Mixing
Does this diode need to be replaced with a higher current rating diode?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
In an application simulation a fly-back diode 1N4001 current is peaking upto 3A and decays to zero in around 25ms as shown below:
The datasheet says the max forward current is 1A. But it also says 30A for 8.3 ms.
In my case it is a peak decaying from 3A to zero in 25ms.
Do I need to change this diode? What can be an alternative?
diodes
edited 3 hours ago
Neil_UK
71.2k273155
asked 3 hours ago
panic attack
11610
add a comment |Â
up vote
1
down vote
favorite
In an application simulation a fly-back diode 1N4001 current is peaking upto 3A and decays to zero in around 25ms as shown below:
The datasheet says the max forward current is 1A. But it also says 30A for 8.3 ms.
In my case it is a peak decaying from 3A to zero in 25ms.
Do I need to change this diode? What can be an alternative?
diodes
edited 3 hours ago
Neil_UK
71.2k273155
asked 3 hours ago
panic attack
11610
How often does this pulse occur?
– Dave Tweed♦
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In an application simulation a fly-back diode 1N4001 current is peaking upto 3A and decays to zero in around 25ms as shown below:
The datasheet says the max forward current is 1A. But it also says 30A for 8.3 ms.
In my case it is a peak decaying from 3A to zero in 25ms.
Do I need to change this diode? What can be an alternative?
diodes
edited 3 hours ago
Neil_UK
71.2k273155
asked 3 hours ago
panic attack
11610
In an application simulation a fly-back diode 1N4001 current is peaking upto 3A and decays to zero in around 25ms as shown below:
The datasheet says the max forward current is 1A. But it also says 30A for 8.3 ms.
In my case it is a peak decaying from 3A to zero in 25ms.
Do I need to change this diode? What can be an alternative?
diodes
diodes
edited 3 hours ago
Neil_UK
71.2k273155
asked 3 hours ago
panic attack
11610
edited 3 hours ago
Neil_UK
71.2k273155
asked 3 hours ago
panic attack
11610
edited 3 hours ago
Neil_UK
71.2k273155
edited 3 hours ago
Neil_UK
71.2k273155
edited 3 hours ago
Neil_UK
71.2k273155
71.2k273155
asked 3 hours ago
panic attack
11610
asked 3 hours ago
panic attack
11610
asked 3 hours ago
panic attack
11610
11610
How often does this pulse occur?
– Dave Tweed♦
1 hour ago
add a comment |Â
How often does this pulse occur?
– Dave Tweed♦
1 hour ago
How often does this pulse occur?
– Dave Tweed♦
1 hour ago
How often does this pulse occur?
– Dave Tweed♦
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
In the case of pulse loading to a component, it's the $I^2t$ that matters. This is a measure of the heat that's dumped essentially instantaneously into the device raising its temperature. 30A peak half sinewave pulse for 8.3mS is an $I^2t$ of about 3.7.
You can either integrate your graph properly, or do a simpler conservative upper bound and say that it's less than a full 3A flowing continuously for 25mS, which is an $I^2t$ of 0.225. This figure is so much less than 3.7 that's it's clear it will handle your proposed pulse with no trouble.
That rating is for a single pulse, with the diode cooling down between pulses. Is your flyback diode going to be hit repeatedly in a time less than the cooling down time? If so, you need to compute the average power that those pulses deliver, and make sure that's well below the continuous power rating.
answered 3 hours ago
Neil_UK
71.2k273155
No it will not be repetitive. Juts turn on and off application.
– panic attack
3 hours ago
@neil_uk: AFAIK, it's VIt that matters. I^2 t is correct for resistors, but only correct in a diode that's acting substantially like a resistor rather than showing a more or less constant voltage.
– TimWescott
2 hours ago
add a comment |Â
up vote
1
down vote
The issue is how hot do various bits of the device get. Continuous current ratings are calculated based on the total thermal resistance of the device to ambient. For pulses, the thermal time constants of the device matters, and for complicated devices (or critical applications) there may be more than one time constant to worry about.
You can either try to find such data for the 1N4001 (and hope that it's the same from brand to brand), or you can figure that if it'll handle 30A for 8.3ms (250mJ, if you assume a 1V drop), it should have no problem with 3A for 25ms (75mJ).
