Hi guys.

I know that to drive a high side MOSFET to switch on, you need to apply a voltage that is about 10-15V higher than the voltage connected to its drain. What I do not understand is how this voltage is generated via a bootstrap capacitor? The only thing I can think of is some kind of back EMF from an inductor, just like a boost converter circuit works. But I am rather confused. Any help is appreciated.

Dan

 

Hi guys.

I know that to drive a high side MOSFET to switch on, you need to apply a voltage that is about 10-15V higher than the voltage connected to its drain. What I do not understand is how this voltage is generated via a bootstrap capacitor? The only thing I can think of is some kind of back EMF from an inductor, just like a boost converter circuit works. But I am rather confused. Any help is appreciated.

Dan

No, for a N mosfet the voltage is 10 - 15 higher than the source voltage. But never higher than 10 - 15 volts measured to ground/common.

The boot strap capacitor is charged on one of its plates to 10 -15 volts or gate voltage, when the other plate is at ground/common level when the mosfet is off. As the source voltage rises the voltage on that side of the cap does too, that causes the other side of the cap to keep the gate voltage at the correct level.

To do all of this in an easy manner a "high side gate driver" is used.

 

"A N-MOSFET/IGBT needs a significantly positive charge (VGS > VDS + Vth) applied to the gate in order to turn on."
http://en.wikipedia.org/wiki/Bootstrapping_(electronics)#Driving_MOS_transistors

"1. GATE DRIVE REQUIREMENTS OF HIGH-SIDE DEVICES....
....1. Gate voltage must be 10 V to 15 V higher than the source voltage. Being a high-side switch, such gate voltage would have to be higher than the rail voltage, which is frequently the highest voltage available in the system."
http://www.irf.com/technical-info/appnotes/an-978.pdf, page 2

Anyone got any idea how a bootstrap circuit achieves this?

 

I'm not completely sure on this

The capacitor should be hooked with the cathode on the voltage you have at the drain of the MOSFET (which should be hooked to the Vs port on the MOSFET driver) and the anode hooked to another supply(which should be hooked to the Vcc port of the MOSFET driver which will place that voltage on the Vb port).

If your source voltage were say 20V, and your bootstrap supply were 15V your capacitor would be storing 35V. The MOSFET driver would apply this voltage to the gate when it receives a high input.

I only learned the concept last Thursday, but I'm fairly sure that's how it works.
Feel free to correct me if I'm wrong

 

I'm not completely sure on this

The capacitor should be hooked with the cathode on the voltage you have at the drain of the MOSFET (which should be hooked to the Vs port on the MOSFET driver) and the anode hooked to another supply(which should be hooked to the Vcc port of the MOSFET driver which will place that voltage on the Vb port).

If your source voltage were say 20V, and your bootstrap supply were 15V your capacitor would be storing 35V. The MOSFET driver would apply this voltage to the gate when it receives a high input.

I only learned the concept last Thursday, but I'm fairly sure that's how it works.
Feel free to correct me if I'm wrong

You also have it wrong.

On a N mosfet the DRAIN is the input terminal. The SOURCE is the output terminal.

The capacitor only uses the gate voltage, which for most mosfets is 10-15 volts. Even though the capacitor may have a higher voltage on its other side when the source is "on", the capacitor acts like a shield or barricade to the source voltage. As the source voltage rises it keeps the gate voltage constant on the gate.

The source must be turned off every so often to recharge the gate side of the capacitor.

 

Badjr, thanks for the help but I built the H-bridge PWM circuit no problem, I'm just after an explanation of how the bootstrap circuit works. There is no bootstrap supply. The supply is the most positive voltage in the circuit which is in fact what the drain terminal of the high-side n-channel mosfet is connected to. You need to exceed that most-positive voltage by about 15V to drive the mosfet into saturation. I am juist curious as to how the bootstrap circuit creates a voltage potential which is 15V higer than the highest supply potential in the circuit.

Shortbus, saying 'You also have it wrong.' to badjr implies that I have said something wrong, which I have not. How does a capacitor 'use' voltage? How can a source terminal be 'on'? How does a capacitor barricade voltage? I'm sorry if I'm coming across pedantic and I kiiind of understand what you are getting at - but I am still just as confused as to how the bootstrap circuit works as I was when I started this thread. I will try and find a clear explanation and post it here for anyone in the future searching for an answer in this forum. I appreaciate your efforts though.

