High-side MOSFET gate driver supply bootstrapping

[HwAoRrDk]

I'm designing a circuit to control a couple of back-to-back (common source) N-channel MOSFETs in a high-side switching configuration, and I'm thinking of using a MIC5014 gate driver to switch the FETs.

I figured I can't supply the V+ pin of the MIC5014 from my 5V rail, as that won't give a high enough gate voltage (something around 7V according to the chart in the datasheet), so I'll need to use the main 12V drain supply. There is an application example on page 6 of the datasheet, entitled 'Bootstrapped High Side Driver', but it confuses me.



The description of this is as follows:

The turn-on time of a high side driver can be improved to faster than 40us by bootstrapping the supply with the MOSFET source. The Schottky barrier diode prevents the supply pin from dropping more than 200mV below the drain supply and improves turn-on time. Since the supply current in the "off" state is only a small leakage, the 100nF bypass capacitor tends to remain charged for several seconds after the MIC5014/15 is turned off. Faster speeds can be obtained at the expense of supply voltage (the overvoltage shutdown will turn the part off when the bootstrapping action pulls the supply pin above 35V) by using a larger capacitor at the junction of the two 1N4001 diodes. In a PWM application (this circuit can be used for either PWM'ed or continuously energized loads), the chip supply is sustained at a higher potential than the system supply, which improves switching time.

It seems to me that the diagram is illustrating two different supply bootstrapping methods, because it doesn't make sense to me that you would do things exactly as-is. One with the schottky diode and 100nF capacitor, and another with the pair of 1N4001s and the 1uF cap.

Is that anyone else's impression?

Another thing that concerns me is that in other places in the datasheet, they recommend a 10uF cap on the V+, but this particular example doesn't use one.

[BrianHG]

I do not see what you are talking about in the data sheet.
Note that there are 2 different ICs, a mic5014 and a mic5015.  The bootstrap circuit you are showing us is using the mic5015.

Figure #3 and only figure #3 is bootstrapped to improve turn on speed by externally boosting the V+ instead of relying solely on the IC's internal charge pump.  All other circuits use the internal charge pump to generate the +12v above the source voltage reference input pin on the IC.

The 10uf cap everywhere else is purely a VCC decoupling cap.
 

[BrianHG]

I figured I can't supply the V+ pin of the MIC5014 from my 5V rail, as that won't give a high enough gate voltage (something around 7V according to the chart in the datasheet), so I'll need to use the main 12V drain supply. There is an application example on page 6 of the datasheet, entitled 'Bootstrapped High Side Driver', but it confuses me.

That's 'gate enhancement' above the 5v'.  Meaning with a 5v supply, the gate will get 12v.  See figure 11.
Try tying the 'source' pin in that example circuit to the +5v as well.  The gate output should increase to ~17v.

But, if your only concern is powering a 5v mosfet rail, why not just use a logic level p-channel mosfet?
The whole purpose of this IC is so you can drive highside N-channel mosfets up with a 30v rail.

[HwAoRrDk]

That's 'gate enhancement' above the 5v'.  Meaning with a 5v supply, the gate will get 12v.  See figure 11.

Ohhh, I see. I misunderstood, I thought it was an absolute value.

Still, it's probably best not to add more load to my 5V regulator when I could be running the gate driver from the 12V supply.

Note that there are 2 different ICs, a mic5014 and a mic5015.  The bootstrap circuit you are showing us is using the mic5015.

Huh, must be a typo; figure 3's little illustrated waveform on the input pin shows 'on' as high state, which corresponds to the 5014. The only difference between the 5014 and 5015 is that the input pin logic is inverted.

Figure #3 and only figure #3 is bootstrapped to improve turn on speed by externally boosting the V+ instead of relying solely on the IC's internal charge pump.  All other circuits use the internal charge pump to generate the +12v above the source voltage reference input pin on the IC.

So, the fig. 3 circuit is meant to be considered in it's entirety? The wording of the description of it is pretty confusing and made it sound to me like they were talking about two separate things.

I tried simulating the bootstrap part, and I think I understand how it works now. The 1uF cap together with the 1N4001 diodes acts as an external 'pump', and the 100nF cap continually stores the boosted voltage. So, this helps because the MIC5014 won't have to do a 'ramp up' with it's own charge pump? And I assume this is what enables higher switching speeds, because the higher gate voltage is always available right from the word go.

Hmm, I don't really need higher switching speed. I'm not PWM-ing or anything. So, this circuit would probably be a waste of time, right? Although, I guess it would mean the power FETs spend less time in the linear region on turn-on, but I don't know whether that's really of any significance with my usage scenario (switching on/off at most maybe once every 2 seconds).

But, if your only concern is powering a 5v mosfet rail, why not just use a logic level p-channel mosfet?
The whole purpose of this IC is so you can drive highside N-channel mosfets up with a 30v rail.

I can't use a P-channel MOSFET. I'm driving one winding of a motor that has a dual windings (two speeds), and when the motor is driven from one winding, it generates a voltage on the other winding that's higher than the supply voltage. P-channel FETs get turned on uncontrollably when that happens, hence the back-to-back N-channel diodes which block it.

[BrianHG]

That's 'gate enhancement' above the 5v'.  Meaning with a 5v supply, the gate will get 12v.  See figure 11.

Ohhh, I see. I misunderstood, I thought it was an absolute value.

Still, it's probably best not to add more load to my 5V regulator when I could be running the gate driver from the 12V supply.

You actually 'must' run the IC' V+ from the save voltage as the Drain on your mosfet when running the IC in a highside configuration.  Otherwise it's internal circuit will short out as the 'source' input in the IC goes above the IC's V+.

The only way you can run the IC's V+ at 5v is in the low side configuration where the 'source' input pin stays at 0v, but again, use a higher supply voltage.