If you study this data sheet from Diodes, Inc., you'll see that they have charts on voltage drop vs. current (which you need to determine power vs. current, which isn't really proportional to either I or I^2), current vs. ambient temperature which you can use to deduce the thermal resistance from case to ambient*, and curent vs. number of repetitive cycles at 60Hz, from which you can get a notion of the thermal time constant from the junction to ambient. From all of that, you can determine whether you're safe.
But to reiterate -- you should be fine. Look at Figure 3, and figure that your 25ms is less than two full cycles at 60Hz, and figure that they're allowing you more than 20A peak for a half sine-wave. The average current of a half sine wave is somewhere around 0.68 * peak (it's 2/pi or something -- I'd have to calculate it to remember), so that works out to 15A for 33ms. Even being really fast and loose with the numbers, it's obvious that you'll be fine with 3A for 25ms, unless your ambient temperature is way closer to 175C than it is to 65C (see their figure 1).
* Said thermal path is through the leads, see Note 3 to the maximum ratings table.
answered 21 mins ago
TimWescott
72916
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
In the case of pulse loading to a component, it's the $I^2t$ that matters. This is a measure of the heat that's dumped essentially instantaneously into the device raising its temperature. 30A peak half sinewave pulse for 8.3mS is an $I^2t$ of about 3.7.
You can either integrate your graph properly, or do a simpler conservative upper bound and say that it's less than a full 3A flowing continuously for 25mS, which is an $I^2t$ of 0.225. This figure is so much less than 3.7 that's it's clear it will handle your proposed pulse with no trouble.
That rating is for a single pulse, with the diode cooling down between pulses. Is your flyback diode going to be hit repeatedly in a time less than the cooling down time? If so, you need to compute the average power that those pulses deliver, and make sure that's well below the continuous power rating.
answered 3 hours ago
Neil_UK
71.2k273155
No it will not be repetitive. Juts turn on and off application.
– panic attack
3 hours ago
@neil_uk: AFAIK, it's VIt that matters. I^2 t is correct for resistors, but only correct in a diode that's acting substantially like a resistor rather than showing a more or less constant voltage.
– TimWescott
2 hours ago
add a comment |Â
up vote
4
down vote
In the case of pulse loading to a component, it's the $I^2t$ that matters. This is a measure of the heat that's dumped essentially instantaneously into the device raising its temperature. 30A peak half sinewave pulse for 8.3mS is an $I^2t$ of about 3.7.
You can either integrate your graph properly, or do a simpler conservative upper bound and say that it's less than a full 3A flowing continuously for 25mS, which is an $I^2t$ of 0.225. This figure is so much less than 3.7 that's it's clear it will handle your proposed pulse with no trouble.
That rating is for a single pulse, with the diode cooling down between pulses. Is your flyback diode going to be hit repeatedly in a time less than the cooling down time? If so, you need to compute the average power that those pulses deliver, and make sure that's well below the continuous power rating.
answered 3 hours ago
Neil_UK
71.2k273155
No it will not be repetitive. Juts turn on and off application.
– panic attack
3 hours ago
@neil_uk: AFAIK, it's VIt that matters. I^2 t is correct for resistors, but only correct in a diode that's acting substantially like a resistor rather than showing a more or less constant voltage.
– TimWescott
2 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
In the case of pulse loading to a component, it's the $I^2t$ that matters. This is a measure of the heat that's dumped essentially instantaneously into the device raising its temperature. 30A peak half sinewave pulse for 8.3mS is an $I^2t$ of about 3.7.
You can either integrate your graph properly, or do a simpler conservative upper bound and say that it's less than a full 3A flowing continuously for 25mS, which is an $I^2t$ of 0.225. This figure is so much less than 3.7 that's it's clear it will handle your proposed pulse with no trouble.
That rating is for a single pulse, with the diode cooling down between pulses. Is your flyback diode going to be hit repeatedly in a time less than the cooling down time? If so, you need to compute the average power that those pulses deliver, and make sure that's well below the continuous power rating.
answered 3 hours ago
Neil_UK
71.2k273155
In the case of pulse loading to a component, it's the $I^2t$ that matters. This is a measure of the heat that's dumped essentially instantaneously into the device raising its temperature. 30A peak half sinewave pulse for 8.3mS is an $I^2t$ of about 3.7.
