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Join Date: Jun 2018
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Proper Power Supply for this Gate Driver IC

07/06/2018 9:44 PM

Hi,
I will use the gate driver configuration shown below:
Gate Driver datasheet
MOSFET Datasheet

As you can see, the 5V and 12V are isolated. I chose this AC/DC converter for the 5V supply. It's 1 Output 5V/3A. It will supply a STM32 and a Raspberry Pi.
I decided to step up the 5V to 12V using this Isolated Module DC DC Converter.
That module is just 1W/ 84mA max. The 12V source will supply an average current of I = Qxfs, where Q is the total charge of the mosfet and fs my pwm frequency. According to my mosfet datasheet the total charge is 160nC, my pwm frequency is 10 kHz. So I = 160nC * 10000 = 1.6 mA... am I right?

Can I say that the step up module will effectively supply the peak currents to charge gate capacitance using decoupling capacitors?

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#1

Re: Proper Power Supply for this Gate Driver IC

07/08/2018 10:25 AM

Read page 16 of the gate driver datasheet:

High level output peak current (Minimum=6A, Typical=10A) Low level output peak current (Minimum=6A, Typical=9.4A)

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#2

Re: Proper Power Supply for this Gate Driver IC

07/08/2018 1:59 PM

Obviously a 84 mA rating is adequate for 1.6 mA mean current.

The capacitor across the supply of the driver 12 V side must be able to supply the mosfet gate charge during the switch-on operation.

Suppose you are switching on in 1 microsecond, so as to reduce EMC possibilities.

Further, the 6 amp maximum of the driver is drawn for 1 microsecond and the supply bypass capacitor is 1 microFarad - as shown on your circuit, typically a high permitivity ceramic capacitor, e.g. AVX make.

A 1 μF capacitor will have a theoretical impedance of about 0.15 ohm at 1 MHz. Practical capacitors also have an ESR [effective series resistance] of about 0.1 0hm. Being pessimistic, that is 0.25 ohm, which will drop 6 x 0.25 = 1.5 volts - practically, the current will decay exponentially. That is not bad for a 12V supply. Some 7V is ample to turn on your Mosfet at 50 amps.

One could look from the capacitor discharge point of view - suppose the voltage drops 1 volt over 1 μs at 6amps - that is dV/dt = 1V/10-6 sec = 106 V/second.

For a capacitor, q = CV, hence i = CdV/dt - so you have 6 amps = C.106 - from which C = 6.10-6 i.e. 6 μF. On the other hand, 6A for 10μs is 6 μCoulombs - which is a lot more than 0.11 μCoulombs gate charge specified for your Mosfet. 1 μF would give 1 μCoulomb with 1 volt drop - which looks adequate. Again you have an ESR volt drop 6a x 0.1 ohm = 0.6V.

Inductance would have an effect with really fast switching, but SMD capacitors have inductance of a few nH - you have more effect from your wiring, a 1 turn 10mm diameter wire loop is 40 nH; 2mm wide tracks on opposite sides of 1.6mm thick PCB are ~3.5 nH/cm.

In summary, so long as that 1 μF capacitor is a low impedance ceramic bypass type close to the driver IC and you have a sensible wiring layout, you should have no trouble.

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#3
In reply to #2

Re: Proper Power Supply for this Gate Driver IC

07/09/2018 5:18 PM

Thank you. That's what I thought about placing the 1uF. I just didn't understand why some people told me that the 12V step-up isolated module provides too low current capability (84 mA max). But the point is that the 12V supply doesn't have to supply actually 6 to 10A peaks. (it's insane for this application to look for a 12V supply rated at 10A). So the concept here to understand is that the peaks currents are provided by the capacitor, not by the power supply as a whole (as far as I understand).

There are other ways for me to obtain the 12V. It's not mandatory to step-up the 5V. It's just cheapier. I could also look for another module, let's say, IRM-15-12. But I have to consider PCB space and part costs.

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#4
In reply to #3

Re: Proper Power Supply for this Gate Driver IC

07/09/2018 8:22 PM

OK - I assume you noticed that with the 12V supply & 10 ohm resistor in "pull-up" output to gate, the peak current must be 1/5 of the 6 amp assumed (minimum capability in data sheet).

You are correct that peak currents come from supply bypass capacitors - this is why lots of ICs need such capacitors close to them since they draw high peak currents when switching (the bipolar NE555 timer [view its data sheet] & many TTL logic chips do this).

I looked at 5V- 12V converter data sheet. It gives max no-load voltage as +25% = 15V which is OK for driver IC & if capacitors are 16V type. Maybe you have to add +5% for input 5V tolerance. It mentions test with 0.1 microfarad capacitor across output without being clear if this is internal or you are expected to provide one external.

It is not clear how much noise may go back from converter into your Raspberry via the 5V supply.

