Diode in opposite direction?Trying to understand flyback diodes and general diode orientationDiode parallel to resistor on H-Bridge MOSFETs gatesQuestion about fly-back diodeWhy is this diode here?Unknown diode identification (27 XD or XD 27 markings)Diode pin labels swapped in this circuit?Why do diode rings multiply?How to solve this diode circuit?NPN driving PNP, Diode from NPN Base to PNP Collector?confused about diode polarity vs circuit polarity
Bob has never been a M before
I'm in charge of equipment buying but no one's ever happy with what I choose. How to fix this?
Should a half Jewish man be discouraged from marrying a Jewess?
How can a jailer prevent the Forge Cleric's Artisan's Blessing from being used?
Books on the History of math research at European universities
Simulating a probability of 1 of 2^N with less than N random bits
Indicating multiple different modes of speech (fantasy language or telepathy)
The One-Electron Universe postulate is true - what simple change can I make to change the whole universe?
Calculating the number of days between 2 dates in Excel
The most efficient algorithm to find all possible integer pairs which sum to a given integer
My boss asked me to take a one-day class, then signs it up as a day off
Would it be legal for a US State to ban exports of a natural resource?
I2C signal and power over long range (10meter cable)
No idea how to draw this using tikz
For airliners, what prevents wing strikes on landing in bad weather?
How to color a zone in Tikz
What if somebody invests in my application?
Is a naturally all "male" species possible?
Why isn't KTEX's runway designation 10/28 instead of 9/27?
Greatest common substring
Can a Bard use an arcane focus?
A known event to a history junkie
Pronouncing Homer as in modern Greek
Have I saved too much for retirement so far?
Diode in opposite direction?
Trying to understand flyback diodes and general diode orientationDiode parallel to resistor on H-Bridge MOSFETs gatesQuestion about fly-back diodeWhy is this diode here?Unknown diode identification (27 XD or XD 27 markings)Diode pin labels swapped in this circuit?Why do diode rings multiply?How to solve this diode circuit?NPN driving PNP, Diode from NPN Base to PNP Collector?confused about diode polarity vs circuit polarity
$begingroup$
Its really bothering me that the diode is shown in the opposite direction here and i don't understand why its been put like that? the input voltage will come from the left side of the circuit then why is the diode's cathode connected to the output of the op amp? does the polarity even matter here?
diodes radio receiver
asked yesterday
HaidyEHaidyE
304
$endgroup$
add a comment |
$begingroup$
Its really bothering me that the diode is shown in the opposite direction here and i don't understand why its been put like that? the input voltage will come from the left side of the circuit then why is the diode's cathode connected to the output of the op amp? does the polarity even matter here?
diodes radio receiver
asked yesterday
HaidyEHaidyE
304
$endgroup$
add a comment |
$begingroup$
Its really bothering me that the diode is shown in the opposite direction here and i don't understand why its been put like that? the input voltage will come from the left side of the circuit then why is the diode's cathode connected to the output of the op amp? does the polarity even matter here?
diodes radio receiver
asked yesterday
HaidyEHaidyE
304
$endgroup$
Its really bothering me that the diode is shown in the opposite direction here and i don't understand why its been put like that? the input voltage will come from the left side of the circuit then why is the diode's cathode connected to the output of the op amp? does the polarity even matter here?
diodes radio receiver
diodes radio receiver
asked yesterday
HaidyEHaidyE
304
asked yesterday
HaidyEHaidyE
304
asked yesterday
HaidyEHaidyE
304
asked yesterday
HaidyEHaidyE
304
asked yesterday
HaidyEHaidyE
304
304
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In this particular circuit application, the demodulation of amplitude modulated RF, no, the polarity of the diode does not matter. If you reverse the diode, you simply track the positive part of the envelope rather than the negative part. Either will give you the demodulated signal.
answered yesterday
Neil_UKNeil_UK
78k284179
$endgroup$
$begingroup$
will the gain not make a difference then? we use three op amps.
$endgroup$
– HaidyE
yesterday
$begingroup$
@HaidyE no, I don't understand your comment. Neither gain nor number of opamps used makes a difference to the fact that inverting the polarity of the diode will not have a significant effect on the audio recovered from the RF signal, except that is, the polarity of the recovered signal.
