High current, low voltage capacitors
$begingroup$
I'm putting together a high current (200A+), low voltage (<5v) rectifier circuit.
I have sufficient bridge rectifiers, but to smooth out the waveform I need a strong enough capacitor bank. Besides opting for capacitors rated at a voltage of 10vdc (double the 5v just to be safe), what other stats or capacitor specs should I be looking for? Just get three one with the highest max operating temperature? The AC wave will be 50hz.
capacitor high-current bridge-rectifier
$endgroup$
|
show 1 more comment
$begingroup$
I'm putting together a high current (200A+), low voltage (<5v) rectifier circuit.
I have sufficient bridge rectifiers, but to smooth out the waveform I need a strong enough capacitor bank. Besides opting for capacitors rated at a voltage of 10vdc (double the 5v just to be safe), what other stats or capacitor specs should I be looking for? Just get three one with the highest max operating temperature? The AC wave will be 50hz.
capacitor high-current bridge-rectifier
$endgroup$
3
$begingroup$
Low-voltage high-amperage power sources are not designed as plain AC rectifiers. They are made as "AC-DC converters", where AC gets rectified to high-voltage DC, the DC gets converted into high-frequency "chopped" signal, and then transformed into low-voltage, and then rectified using synchronous active rectifiers. The smoothing caps then operate at a very high frequency and don't need to be of extremely high values. You need to seriously reconsider your design approach, unless you are making some welding apparatus.
$endgroup$
– Ale..chenski
5 hours ago
1
$begingroup$
In addition to what Ale..chenski said, you have discovered that at high currents and low frequency you need an unreasonable amount of smoothing capacitance since the charge consumed is so high but the refresh rate is so low. The ripple will be enormous even with massive caps. Refresh a few thousand times faster and your capacitors can be a few thousand times smaller. Capacitors have a limit on ripple current anyways and such high currents and low frequencies will produce enormous ripple currents that will overheat and blow caps.
$endgroup$
– Toor
5 hours ago
$begingroup$
what is your ripple spec 10% V? with 200A RMS
$endgroup$
– Sunnyskyguy EE75
3 hours ago
1
$begingroup$
@SunnyskyguyEE75 haven't considered that, just want to minimise what ripple there is. If it's not worth the effort, I may just leave it as AC. The purpose is to heat a piece of carbon or stainless to serve as a heating element. Was considering rectifying the AC signal to minimise any eddy currents.
$endgroup$
– Chris
3 hours ago
1
$begingroup$
You can use a 6V 700A CCA lead acid battery
$endgroup$
– Sunnyskyguy EE75
2 hours ago
|
show 1 more comment
$begingroup$
I'm putting together a high current (200A+), low voltage (<5v) rectifier circuit.
I have sufficient bridge rectifiers, but to smooth out the waveform I need a strong enough capacitor bank. Besides opting for capacitors rated at a voltage of 10vdc (double the 5v just to be safe), what other stats or capacitor specs should I be looking for? Just get three one with the highest max operating temperature? The AC wave will be 50hz.
capacitor high-current bridge-rectifier
$endgroup$
I'm putting together a high current (200A+), low voltage (<5v) rectifier circuit.
I have sufficient bridge rectifiers, but to smooth out the waveform I need a strong enough capacitor bank. Besides opting for capacitors rated at a voltage of 10vdc (double the 5v just to be safe), what other stats or capacitor specs should I be looking for? Just get three one with the highest max operating temperature? The AC wave will be 50hz.
capacitor high-current bridge-rectifier
capacitor high-current bridge-rectifier
asked 5 hours ago
ChrisChris
244
244
3
$begingroup$
Low-voltage high-amperage power sources are not designed as plain AC rectifiers. They are made as "AC-DC converters", where AC gets rectified to high-voltage DC, the DC gets converted into high-frequency "chopped" signal, and then transformed into low-voltage, and then rectified using synchronous active rectifiers. The smoothing caps then operate at a very high frequency and don't need to be of extremely high values. You need to seriously reconsider your design approach, unless you are making some welding apparatus.