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#5
In reply to #4

Re: Proper Power Supply for this Gate Driver IC

07/09/2018 8:57 PM

Would you consider "overrated" using two independents AC/DC converters? For example, IRM-15-5 for the 5V and IRM-15-12 for the 12V?

The reason why I'm using those AC/DC modules is because the project is fed from the main (120VAC/60hz) and I don't have the time to design a switching power supply.

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#6
In reply to #5

Re: Proper Power Supply for this Gate Driver IC

07/10/2018 5:12 AM

I would go with AC to 5V & 5VDC to 12V DC, but make sure input to DC/DC is bypassed by 0.1 μF ceramic & 100μF aluminium electrolytic right at its input. You would need a lot more space to have two AC/DC PSU. As you have realised, actual load of DC/DC will be negligible. You would keep output of DC-DC away from Raspberry anyhow, because it is connected to source of MOSFET blasting away at 150V/30amp/10kHz square with < 1 μs risetime.

The AC to DC PSU does have EMC certification with the usual condition that the maker of final product is responsible for EMC of final product into which PSU is built.

It is worth asking maker of DC/DC what noise/EMC testing was done & results they got.

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#7
In reply to #6

Re: Proper Power Supply for this Gate Driver IC

07/10/2018 11:24 PM

Just a brief question that has come to my mind. The power rating for that 10 ohms gate resistor can be 1/4W, right? I mean, the peak pulses are short. If I say that P = (Ipeak^2)*Rg, it would throw a high power, but as it's a short pulse, a low power rating resistor can handle it, right?

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#8
In reply to #7

Re: Proper Power Supply for this Gate Driver IC

07/11/2018 12:46 AM

The power dissipated by a resistor depends on the rms current, usually calculated for a period (easy) and by the value of the resistor at the switching frequency (that would need to be available in the datasheet).

If you have neved heard of the frequency response of a resistor, better oversize the resistor by a factor or 10 in terms of power.

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#9
In reply to #7

Re: Proper Power Supply for this Gate Driver IC

07/11/2018 7:09 AM

Do not speculate - calculate & ask manufacturer!

Crudely, 12V/10 ohm = 1.2 amps, for 10-6 sec I2Rt = 1.44.10.10-6 = 14.4.10-6J, but it is 104 times sec = 14.4.10-2J/s = 0.14W. But a 10 ohm 0.1W resistor has rated current 0.1 amp and you are going 10x & 100 times rated power.

I once had to reverse engineer a battery standby unit, one fault to fix was failures of the battery test resistor an Arcol metalclad one. If I remember right, they recommended no more than 2x rating watts for 10 seconds. When you consider that wirewound resistors have maximum temperature about 300'C, thermal expansion forces must be considerable.

A more correct power is that charging a capacitor from a DC voltage loses as much energy in resistor as is stored in capacitor - J = 1/2CV2 = 0.5QV since Q= CV. Your charge is ~0.15.10-6 C so energy is 0.15.10-6 .12/2 = 0.9.10-6@ 104times/s = 0.9 x 10-2 W.

I would use 0.5W resistor, only 5x rated current. If I was building commercial, I would check with resistor supplier on pulse rating. No prizes for having a high failure rate after a few months use. The weak point in resistors is the connections at ends to copper & overcurrent concentrates loss there. It is energy & overheating which destroys components - I2Rt versus thermal capacity in this case, and repeated fast expansion/temperature rise is destructive.

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#10
In reply to #9

Re: Proper Power Supply for this Gate Driver IC

07/11/2018 6:40 PM

The same would apply for the discharging resistor, as I have separate source/sink configuration. According to the one recommended in the datasheet, i.e, 3.3 ohms, the discharging current is 12/3.3 = 3.64A. As my PWM frequency is 10 kHz, that is, 100 uS, I think 0.5W resistors are okay, as I far as I understand from your calculations. I will look into them in deep.

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#11
In reply to #10

Re: Proper Power Supply for this Gate Driver IC

07/12/2018 6:24 PM

This is a "continuous pulse" rating chart for a thin film SMD resistor from a reputable supplier, Vishay....

Text under chart reads.....

Maximum pulse load, continuous pulses; applicable if P <=(Pamb) and Û < Ûmax.;
for permissible resistance change equivalent to 8000 h operation in standard operation mode.

This means that so long as the average power is <= rated power at ambient, the charted pulse length & power is permissible so long as resistance change is acceptable.

Note that the smallest size resistor, 0402, although it meets continuous dissipation of 100mW [10Wx 10-6 J x 104/sec] is not suitable for pulses of > 2 watts. Size 0603 is OK for 4 W, 0805 for 10 watts. Also, if you were using the full 12V/10amp 120W of the driver, even 1206 size is not adequate. Thick film resistors have better pulse ratings, the thicker resistance has more mass to absorb pulses of heat.

