = 7.2
The static power dissipation is 7.2W

Next the power lost while the device is switching (dynamic power dissipation):
Pdyn = Pon + Poff (power is lost while switching on and while switching off)
Pon = 1/2 (Vds * Ids) * Ton * f (Ton is the time spent switching on)
Poff = 1/2 (Vds * Ids) * Toff * f (Toff is the time spent switching off)

Ton = Q/Ig(on) (time is charge divided by current)
Toff = Q/Ig(off)

We assume a constant current because the majority of the switching occurs while Vgs is at Vgs(th)

Ig(on) = (Vgs - Vgs(th)) / Rg (assuming that the gate is pulled to Vgs)
Ig(off) = Vgs(th) / Rg (assuming that the gate is pulled to the source potential)

Using the assumption Vgs(th) = Vgs/2, Vgs - Vgs(th) = Vgs(th) and thus Ig(on) = Ig(off), and Ton = Toff, and Pon = Poff,

Pdyn = (Vds * Vgs) * Poff * f (because (Pon + Poff)/2 = Poff)

Ig(off) = 6 / 10000 = 0.0006A (0.6mA)

Toff = 6E-8 / 6E-4 = 1E-4 (0.1ms)

Pdyn = (3 * 20) * 1E-4 * 1000 = 60 * 0.1 = 6W

Thus the dynamic power is 6W

The total power is 6 + 7.2 = 13.2W per mosfet

If the driver (which is already supplying 2.4mA) can supply an additional 3.6mA, then you will have sufficient drive for the additional mosfets.

Note that these calculations are based on the assumption that the driver has zero impedance along with the other voltage and current assumptions above.

The fact that I have found the static and dynamic power to be roughly equal is a reasonable sign that this set of assumptions could bear some relationship to a reality somewhere, perhaps even yours. It would be unusual to find an actual, well designed high power case where the static and dynamic power were more than an order of magnitude apart given that the device is continually switching.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Ideally each mosfet should have a Low Value (0.1 Ohms) SOURCE RESISTOR, to even out possible difference between the mosfets

The beauty of mosfets is that they (source resistors) are not required. Rds increases with temperature. As long as the devices are reasonably matched, they will share current quite well.

#### Fish4Fun

##### So long, and Thanks for all the Fish!
Aug 27, 2013
481
I would suspect the large value gate resistors are to limit di/dt and are an important part of the circuit. I would also ass-u-me that the switching frequency is VERY low, perhaps as low as 0.01hz to 1hz if the mosfets are being used used to switch the charging current on/off. I am ass-u-me-ing that during the off duty cycle the battery voltage is being "read" by an ADC, and that some "settling time" is required by the batteries (chemical processes in the time domains more generally associated with mosfets are AKA: 'explosions' ;-) ).

A word of caution to the OP about the proposed modifications: BE VERY CAREFUL you FULLY UNDERSTAND the operation of the charger before proceeding!

All batteries:
Code:
Lead-Acid

Zink-Carbon aka Alkaline batteries

Nicklel Batteries:
Nickel-metal-hydride aka NiMH,
Nickel-iron aka NiFe,
Nickel-zinc aka NiZn,
Nickel-hydrogen aka NiH

and

Lithium Batteries:
Lithium Cobalt Oxide,
Lithium Manganese Oxide,
Lithium Iron Phosphate,
Lithium Nickel Manganese Cobalt Oxide,
Lithium Nickel Cobalt Aluminum Oxide,
Lithium Titanate)

Have VERY specific rates at which they can be safely charged and/or discharged. Exceeding the recommended charge rates will **best case** result in premature cell failure .... and in many cases can result in catastrophic failure (READ: EXPLOSION).

One final thought: If this charger already uses 4 * 50A mosfets I would ass-u-me that the charger is already outputting (AT LEAST) 20A....If you increase this by 16 more outputs this would imply a total output of (AT LEAST) 100A.....This amount of current is typically reserved for welding, automotive starters and industrial machines and requires careful safety precautions to prevent FIRE. A source for 100A+ of current is not readily available on the consumer market (except, of course welding machines) , so I must ass-u-me the power source for the charger is an automotive/marine/RV battery....please note: "Deep-Cycle" lead-acid batteries are not designed for high rates of discharge, and exceeding the recommended discharge rate can permanently damage the battery/cause a FIRE or EXPLOSION.

Good Luck!

Fish

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