$endgroup$
– Ale..chenski
5 hours ago
1
$begingroup$
In addition to what Ale..chenski said, you have discovered that at high currents and low frequency you need an unreasonable amount of smoothing capacitance since the charge consumed is so high but the refresh rate is so low. The ripple will be enormous even with massive caps. Refresh a few thousand times faster and your capacitors can be a few thousand times smaller. Capacitors have a limit on ripple current anyways and such high currents and low frequencies will produce enormous ripple currents that will overheat and blow caps.
$endgroup$
– Toor
5 hours ago
$begingroup$
what is your ripple spec 10% V? with 200A RMS
$endgroup$
– Sunnyskyguy EE75
3 hours ago
1
$begingroup$
@SunnyskyguyEE75 haven't considered that, just want to minimise what ripple there is. If it's not worth the effort, I may just leave it as AC. The purpose is to heat a piece of carbon or stainless to serve as a heating element. Was considering rectifying the AC signal to minimise any eddy currents.
$endgroup$
– Chris
3 hours ago
1
$begingroup$
You can use a 6V 700A CCA lead acid battery
$endgroup$
– Sunnyskyguy EE75
2 hours ago
|
show 1 more comment
3
$begingroup$
Low-voltage high-amperage power sources are not designed as plain AC rectifiers. They are made as "AC-DC converters", where AC gets rectified to high-voltage DC, the DC gets converted into high-frequency "chopped" signal, and then transformed into low-voltage, and then rectified using synchronous active rectifiers. The smoothing caps then operate at a very high frequency and don't need to be of extremely high values. You need to seriously reconsider your design approach, unless you are making some welding apparatus.
$endgroup$
– Ale..chenski
5 hours ago
1
$begingroup$
In addition to what Ale..chenski said, you have discovered that at high currents and low frequency you need an unreasonable amount of smoothing capacitance since the charge consumed is so high but the refresh rate is so low. The ripple will be enormous even with massive caps. Refresh a few thousand times faster and your capacitors can be a few thousand times smaller. Capacitors have a limit on ripple current anyways and such high currents and low frequencies will produce enormous ripple currents that will overheat and blow caps.
$endgroup$
– Toor
5 hours ago
$begingroup$
what is your ripple spec 10% V? with 200A RMS
$endgroup$
– Sunnyskyguy EE75
3 hours ago
1
$begingroup$
@SunnyskyguyEE75 haven't considered that, just want to minimise what ripple there is. If it's not worth the effort, I may just leave it as AC. The purpose is to heat a piece of carbon or stainless to serve as a heating element. Was considering rectifying the AC signal to minimise any eddy currents.
$endgroup$
– Chris
3 hours ago
1
$begingroup$
You can use a 6V 700A CCA lead acid battery
$endgroup$
– Sunnyskyguy EE75
2 hours ago
3
3
$begingroup$
Low-voltage high-amperage power sources are not designed as plain AC rectifiers. They are made as "AC-DC converters", where AC gets rectified to high-voltage DC, the DC gets converted into high-frequency "chopped" signal, and then transformed into low-voltage, and then rectified using synchronous active rectifiers. The smoothing caps then operate at a very high frequency and don't need to be of extremely high values. You need to seriously reconsider your design approach, unless you are making some welding apparatus.
$endgroup$
– Ale..chenski
5 hours ago
$begingroup$
Low-voltage high-amperage power sources are not designed as plain AC rectifiers. They are made as "AC-DC converters", where AC gets rectified to high-voltage DC, the DC gets converted into high-frequency "chopped" signal, and then transformed into low-voltage, and then rectified using synchronous active rectifiers. The smoothing caps then operate at a very high frequency and don't need to be of extremely high values. You need to seriously reconsider your design approach, unless you are making some welding apparatus.