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#12
In reply to #6

Re: Proper Power Supply for this Gate Driver IC

07/13/2018 9:26 AM

I asked the manufacturer of the gate driver (Infineon) about the 12V converter that I selected. This is what they told me:

"Your numbers look oK, but I would recommend that you choose another DC-DC converter based off of the Ig charging peak value of 1.2A.

I believe you used the Ig (average) value of 1.6mA to come up with the R1SE-0512/H2-R converter that is spec'd @ an output current of 84mA.

However, in actuality, the formula that you used to come up with Ig_mean will actually give you a value that is smaller than the actual value of Ig_mean because Iavg is actually calculated by:

(Qgc + Qge)/tsw

as you can see in this app note.

Therefore, the Fsw value that you use to calculate Ig_mean is actually smaller than the real actual Fsw because tsw is smaller than 1/Fsw as you can see in Fig 2 of the same app note. "

Then, they suggested this:

120VDC to 12V acdc --> 12V to 5V DCDC is our recommendation which is also used on our board I mentioned before.

As for my point of view, 120VDC to 12V acdc --> 12V to 5V DCDC (isolated) is way too expensive.

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#13
In reply to #12

Re: Proper Power Supply for this Gate Driver IC

07/14/2018 6:43 AM

It has been said "brawn baffles brain"!

It does not help that the application note is about driving IGBT, so Cgc corresponds to Cgd for MOSFET & Cge to Cgs {collector in place of drain, emitter in place of source}.

If you read the note carefully, you will realise that Iavg is not the DC average current but the average gate current during the switch-on time.

If you look at datasheet for your Mosfet, figure 5 typical sample 25'C, you will see that conduction starts at about 3.5V Vgs, at 4.5V you have ~40 amps. It is this range which corresponds to "plateau voltage" in the app. note, Vge on Fig.2 diagram. Note that in real life there is a range for Vge [Vgs] - Table 5, threshold voltage in Mosfet data sheet.

Vge could be approximated by a constant value[see Fig.2] of 4V for your FET.

Turn-on is considered in two stages..

  1. Getting gate voltage up to Vge, drain [collector] current will not flow until volts reach that level - this is usually called "delay time" in data sheets.
  2. Once drain conduction begins, negative feedback makes the circuit behave as an integrator [change of Vgs is opposed by negative feedback] , gate voltage stays approximately constant at Vge, while current into the drain source capacitance determines its dV/dt, which is drain-source dV/dt and sets turn-on time.

Taking rough values of 0.9A for charging current & 10-8 F for input capacitance (~10000pF in data sheet Fig.7), you have 0.9 = 10-8 .dV/dt [N.B. Q = CV, dQ/dt = i = CdV/dt] - hence dV/dt = 90.106 V/s, so it will take ~4.5/90 = 0.05μs to get to a plateau of 4.5V.

The integrating stage is determined by gate-drain capacity Crss in Fig.7 of data sheet make it 1000 pF, with mean charging current 0.5 amp gives 500V/μs, from which 160 fall of Vd takes 3μs. so the total turn-on time is ~3 μs. If you look at Fig.7, you will see that the capcitances fall sharply with voltage - it is really a matter of charge rather than capacitance.

Even this over-estimate only indicates ~1 amp for a few μs to turn on, 100μs period, 1 amp x 3/100 = 30mA mean current. If it worries you, add an electrolytic cap across your 12V supply, as an extra reservoir.

This does assume you turn FET on every 100μs & vary the ON time, I think that is what you intend.

One might speculate that your query was given to the "office junior" or "new starter" for a quick reply.

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#14
In reply to #13

Re: Proper Power Supply for this Gate Driver IC

07/14/2018 11:37 AM

Actually I disagree with Infineon's answer. I just gave it a little of relevance since it was a "manufacturer response". They didn't even gave a solid answer why my 84mA converter is not suitable. They just say that they recommend choosing a 12V supply based off the peak current. But again, that's doesn't make much sense. What if I don't limit the peak current? Do I have to choose a supply with 10A capability ?

Your analysis looks more convincing. I will come back to the idea of using the RECOM DC/DC converter, 84mA/1W. Although it is an unregulated converter, I think it can do the task.

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#15
In reply to #14

Re: Proper Power Supply for this Gate Driver IC

07/14/2018 12:23 PM

XaviPacheco,

Yes, the unregulated converter looks fine. If the no-load "peak-up" of voltage were a problem, a 5% or 10% resistor "ballast" load would keep it down.

Good luck with your project!

67model

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#16
In reply to #15

Re: Proper Power Supply for this Gate Driver IC

07/14/2018 12:36 PM

Thank you. As a feedback, I will upload my results once I test the PCB prototype.

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