$endgroup$
– Neil_UK
yesterday
$begingroup$
@HaidyE Your RF "Amp" (between L1C2 and Diode) is an op-amp? A large portion of available opamps haven't much gain available for radio-frequency signals. Audio signals after the detector stage are considered low-frequency, and op-amps have lots of useful gain. That first amplifier really needs a large GBW product (gain x bandwidth)...if you use an op-amp, select carefully for this spec.
$endgroup$
– glen_geek
yesterday
2
$begingroup$
The RF amp may have a open collector output, so the diode would have to have its cathode facing the IC.
$endgroup$
– Sparky256
yesterday
$begingroup$
i have used 3 op amps with a gain of 5 as i have mentioned before
$endgroup$
– HaidyE
yesterday
|
show 2 more comments
$begingroup$
At the point just before the first amp, the radio waves have been filtered to a particular frequency by L1, C1 & C2. That first amplifier is not an op-amp, it is an RF gain amplifier. It amplifies the incoming signal by a number of dB. The signal before and after will be an AC signal, equally biased around ground.
The diode or detector gets rid of one half of the signal (either the positive half or the negative half, depending on which way the diode is). Description of AM Detector.
The next stage filters off the carrier signal with a low-pass filter.
As the next amp, an op-amp has a bipolar power supply so it can handle either the positive or negative signal. It amplifies the sound waveform for the speaker.
edited yesterday
SamGibson
11.5k41739
answered yesterday
TpKnetTpKnet
913
New contributor
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The signal into the diode is a (presumably) AM (amplitude modulated) signal. The variations in amplitude are what is of interest to the listener and are what the amplifier chain is seeking to recover.
As shown the diode rectifies negative going halves of the signal.
If reversed it would rectify positive going half cycles.
Either way, Cd provides a filter that smooths out (and so removes) the RF variations and results in a voltage that varies with the amplitude of the incoming signal. As shown you get negative variations which are smoothed. Reverse the diode and you would get positive going variations. The two are the same except inverted.
In either case the resultant "envelope" is AC coupled by Cb, and is DC ground referenced by Rb. So EITHER way "Amplifier" "sees" an AC signal centred around ground. This is amplified and, again, AC coupled vi C3 to the headphones.
So, either way the result is much the same to the end user.
answered 23 hours ago
Russell McMahonRussell McMahon
117k9165295
$endgroup$
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%2f428844%2fdiode-in-opposite-direction%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In this particular circuit application, the demodulation of amplitude modulated RF, no, the polarity of the diode does not matter. If you reverse the diode, you simply track the positive part of the envelope rather than the negative part. Either will give you the demodulated signal.
answered yesterday
Neil_UKNeil_UK
78k284179
$endgroup$
$begingroup$
will the gain not make a difference then? we use three op amps.
$endgroup$
– HaidyE
yesterday
$begingroup$
@HaidyE no, I don't understand your comment. Neither gain nor number of opamps used makes a difference to the fact that inverting the polarity of the diode will not have a significant effect on the audio recovered from the RF signal, except that is, the polarity of the recovered signal.
$endgroup$
– Neil_UK
yesterday
$begingroup$
@HaidyE Your RF "Amp" (between L1C2 and Diode) is an op-amp? A large portion of available opamps haven't much gain available for radio-frequency signals. Audio signals after the detector stage are considered low-frequency, and op-amps have lots of useful gain. That first amplifier really needs a large GBW product (gain x bandwidth)...if you use an op-amp, select carefully for this spec.
$endgroup$
– glen_geek
yesterday
2
$begingroup$
The RF amp may have a open collector output, so the diode would have to have its cathode facing the IC.
$endgroup$
– Sparky256
yesterday
$begingroup$
i have used 3 op amps with a gain of 5 as i have mentioned before
$endgroup$
– HaidyE
yesterday
|
show 2 more comments
$begingroup$
In this particular circuit application, the demodulation of amplitude modulated RF, no, the polarity of the diode does not matter. If you reverse the diode, you simply track the positive part of the envelope rather than the negative part. Either will give you the demodulated signal.
answered yesterday
Neil_UKNeil_UK
78k284179
$endgroup$
$begingroup$
will the gain not make a difference then? we use three op amps.
$endgroup$
– HaidyE
yesterday
$begingroup$
@HaidyE no, I don't understand your comment. Neither gain nor number of opamps used makes a difference to the fact that inverting the polarity of the diode will not have a significant effect on the audio recovered from the RF signal, except that is, the polarity of the recovered signal.