 

Here's the bootstrap sequence for an N-MOSFET with a source follower load:

1. Connect one end of the bootstrap capacitor to ground potential (or have it connected to the MOSFET source with the MOSFET off and the source at ground potential). The other side of the capacitor is charged to near +15V, typically through a diode connected to a +15V supply.

2. To turn the transistor on, connect the plus side of the capacitor to the MOSFET gate using a switch. This will turn the MOSFET on.

3. As the source voltage starts to rise, so will the negative side of the capacitor connected to the source. This will cause the plus side of the capacitor voltage to rise an equal amount, keeping the gate-source voltage at a near constant 15V.

4. This process continues until the MOSFET reaches the drain supply voltage and is fully on. At that point the gate-source voltage is still near+15V, and above the supply voltage.

5. The transistor is turned off by opening the capacitor switch and connecting the gate to the source with another switch.

Make sense?

 

Shortbus, saying 'You also have it wrong.' to badjr implies that I have said something wrong, which I have not. How does a capacitor 'use' voltage? How can a source terminal be 'on'? How does a capacitor barricade voltage? I'm sorry if I'm coming across pedantic and I kiiind of understand what you are getting at - but I am still just as confused as to how the bootstrap circuit works as I was when I started this thread. I will try and find a clear explanation and post it here for anyone in the future searching for an answer in this forum. I appreaciate your efforts though.

Both of you said that the gate voltage needs to be higher than the drain voltage. This is just plain wrong, no other way around it!!! The gate voltage is listed on the data sheet as Vgs, that means voltage, gate to SOURCE!!!!! The drain voltage has nothing to do with the gate voltage. The gate voltage also has a maximum value on the data sheet. This varies between mosfets but is in the ~ 20-30 volt range.

Since you were having a problem with my explanation I tried to put it in terms you "might" understand. A capacitor IS a barrier to DC voltage, and this is the type of voltage being used in these circuits. It is a barrier to different DC voltages on each side. It does store voltages for use at a later time.

The source terminal is "on" when the gate voltage is high enough.

I assumed if you were asking these questions you knew little about the workings and needed a simple answer to learn from. Crutschow's answer in different wording is what I explained.

If you are still having trouble understanding this, this link, page 17, shows a graphic of the bootstrap working - http://www.irf.com/product-info/audio/classdtutorial.pdf

 

Both of you said that the gate voltage needs to be higher than the drain voltage. This is just plain wrong, no other way around it!!! The gate voltage is listed on the data sheet as Vgs, that means voltage, gate to SOURCE!!!!! The drain voltage has nothing to do with the gate voltage. The gate voltage also has a maximum value on the data sheet. This varies between mosfets but is in the ~ 20-30 volt range.
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I believe they are referring to the case for a source follower, high side driver, where the gate voltage needs to be higher than the drain voltage when the MOSFET is fully on (since at that point the drain voltage is essentially equal to the source voltage). That's the whole reason for the bootstrap driver.

 

shortbus, I proved that when an N-Channel MOSFET is used as a high side driver, the gate voltage *needs to be higher than the drain* in my second post. I understand correct, technical explanations - no offence was intended. However your explanation did neither answer my question (because your fundemental understanding of my question was not complete) and the terminology you used was not clear.

Crutschow, thanks for explaining, but I not 100% sure on step 2. How does the high side capacitor initially get turned on when you just apply 15 volts to the gate (before the capacitor has had chance to reach a potential betond the drain voltage)? I understand the princible of the capacitor voltage rising when you apply a voltage to the grounded side now though, quite a clever concept!

 

I believe they are referring to the case for a source follower, high side driver, where the gate voltage needs to be higher than the drain voltage when the MOSFET is fully on (since at that point the drain voltage is essentially equal to the source voltage). That's the whole reason for the bootstrap driver. Note step 4 in my explanation.

So your saying that, for a example, if the drain has 100V on it then you need around 110-115V on the gate to completely turn on the mosfet? I think you will find that you will blow up your mosfet.

This link from Texas Inst explains it better than I can - http://www.ti.com/lit/ml/slup169/slup169.pdf Starting on page 24 where it says "bootstrap switching action".

The bootstrap capacitor is only charged to the gate voltage (10-15V) on one side. The other side is used as a "reference", 0V when source is off(so other plate of cap can get charged to gate voltage), and as voltage on source rises to equal drain voltage it keeps gate supplied with the same voltage it was originally charged to(10-15V).

If the gate was able to withstand the turned on source voltage there would be no reason to use a bootstrap capacitor. It (the gate) could just be connected to the source.