$endgroup$
– Ale..chenski
5 hours ago
1
1
$begingroup$
In addition to what Ale..chenski said, you have discovered that at high currents and low frequency you need an unreasonable amount of smoothing capacitance since the charge consumed is so high but the refresh rate is so low. The ripple will be enormous even with massive caps. Refresh a few thousand times faster and your capacitors can be a few thousand times smaller. Capacitors have a limit on ripple current anyways and such high currents and low frequencies will produce enormous ripple currents that will overheat and blow caps.
$endgroup$
– Toor
5 hours ago
$begingroup$
In addition to what Ale..chenski said, you have discovered that at high currents and low frequency you need an unreasonable amount of smoothing capacitance since the charge consumed is so high but the refresh rate is so low. The ripple will be enormous even with massive caps. Refresh a few thousand times faster and your capacitors can be a few thousand times smaller. Capacitors have a limit on ripple current anyways and such high currents and low frequencies will produce enormous ripple currents that will overheat and blow caps.
$endgroup$
– Toor
5 hours ago
$begingroup$
what is your ripple spec 10% V? with 200A RMS
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
what is your ripple spec 10% V? with 200A RMS
$endgroup$
– Sunnyskyguy EE75
3 hours ago
1
1
$begingroup$
@SunnyskyguyEE75 haven't considered that, just want to minimise what ripple there is. If it's not worth the effort, I may just leave it as AC. The purpose is to heat a piece of carbon or stainless to serve as a heating element. Was considering rectifying the AC signal to minimise any eddy currents.
$endgroup$
– Chris
3 hours ago
$begingroup$
@SunnyskyguyEE75 haven't considered that, just want to minimise what ripple there is. If it's not worth the effort, I may just leave it as AC. The purpose is to heat a piece of carbon or stainless to serve as a heating element. Was considering rectifying the AC signal to minimise any eddy currents.
$endgroup$
– Chris
3 hours ago
1
1
$begingroup$
You can use a 6V 700A CCA lead acid battery
$endgroup$
– Sunnyskyguy EE75
2 hours ago
$begingroup$
You can use a 6V 700A CCA lead acid battery
$endgroup$
– Sunnyskyguy EE75
2 hours ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
First make a rough spec for ripple from C and ESR.
For $C=I_C*dt/dV$
with Vc=5V and let ripple=5% or dV=250mV(pp) and dt= 10ms ( 100Hz) with Ic=200A
$C=200A*10ms/250V=8mF$
let dV=2% 100mV from cap
ESR=V/ESR=0.1V/200A=500 μΩ
For a total ripple of 250mV+100mV=350mVpp
ESR*C=T= 500 μΩ * 8mF = 4 μs
Now check Electric Double Layer Capacitors (EDLC), Supercapacitors specs. anything close rated for >5V
8.2uF 50mΩ TOO HIGH by 100x even at $16
Now you know this concept is going to be more expensive than a 1kW PC SMPS PSU.
$endgroup$
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%2f423103%2fhigh-current-low-voltage-capacitors%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
$begingroup$
First make a rough spec for ripple from C and ESR.
For $C=I_C*dt/dV$
with Vc=5V and let ripple=5% or dV=250mV(pp) and dt= 10ms ( 100Hz) with Ic=200A
$C=200A*10ms/250V=8mF$
let dV=2% 100mV from cap
ESR=V/ESR=0.1V/200A=500 μΩ
For a total ripple of 250mV+100mV=350mVpp
ESR*C=T= 500 μΩ * 8mF = 4 μs
Now check Electric Double Layer Capacitors (EDLC), Supercapacitors specs. anything close rated for >5V
8.2uF 50mΩ TOO HIGH by 100x even at $16
Now you know this concept is going to be more expensive than a 1kW PC SMPS PSU.
$endgroup$
add a comment |
$begingroup$
First make a rough spec for ripple from C and ESR.