$endgroup$
– Neil_UK
yesterday
$begingroup$
@HaidyE Your RF "Amp" (between L1C2 and Diode) is an op-amp? A large portion of available opamps haven't much gain available for radio-frequency signals. Audio signals after the detector stage are considered low-frequency, and op-amps have lots of useful gain. That first amplifier really needs a large GBW product (gain x bandwidth)...if you use an op-amp, select carefully for this spec.
$endgroup$
– glen_geek
yesterday
2
$begingroup$
The RF amp may have a open collector output, so the diode would have to have its cathode facing the IC.
$endgroup$
– Sparky256
yesterday
$begingroup$
i have used 3 op amps with a gain of 5 as i have mentioned before
$endgroup$
– HaidyE
yesterday
|
show 2 more comments
$begingroup$
In this particular circuit application, the demodulation of amplitude modulated RF, no, the polarity of the diode does not matter. If you reverse the diode, you simply track the positive part of the envelope rather than the negative part. Either will give you the demodulated signal.
answered yesterday
Neil_UKNeil_UK
78k284179
$endgroup$
In this particular circuit application, the demodulation of amplitude modulated RF, no, the polarity of the diode does not matter. If you reverse the diode, you simply track the positive part of the envelope rather than the negative part. Either will give you the demodulated signal.
answered yesterday
Neil_UKNeil_UK
78k284179
answered yesterday
Neil_UKNeil_UK
78k284179
answered yesterday
Neil_UKNeil_UK
78k284179
answered yesterday
Neil_UKNeil_UK
78k284179
78k284179
$begingroup$
will the gain not make a difference then? we use three op amps.
$endgroup$
– HaidyE
yesterday
$begingroup$
@HaidyE no, I don't understand your comment. Neither gain nor number of opamps used makes a difference to the fact that inverting the polarity of the diode will not have a significant effect on the audio recovered from the RF signal, except that is, the polarity of the recovered signal.
$endgroup$
– Neil_UK
yesterday
$begingroup$
@HaidyE Your RF "Amp" (between L1C2 and Diode) is an op-amp? A large portion of available opamps haven't much gain available for radio-frequency signals. Audio signals after the detector stage are considered low-frequency, and op-amps have lots of useful gain. That first amplifier really needs a large GBW product (gain x bandwidth)...if you use an op-amp, select carefully for this spec.
$endgroup$
– glen_geek
yesterday
2
$begingroup$
The RF amp may have a open collector output, so the diode would have to have its cathode facing the IC.
$endgroup$
– Sparky256
yesterday
$begingroup$
i have used 3 op amps with a gain of 5 as i have mentioned before
$endgroup$
– HaidyE
yesterday
|
show 2 more comments
$begingroup$
will the gain not make a difference then? we use three op amps.
$endgroup$
– HaidyE
yesterday
$begingroup$
@HaidyE no, I don't understand your comment. Neither gain nor number of opamps used makes a difference to the fact that inverting the polarity of the diode will not have a significant effect on the audio recovered from the RF signal, except that is, the polarity of the recovered signal.
$endgroup$
– Neil_UK
yesterday
$begingroup$
@HaidyE Your RF "Amp" (between L1C2 and Diode) is an op-amp? A large portion of available opamps haven't much gain available for radio-frequency signals. Audio signals after the detector stage are considered low-frequency, and op-amps have lots of useful gain. That first amplifier really needs a large GBW product (gain x bandwidth)...if you use an op-amp, select carefully for this spec.
$endgroup$
– glen_geek
yesterday
2
$begingroup$
The RF amp may have a open collector output, so the diode would have to have its cathode facing the IC.
$endgroup$
– Sparky256
yesterday
$begingroup$
i have used 3 op amps with a gain of 5 as i have mentioned before
$endgroup$
– HaidyE
yesterday
$begingroup$
will the gain not make a difference then? we use three op amps.
$endgroup$
– HaidyE
yesterday
$begingroup$
will the gain not make a difference then? we use three op amps.
$endgroup$
– HaidyE
yesterday
$begingroup$
@HaidyE no, I don't understand your comment. Neither gain nor number of opamps used makes a difference to the fact that inverting the polarity of the diode will not have a significant effect on the audio recovered from the RF signal, except that is, the polarity of the recovered signal.
