Is turning off MOSFET too fast, bad?
$begingroup$
As far as i understand turning off a MOSFET too fast, is bad, as high drain-source dv/dt can cause ringing through:
a) biasing Parasitic NPN and:
b) charging the GS capacitor through DG capacitor, and thus turning on the MOSFET.
1- is my understanding correct?
2- if yes, then it won't be an issue in the circuit below, which uses a diode to turn off the MOSFET as fast as possible?
3- is that diode even necessary, considering there is only one low side MOSFET, thus no chance of shoot-through?
mosfet driver
asked Jan 2 at 2:34
SudoerSudoer
19512
$endgroup$
add a comment |
$begingroup$
As far as i understand turning off a MOSFET too fast, is bad, as high drain-source dv/dt can cause ringing through:
a) biasing Parasitic NPN and:
b) charging the GS capacitor through DG capacitor, and thus turning on the MOSFET.
1- is my understanding correct?
2- if yes, then it won't be an issue in the circuit below, which uses a diode to turn off the MOSFET as fast as possible?
3- is that diode even necessary, considering there is only one low side MOSFET, thus no chance of shoot-through?
mosfet driver
asked Jan 2 at 2:34
SudoerSudoer
19512
$endgroup$
$begingroup$
Also depends on the (unstated) load. Does that have an inductive component? Turning the MOSFET OFF too slowly is also problematic because the MOSFET dissipates power during this time.
$endgroup$
– Brian Drummond
Jan 2 at 13:00
add a comment |
$begingroup$
As far as i understand turning off a MOSFET too fast, is bad, as high drain-source dv/dt can cause ringing through:
a) biasing Parasitic NPN and:
b) charging the GS capacitor through DG capacitor, and thus turning on the MOSFET.
1- is my understanding correct?
2- if yes, then it won't be an issue in the circuit below, which uses a diode to turn off the MOSFET as fast as possible?
3- is that diode even necessary, considering there is only one low side MOSFET, thus no chance of shoot-through?
mosfet driver
asked Jan 2 at 2:34
SudoerSudoer
19512
$endgroup$
As far as i understand turning off a MOSFET too fast, is bad, as high drain-source dv/dt can cause ringing through:
a) biasing Parasitic NPN and:
b) charging the GS capacitor through DG capacitor, and thus turning on the MOSFET.
1- is my understanding correct?
2- if yes, then it won't be an issue in the circuit below, which uses a diode to turn off the MOSFET as fast as possible?
3- is that diode even necessary, considering there is only one low side MOSFET, thus no chance of shoot-through?
mosfet driver
mosfet driver
asked Jan 2 at 2:34
SudoerSudoer
19512
asked Jan 2 at 2:34
SudoerSudoer
19512
asked Jan 2 at 2:34
SudoerSudoer
19512
asked Jan 2 at 2:34
SudoerSudoer
19512
asked Jan 2 at 2:34
SudoerSudoer
19512
19512
$begingroup$
Also depends on the (unstated) load. Does that have an inductive component? Turning the MOSFET OFF too slowly is also problematic because the MOSFET dissipates power during this time.
$endgroup$
– Brian Drummond
Jan 2 at 13:00
add a comment |
$begingroup$
Also depends on the (unstated) load. Does that have an inductive component? Turning the MOSFET OFF too slowly is also problematic because the MOSFET dissipates power during this time.
$endgroup$
– Brian Drummond
Jan 2 at 13:00
$begingroup$
Also depends on the (unstated) load. Does that have an inductive component? Turning the MOSFET OFF too slowly is also problematic because the MOSFET dissipates power during this time.
$endgroup$
– Brian Drummond
Jan 2 at 13:00
$begingroup$
Also depends on the (unstated) load. Does that have an inductive component? Turning the MOSFET OFF too slowly is also problematic because the MOSFET dissipates power during this time.
$endgroup$
– Brian Drummond
Jan 2 at 13:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
EDIT: Not sure what you are looking for that is not already well documented. The MOSFET should always have a low value resistor right at the gate. Not only does it limit rise and fall times to prevent ringing and EMI emissions it prevents the gate from picking up stray RF which can modulate the drain current so that it is no longer producing a clean pulse of current.
As far as stray inductance goes that it usually due to poor layout. The driver IC, the MOSFET and the load should form a tight triangle with short fat traces for all leads except the one to the gate, which can be very narrow as long as it is short.
answered Jan 2 at 3:37
Sparky256Sparky256
12k21636
$endgroup$
$begingroup$
Thanks. can you elaborate on the first part of your answer please? electronics.stackexchange.com/questions/414805/…
$endgroup$
– Sudoer
Jan 2 at 9:51
1
$begingroup$
Added EDIT section to cover more details. If you want the math and other esoteric details please read the thousands of articles on designing with MOSFETs.
$endgroup$
– Sparky256
Jan 2 at 16:10
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
return StackExchange.using("schematics", function () {
StackExchange.schematics.init();
});
}, "cicuitlab");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "135"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f414759%2fis-turning-off-mosfet-too-fast-bad%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
EDIT: Not sure what you are looking for that is not already well documented. The MOSFET should always have a low value resistor right at the gate. Not only does it limit rise and fall times to prevent ringing and EMI emissions it prevents the gate from picking up stray RF which can modulate the drain current so that it is no longer producing a clean pulse of current.