For $C=I_C*dt/dV$
with Vc=5V and let ripple=5% or dV=250mV(pp) and dt= 10ms ( 100Hz) with Ic=200A
$C=200A*10ms/250V=8mF$
let dV=2% 100mV from cap
ESR=V/ESR=0.1V/200A=500 μΩ
For a total ripple of 250mV+100mV=350mVpp
ESR*C=T= 500 μΩ * 8mF = 4 μs
Now check Electric Double Layer Capacitors (EDLC), Supercapacitors specs. anything close rated for >5V
8.2uF 50mΩ TOO HIGH by 100x even at $16
Now you know this concept is going to be more expensive than a 1kW PC SMPS PSU.
$endgroup$
add a comment |
$begingroup$
First make a rough spec for ripple from C and ESR.
For $C=I_C*dt/dV$
with Vc=5V and let ripple=5% or dV=250mV(pp) and dt= 10ms ( 100Hz) with Ic=200A
$C=200A*10ms/250V=8mF$
let dV=2% 100mV from cap
ESR=V/ESR=0.1V/200A=500 μΩ
For a total ripple of 250mV+100mV=350mVpp
ESR*C=T= 500 μΩ * 8mF = 4 μs
Now check Electric Double Layer Capacitors (EDLC), Supercapacitors specs. anything close rated for >5V
8.2uF 50mΩ TOO HIGH by 100x even at $16
Now you know this concept is going to be more expensive than a 1kW PC SMPS PSU.
$endgroup$
First make a rough spec for ripple from C and ESR.
For $C=I_C*dt/dV$
with Vc=5V and let ripple=5% or dV=250mV(pp) and dt= 10ms ( 100Hz) with Ic=200A
$C=200A*10ms/250V=8mF$
let dV=2% 100mV from cap
ESR=V/ESR=0.1V/200A=500 μΩ
For a total ripple of 250mV+100mV=350mVpp
ESR*C=T= 500 μΩ * 8mF = 4 μs
Now check Electric Double Layer Capacitors (EDLC), Supercapacitors specs. anything close rated for >5V
8.2uF 50mΩ TOO HIGH by 100x even at $16
Now you know this concept is going to be more expensive than a 1kW PC SMPS PSU.
answered 2 hours ago
Sunnyskyguy EE75Sunnyskyguy EE75
66.6k22397
66.6k22397
add a comment |
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%2f423103%2fhigh-current-low-voltage-capacitors%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
3
$begingroup$
Low-voltage high-amperage power sources are not designed as plain AC rectifiers. They are made as "AC-DC converters", where AC gets rectified to high-voltage DC, the DC gets converted into high-frequency "chopped" signal, and then transformed into low-voltage, and then rectified using synchronous active rectifiers. The smoothing caps then operate at a very high frequency and don't need to be of extremely high values. You need to seriously reconsider your design approach, unless you are making some welding apparatus.
$endgroup$
– Ale..chenski
5 hours ago
1
$begingroup$
In addition to what Ale..chenski said, you have discovered that at high currents and low frequency you need an unreasonable amount of smoothing capacitance since the charge consumed is so high but the refresh rate is so low. The ripple will be enormous even with massive caps. Refresh a few thousand times faster and your capacitors can be a few thousand times smaller. Capacitors have a limit on ripple current anyways and such high currents and low frequencies will produce enormous ripple currents that will overheat and blow caps.
$endgroup$
– Toor
5 hours ago
$begingroup$
what is your ripple spec 10% V? with 200A RMS
$endgroup$
– Sunnyskyguy EE75
3 hours ago
1
$begingroup$
@SunnyskyguyEE75 haven't considered that, just want to minimise what ripple there is. If it's not worth the effort, I may just leave it as AC. The purpose is to heat a piece of carbon or stainless to serve as a heating element. Was considering rectifying the AC signal to minimise any eddy currents.
$endgroup$
– Chris
3 hours ago
1
$begingroup$
You can use a 6V 700A CCA lead acid battery
$endgroup$
– Sunnyskyguy EE75
2 hours ago