 

So your saying that, for a example, if the drain has 100V on it then you need around 110-115V on the gate to completely turn on the mosfet? I think you will find that you will blow up your mosfet.

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You are misunderstanding my description. After the MOSFET is fully turned on, the gate voltage to ground, will be at 115V if the drain (and source) is at 100V, but the gate-source voltage is still 15V, which is the critical voltage. The voltage won't blow the MOSFET, since no part of the MOSFET is at ground potential.

 

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Crutschow, thanks for explaining, but I not 100% sure on step 2. How does the high side capacitor initially get turned on when you just apply 15 volts to the gate (before the capacitor has had chance to reach a potential betond the drain voltage)? I understand the princible of the capacitor voltage rising when you apply a voltage to the grounded side now though, quite a clever concept!

Note that in Step 1 the capacitor is charged to 15V. Thus when you connect the capacitor to the gate is step 2, the gate-source voltage will immediately go to 15V and the MOSFET will start to turn on. At this instant the drain voltage is still at the power supply voltage, but it's the gate-source voltage, not the gate-drain voltage that determines the transistor operation.

 

You are misunderstanding my description. After the MOSFET is fully turned on, the gate voltage to ground, will be at 115V if the drain (and source) is at 100V, but the gate-source voltage is still 15V, which is the critical voltage. The voltage won't blow the MOSFET, since no part of the MOSFET is at ground potential.

I hate to keep this going but I still think your wrong. You and I had a similar discussion in another thread a while back. Here are some drawings to show my point of view, because I have trouble conveying information verbally I guess. The pictures are from this data sheet - http://www.irf.com/product-info/datasheets/data/ir2110.pdf I don't know how to make the images appear full size on the page, sorry.

 

The gate-source voltage starts out at 15V, applied from the boot-strap capacitor when the N-MOSFET starts to turn on, and never exceeds 15V during the turn-on until the turn-on is completed. When the high-side N-MOSFET is fully on, the gate voltage is at 115V (with respect to ground) while the drain and source are at 100V (with respect to ground). Thus the gate voltage is still 15V above the source/drain voltage and nothing is being stressed.

So what part of that do you think is wrong? I'm at a loss to further explain it.

 

Hi all. I'm new to allaboutcircuits. I hope you don't mind me butting in.
shortbus, I think I can explain the confusion.

These comments saying that the gate must be brought ~10V above the drain voltage apply only to the case where the MOSFET has its drain connected to a positive supply rail, and its source driving the load. This is the "source follower" configuration, similar to an emitter follower.

It's true that a MOSFET responds to the voltage between its gate and its source, and the drain voltage doesn't come into it. BUT, when you saturate a MOSFET in this configuration, it pulls its source up to its drain, so its source voltage will be equal to the positive supply voltage (minus a bit due to the MOSFET's ON resistance and the load current).

If you want to fully saturate a MOSFET whose drain is connected to a positive supply rail, you need to provide a gate voltage that is HIGHER than that positive supply rail. So when the MOSFET pulls its source up to that positive supply rail, there is still the required amount of gate-source bias voltage to keep the MOSFET saturated.

If you don't bring the gate voltage higher than the drain voltage, the MOSFET will not saturate. Its source voltage will be, say, 5V lower than its gate voltage, and there will be +5V of gate-source bias. The source will not fully pull up to the positive rail that feeds the drain, and the MOSFET will remain in its linear region.

Does this make it clear?

 

KrisBlueNZ, I know how the mosfet works and also how the driver operates. What I'm trying to get at is that the gate voltage never goes above its original voltage, in this example 15V. The boot strap cap is a "virtual or floating" ground point. One side is at, again in this example, 15V the other at 100v. The 15V side does not measure 115 volts to circuit ground only 15V.

 

KrisBlueNZ, I know how the mosfet works and also how the driver operates. What I'm trying to get at is that the gate voltage never goes above its original voltage, in this example 15V. The boot strap cap is a "virtual or floating" ground point. One side is at, again in this example, 15V the other at 100v. The 15V side does not measure 115 volts to circuit ground only 15V.

That is incorrect. It's not the gate voltage that never goes above 15V, it's the gate-source voltage. The "virtual or floating" ground point on the capacitor is the transistor source and that keeps the relative gate-source voltage constant. As the source voltage rises, so does the gate voltage. When the transistor is fully turned on, the drain and source are both at 100V measured to ground and the gate, still being 15V higher than the source, will obviously have to measure 115V to ground.

If the gate voltage stayed at only 15V to ground with the source and drain at 100V then the transistor would be well turned off (and likely the gate would short).