As far as stray inductance goes that it usually due to poor layout. The driver IC, the MOSFET and the load should form a tight triangle with short fat traces for all leads except the one to the gate, which can be very narrow as long as it is short.
answered Jan 2 at 3:37
Sparky256Sparky256
12k21636
$endgroup$
$begingroup$
Thanks. can you elaborate on the first part of your answer please? electronics.stackexchange.com/questions/414805/…
$endgroup$
– Sudoer
Jan 2 at 9:51
1
$begingroup$
Added EDIT section to cover more details. If you want the math and other esoteric details please read the thousands of articles on designing with MOSFETs.
$endgroup$
– Sparky256
Jan 2 at 16:10
add a comment |
$begingroup$
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
EDIT: Not sure what you are looking for that is not already well documented. The MOSFET should always have a low value resistor right at the gate. Not only does it limit rise and fall times to prevent ringing and EMI emissions it prevents the gate from picking up stray RF which can modulate the drain current so that it is no longer producing a clean pulse of current.
As far as stray inductance goes that it usually due to poor layout. The driver IC, the MOSFET and the load should form a tight triangle with short fat traces for all leads except the one to the gate, which can be very narrow as long as it is short.
answered Jan 2 at 3:37
Sparky256Sparky256
12k21636
$endgroup$
$begingroup$
Thanks. can you elaborate on the first part of your answer please? electronics.stackexchange.com/questions/414805/…
$endgroup$
– Sudoer
Jan 2 at 9:51
1
$begingroup$
Added EDIT section to cover more details. If you want the math and other esoteric details please read the thousands of articles on designing with MOSFETs.
$endgroup$
– Sparky256
Jan 2 at 16:10
add a comment |
$begingroup$
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
EDIT: Not sure what you are looking for that is not already well documented. The MOSFET should always have a low value resistor right at the gate. Not only does it limit rise and fall times to prevent ringing and EMI emissions it prevents the gate from picking up stray RF which can modulate the drain current so that it is no longer producing a clean pulse of current.
As far as stray inductance goes that it usually due to poor layout. The driver IC, the MOSFET and the load should form a tight triangle with short fat traces for all leads except the one to the gate, which can be very narrow as long as it is short.
answered Jan 2 at 3:37
Sparky256Sparky256
12k21636
$endgroup$
Your understanding is correct. To break down the issues a bit more:
Turning the MOSFET ON too fast can cause overshoot and ringing at the drain, hence the use of a low ohms resistor right at the gate. The ultra fast diode is to turn the MOSFET OFF faster than it turns ON, but the diode has internal resistance. What is bad is driving the gate with no resistor at all.
Yes the diode can be removed for the reasons you mentioned-there is no 'competing' MOSFET. If this were a totem pole or push-pull design the diodes would help insure one is turning OFF before the other MOSFET turns on. This would include both sides of an H-Bridge topology.
At high clock rates and wide pulse widths there is a risk of shoot-through for a totem pole design, or in a push-pull design the core can saturate and blow a fuse, or in a worst case scenario the ferrite core can crack.
EDIT: Not sure what you are looking for that is not already well documented. The MOSFET should always have a low value resistor right at the gate. Not only does it limit rise and fall times to prevent ringing and EMI emissions it prevents the gate from picking up stray RF which can modulate the drain current so that it is no longer producing a clean pulse of current.
As far as stray inductance goes that it usually due to poor layout. The driver IC, the MOSFET and the load should form a tight triangle with short fat traces for all leads except the one to the gate, which can be very narrow as long as it is short.
answered Jan 2 at 3:37
Sparky256Sparky256
12k21636
edited Jan 2 at 16:13
answered Jan 2 at 3:37
Sparky256Sparky256
12k21636
answered Jan 2 at 3:37
Sparky256Sparky256
12k21636
answered Jan 2 at 3:37
Sparky256Sparky256
12k21636
12k21636
$begingroup$
Thanks. can you elaborate on the first part of your answer please? electronics.stackexchange.com/questions/414805/…
$endgroup$
– Sudoer
Jan 2 at 9:51
1
$begingroup$
Added EDIT section to cover more details. If you want the math and other esoteric details please read the thousands of articles on designing with MOSFETs.
$endgroup$
– Sparky256
Jan 2 at 16:10
add a comment |
$begingroup$
Thanks. can you elaborate on the first part of your answer please? electronics.stackexchange.com/questions/414805/…
$endgroup$
– Sudoer
Jan 2 at 9:51
1
$begingroup$
Added EDIT section to cover more details. If you want the math and other esoteric details please read the thousands of articles on designing with MOSFETs.
$endgroup$
– Sparky256
Jan 2 at 16:10
$begingroup$
Thanks. can you elaborate on the first part of your answer please? electronics.stackexchange.com/questions/414805/…
$endgroup$
– Sudoer
Jan 2 at 9:51
$begingroup$
Thanks. can you elaborate on the first part of your answer please? electronics.stackexchange.com/questions/414805/…
$endgroup$
– Sudoer
Jan 2 at 9:51
1
1
$begingroup$
Added EDIT section to cover more details. If you want the math and other esoteric details please read the thousands of articles on designing with MOSFETs.
$endgroup$
– Sparky256
Jan 2 at 16:10
$begingroup$
Added EDIT section to cover more details. If you want the math and other esoteric details please read the thousands of articles on designing with MOSFETs.
$endgroup$
– Sparky256
Jan 2 at 16:10
add a comment |
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f414759%2fis-turning-off-mosfet-too-fast-bad%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Also depends on the (unstated) load. Does that have an inductive component? Turning the MOSFET OFF too slowly is also problematic because the MOSFET dissipates power during this time.
$endgroup$
– Brian Drummond
Jan 2 at 13:00