$endgroup$
– Neil_UK
yesterday
$begingroup$
@HaidyE no, I don't understand your comment. Neither gain nor number of opamps used makes a difference to the fact that inverting the polarity of the diode will not have a significant effect on the audio recovered from the RF signal, except that is, the polarity of the recovered signal.
$endgroup$
– Neil_UK
yesterday
$begingroup$
@HaidyE Your RF "Amp" (between L1C2 and Diode) is an op-amp? A large portion of available opamps haven't much gain available for radio-frequency signals. Audio signals after the detector stage are considered low-frequency, and op-amps have lots of useful gain. That first amplifier really needs a large GBW product (gain x bandwidth)...if you use an op-amp, select carefully for this spec.
$endgroup$
– glen_geek
yesterday
$begingroup$
@HaidyE Your RF "Amp" (between L1C2 and Diode) is an op-amp? A large portion of available opamps haven't much gain available for radio-frequency signals. Audio signals after the detector stage are considered low-frequency, and op-amps have lots of useful gain. That first amplifier really needs a large GBW product (gain x bandwidth)...if you use an op-amp, select carefully for this spec.
$endgroup$
– glen_geek
yesterday
2
2
$begingroup$
The RF amp may have a open collector output, so the diode would have to have its cathode facing the IC.
$endgroup$
– Sparky256
yesterday
$begingroup$
The RF amp may have a open collector output, so the diode would have to have its cathode facing the IC.
$endgroup$
– Sparky256
yesterday
$begingroup$
i have used 3 op amps with a gain of 5 as i have mentioned before
$endgroup$
– HaidyE
yesterday
$begingroup$
i have used 3 op amps with a gain of 5 as i have mentioned before
$endgroup$
– HaidyE
yesterday
|
show 2 more comments
$begingroup$
At the point just before the first amp, the radio waves have been filtered to a particular frequency by L1, C1 & C2. That first amplifier is not an op-amp, it is an RF gain amplifier. It amplifies the incoming signal by a number of dB. The signal before and after will be an AC signal, equally biased around ground.
The diode or detector gets rid of one half of the signal (either the positive half or the negative half, depending on which way the diode is). Description of AM Detector.
The next stage filters off the carrier signal with a low-pass filter.
As the next amp, an op-amp has a bipolar power supply so it can handle either the positive or negative signal. It amplifies the sound waveform for the speaker.
edited yesterday
SamGibson
11.5k41739
answered yesterday
TpKnetTpKnet
913
New contributor
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
At the point just before the first amp, the radio waves have been filtered to a particular frequency by L1, C1 & C2. That first amplifier is not an op-amp, it is an RF gain amplifier. It amplifies the incoming signal by a number of dB. The signal before and after will be an AC signal, equally biased around ground.
The diode or detector gets rid of one half of the signal (either the positive half or the negative half, depending on which way the diode is). Description of AM Detector.
The next stage filters off the carrier signal with a low-pass filter.
As the next amp, an op-amp has a bipolar power supply so it can handle either the positive or negative signal. It amplifies the sound waveform for the speaker.
edited yesterday
SamGibson
11.5k41739
answered yesterday
TpKnetTpKnet
913
New contributor
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
At the point just before the first amp, the radio waves have been filtered to a particular frequency by L1, C1 & C2. That first amplifier is not an op-amp, it is an RF gain amplifier. It amplifies the incoming signal by a number of dB. The signal before and after will be an AC signal, equally biased around ground.
The diode or detector gets rid of one half of the signal (either the positive half or the negative half, depending on which way the diode is). Description of AM Detector.
The next stage filters off the carrier signal with a low-pass filter.
As the next amp, an op-amp has a bipolar power supply so it can handle either the positive or negative signal. It amplifies the sound waveform for the speaker.
edited yesterday
SamGibson
11.5k41739
answered yesterday
TpKnetTpKnet
913
New contributor
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
At the point just before the first amp, the radio waves have been filtered to a particular frequency by L1, C1 & C2. That first amplifier is not an op-amp, it is an RF gain amplifier. It amplifies the incoming signal by a number of dB. The signal before and after will be an AC signal, equally biased around ground.
The diode or detector gets rid of one half of the signal (either the positive half or the negative half, depending on which way the diode is). Description of AM Detector.
The next stage filters off the carrier signal with a low-pass filter.