You can either integrate your graph properly, or do a simpler conservative upper bound and say that it's less than a full 3A flowing continuously for 25mS, which is an $I^2t$ of 0.225. This figure is so much less than 3.7 that's it's clear it will handle your proposed pulse with no trouble.
That rating is for a single pulse, with the diode cooling down between pulses. Is your flyback diode going to be hit repeatedly in a time less than the cooling down time? If so, you need to compute the average power that those pulses deliver, and make sure that's well below the continuous power rating.
answered 3 hours ago
Neil_UK
71.2k273155
edited 3 hours ago
answered 3 hours ago
Neil_UK
71.2k273155
answered 3 hours ago
Neil_UK
71.2k273155
answered 3 hours ago
Neil_UK
71.2k273155
71.2k273155
No it will not be repetitive. Juts turn on and off application.
– panic attack
3 hours ago
@neil_uk: AFAIK, it's VIt that matters. I^2 t is correct for resistors, but only correct in a diode that's acting substantially like a resistor rather than showing a more or less constant voltage.
– TimWescott
2 hours ago
add a comment |Â
No it will not be repetitive. Juts turn on and off application.
– panic attack
3 hours ago
@neil_uk: AFAIK, it's VIt that matters. I^2 t is correct for resistors, but only correct in a diode that's acting substantially like a resistor rather than showing a more or less constant voltage.
– TimWescott
2 hours ago
No it will not be repetitive. Juts turn on and off application.
– panic attack
3 hours ago
No it will not be repetitive. Juts turn on and off application.
– panic attack
3 hours ago
@neil_uk: AFAIK, it's VIt that matters. I^2 t is correct for resistors, but only correct in a diode that's acting substantially like a resistor rather than showing a more or less constant voltage.
– TimWescott
2 hours ago
@neil_uk: AFAIK, it's VIt that matters. I^2 t is correct for resistors, but only correct in a diode that's acting substantially like a resistor rather than showing a more or less constant voltage.
– TimWescott
2 hours ago
add a comment |Â
up vote
1
down vote
The issue is how hot do various bits of the device get. Continuous current ratings are calculated based on the total thermal resistance of the device to ambient. For pulses, the thermal time constants of the device matters, and for complicated devices (or critical applications) there may be more than one time constant to worry about.
You can either try to find such data for the 1N4001 (and hope that it's the same from brand to brand), or you can figure that if it'll handle 30A for 8.3ms (250mJ, if you assume a 1V drop), it should have no problem with 3A for 25ms (75mJ).
If you study this data sheet from Diodes, Inc., you'll see that they have charts on voltage drop vs. current (which you need to determine power vs. current, which isn't really proportional to either I or I^2), current vs. ambient temperature which you can use to deduce the thermal resistance from case to ambient*, and curent vs. number of repetitive cycles at 60Hz, from which you can get a notion of the thermal time constant from the junction to ambient. From all of that, you can determine whether you're safe.
But to reiterate -- you should be fine. Look at Figure 3, and figure that your 25ms is less than two full cycles at 60Hz, and figure that they're allowing you more than 20A peak for a half sine-wave. The average current of a half sine wave is somewhere around 0.68 * peak (it's 2/pi or something -- I'd have to calculate it to remember), so that works out to 15A for 33ms. Even being really fast and loose with the numbers, it's obvious that you'll be fine with 3A for 25ms, unless your ambient temperature is way closer to 175C than it is to 65C (see their figure 1).
* Said thermal path is through the leads, see Note 3 to the maximum ratings table.
answered 21 mins ago
TimWescott
72916
add a comment |Â
up vote
1
down vote
The issue is how hot do various bits of the device get. Continuous current ratings are calculated based on the total thermal resistance of the device to ambient. For pulses, the thermal time constants of the device matters, and for complicated devices (or critical applications) there may be more than one time constant to worry about.
You can either try to find such data for the 1N4001 (and hope that it's the same from brand to brand), or you can figure that if it'll handle 30A for 8.3ms (250mJ, if you assume a 1V drop), it should have no problem with 3A for 25ms (75mJ).
If you study this data sheet from Diodes, Inc., you'll see that they have charts on voltage drop vs. current (which you need to determine power vs. current, which isn't really proportional to either I or I^2), current vs. ambient temperature which you can use to deduce the thermal resistance from case to ambient*, and curent vs. number of repetitive cycles at 60Hz, from which you can get a notion of the thermal time constant from the junction to ambient. From all of that, you can determine whether you're safe.