As the next amp, an op-amp has a bipolar power supply so it can handle either the positive or negative signal. It amplifies the sound waveform for the speaker.
edited yesterday
SamGibson
11.5k41739
answered yesterday
TpKnetTpKnet
913
New contributor
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
SamGibson
11.5k41739
edited yesterday
SamGibson
11.5k41739
edited yesterday
SamGibson
11.5k41739
11.5k41739
answered yesterday
TpKnetTpKnet
913
New contributor
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
TpKnetTpKnet
913
answered yesterday
TpKnetTpKnet
913
913
New contributor
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
TpKnet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
The signal into the diode is a (presumably) AM (amplitude modulated) signal. The variations in amplitude are what is of interest to the listener and are what the amplifier chain is seeking to recover.
As shown the diode rectifies negative going halves of the signal.
If reversed it would rectify positive going half cycles.
Either way, Cd provides a filter that smooths out (and so removes) the RF variations and results in a voltage that varies with the amplitude of the incoming signal. As shown you get negative variations which are smoothed. Reverse the diode and you would get positive going variations. The two are the same except inverted.
In either case the resultant "envelope" is AC coupled by Cb, and is DC ground referenced by Rb. So EITHER way "Amplifier" "sees" an AC signal centred around ground. This is amplified and, again, AC coupled vi C3 to the headphones.
So, either way the result is much the same to the end user.
answered 23 hours ago
Russell McMahonRussell McMahon
117k9165295
$endgroup$
add a comment |
$begingroup$
The signal into the diode is a (presumably) AM (amplitude modulated) signal. The variations in amplitude are what is of interest to the listener and are what the amplifier chain is seeking to recover.
As shown the diode rectifies negative going halves of the signal.
If reversed it would rectify positive going half cycles.
Either way, Cd provides a filter that smooths out (and so removes) the RF variations and results in a voltage that varies with the amplitude of the incoming signal. As shown you get negative variations which are smoothed. Reverse the diode and you would get positive going variations. The two are the same except inverted.
In either case the resultant "envelope" is AC coupled by Cb, and is DC ground referenced by Rb. So EITHER way "Amplifier" "sees" an AC signal centred around ground. This is amplified and, again, AC coupled vi C3 to the headphones.
So, either way the result is much the same to the end user.
answered 23 hours ago
Russell McMahonRussell McMahon
117k9165295
$endgroup$
add a comment |
$begingroup$
The signal into the diode is a (presumably) AM (amplitude modulated) signal. The variations in amplitude are what is of interest to the listener and are what the amplifier chain is seeking to recover.
As shown the diode rectifies negative going halves of the signal.
If reversed it would rectify positive going half cycles.
Either way, Cd provides a filter that smooths out (and so removes) the RF variations and results in a voltage that varies with the amplitude of the incoming signal. As shown you get negative variations which are smoothed. Reverse the diode and you would get positive going variations. The two are the same except inverted.
In either case the resultant "envelope" is AC coupled by Cb, and is DC ground referenced by Rb. So EITHER way "Amplifier" "sees" an AC signal centred around ground. This is amplified and, again, AC coupled vi C3 to the headphones.
So, either way the result is much the same to the end user.
answered 23 hours ago
Russell McMahonRussell McMahon
117k9165295
$endgroup$
The signal into the diode is a (presumably) AM (amplitude modulated) signal. The variations in amplitude are what is of interest to the listener and are what the amplifier chain is seeking to recover.
As shown the diode rectifies negative going halves of the signal.
If reversed it would rectify positive going half cycles.
Either way, Cd provides a filter that smooths out (and so removes) the RF variations and results in a voltage that varies with the amplitude of the incoming signal. As shown you get negative variations which are smoothed. Reverse the diode and you would get positive going variations. The two are the same except inverted.
In either case the resultant "envelope" is AC coupled by Cb, and is DC ground referenced by Rb. So EITHER way "Amplifier" "sees" an AC signal centred around ground. This is amplified and, again, AC coupled vi C3 to the headphones.
So, either way the result is much the same to the end user.
answered 23 hours ago
Russell McMahonRussell McMahon
117k9165295
answered 23 hours ago
Russell McMahonRussell McMahon
117k9165295
answered 23 hours ago
Russell McMahonRussell McMahon
117k9165295
answered 23 hours ago
Russell McMahonRussell McMahon
117k9165295
117k9165295
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%2f428844%2fdiode-in-opposite-direction%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