But to reiterate -- you should be fine. Look at Figure 3, and figure that your 25ms is less than two full cycles at 60Hz, and figure that they're allowing you more than 20A peak for a half sine-wave. The average current of a half sine wave is somewhere around 0.68 * peak (it's 2/pi or something -- I'd have to calculate it to remember), so that works out to 15A for 33ms. Even being really fast and loose with the numbers, it's obvious that you'll be fine with 3A for 25ms, unless your ambient temperature is way closer to 175C than it is to 65C (see their figure 1).
* Said thermal path is through the leads, see Note 3 to the maximum ratings table.
answered 21 mins ago
TimWescott
72916
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The issue is how hot do various bits of the device get. Continuous current ratings are calculated based on the total thermal resistance of the device to ambient. For pulses, the thermal time constants of the device matters, and for complicated devices (or critical applications) there may be more than one time constant to worry about.
You can either try to find such data for the 1N4001 (and hope that it's the same from brand to brand), or you can figure that if it'll handle 30A for 8.3ms (250mJ, if you assume a 1V drop), it should have no problem with 3A for 25ms (75mJ).
If you study this data sheet from Diodes, Inc., you'll see that they have charts on voltage drop vs. current (which you need to determine power vs. current, which isn't really proportional to either I or I^2), current vs. ambient temperature which you can use to deduce the thermal resistance from case to ambient*, and curent vs. number of repetitive cycles at 60Hz, from which you can get a notion of the thermal time constant from the junction to ambient. From all of that, you can determine whether you're safe.
But to reiterate -- you should be fine. Look at Figure 3, and figure that your 25ms is less than two full cycles at 60Hz, and figure that they're allowing you more than 20A peak for a half sine-wave. The average current of a half sine wave is somewhere around 0.68 * peak (it's 2/pi or something -- I'd have to calculate it to remember), so that works out to 15A for 33ms. Even being really fast and loose with the numbers, it's obvious that you'll be fine with 3A for 25ms, unless your ambient temperature is way closer to 175C than it is to 65C (see their figure 1).
* Said thermal path is through the leads, see Note 3 to the maximum ratings table.
answered 21 mins ago
TimWescott
72916
The issue is how hot do various bits of the device get. Continuous current ratings are calculated based on the total thermal resistance of the device to ambient. For pulses, the thermal time constants of the device matters, and for complicated devices (or critical applications) there may be more than one time constant to worry about.
You can either try to find such data for the 1N4001 (and hope that it's the same from brand to brand), or you can figure that if it'll handle 30A for 8.3ms (250mJ, if you assume a 1V drop), it should have no problem with 3A for 25ms (75mJ).
If you study this data sheet from Diodes, Inc., you'll see that they have charts on voltage drop vs. current (which you need to determine power vs. current, which isn't really proportional to either I or I^2), current vs. ambient temperature which you can use to deduce the thermal resistance from case to ambient*, and curent vs. number of repetitive cycles at 60Hz, from which you can get a notion of the thermal time constant from the junction to ambient. From all of that, you can determine whether you're safe.
But to reiterate -- you should be fine. Look at Figure 3, and figure that your 25ms is less than two full cycles at 60Hz, and figure that they're allowing you more than 20A peak for a half sine-wave. The average current of a half sine wave is somewhere around 0.68 * peak (it's 2/pi or something -- I'd have to calculate it to remember), so that works out to 15A for 33ms. Even being really fast and loose with the numbers, it's obvious that you'll be fine with 3A for 25ms, unless your ambient temperature is way closer to 175C than it is to 65C (see their figure 1).
* Said thermal path is through the leads, see Note 3 to the maximum ratings table.
answered 21 mins ago
TimWescott
72916
answered 21 mins ago
TimWescott
72916
answered 21 mins ago
TimWescott
72916
answered 21 mins ago
TimWescott
72916
72916
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f402970%2fdoes-this-diode-need-to-be-replaced-with-a-higher-current-rating-diode%23new-answer', 'question_page');
);
Post as a guest
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
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
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
How often does this pulse occur?
– Dave Tweed♦
1 hour ago