Why does a 74HC595N work in LED matrices?
I have checked a lot of LED matrices, and mostly a 74HC595N shift register is used. In some cases a TPIC595B like below.
Example: Driving LED arrays with Arduino
Circuit from above example:
I understand that for sinking a TPIC is needed, because when all columns are given 25 mA, it would result in a total current of 200 mA. The 74HC595N can only handle 70 mA.
As I understand, the 74HC595 controls the rows one by one. However, if all columns of a row are getting a current of 25 mA, will there be 200 mA through a single source pin of the 74HC595? Or should there be max. 70 / 8 = 8.8 mA per LED? The TPIC can handle 150 mA per pin, so that more than enough.
Can the 74HC595 handle 8 LEDs at 20 mA in the above circuit?
led led-matrix 74hc595
asked 2 hours ago
Michel KeijzersMichel Keijzers
5,75992562
add a comment |
I have checked a lot of LED matrices, and mostly a 74HC595N shift register is used. In some cases a TPIC595B like below.
Example: Driving LED arrays with Arduino
Circuit from above example:
I understand that for sinking a TPIC is needed, because when all columns are given 25 mA, it would result in a total current of 200 mA. The 74HC595N can only handle 70 mA.
As I understand, the 74HC595 controls the rows one by one. However, if all columns of a row are getting a current of 25 mA, will there be 200 mA through a single source pin of the 74HC595? Or should there be max. 70 / 8 = 8.8 mA per LED? The TPIC can handle 150 mA per pin, so that more than enough.
Can the 74HC595 handle 8 LEDs at 20 mA in the above circuit?
led led-matrix 74hc595
asked 2 hours ago
Michel KeijzersMichel Keijzers
5,75992562
1
Bit of a duplicate of this question?
– Finbarr
1 hour ago
@Finbarr not a duplicate, but a nice bit of extra information, thanks for the link.
– Michel Keijzers
1 hour ago
add a comment |
I have checked a lot of LED matrices, and mostly a 74HC595N shift register is used. In some cases a TPIC595B like below.
Example: Driving LED arrays with Arduino
Circuit from above example:
I understand that for sinking a TPIC is needed, because when all columns are given 25 mA, it would result in a total current of 200 mA. The 74HC595N can only handle 70 mA.
As I understand, the 74HC595 controls the rows one by one. However, if all columns of a row are getting a current of 25 mA, will there be 200 mA through a single source pin of the 74HC595? Or should there be max. 70 / 8 = 8.8 mA per LED? The TPIC can handle 150 mA per pin, so that more than enough.
Can the 74HC595 handle 8 LEDs at 20 mA in the above circuit?
led led-matrix 74hc595
asked 2 hours ago
Michel KeijzersMichel Keijzers
5,75992562
I have checked a lot of LED matrices, and mostly a 74HC595N shift register is used. In some cases a TPIC595B like below.
Example: Driving LED arrays with Arduino
Circuit from above example:
I understand that for sinking a TPIC is needed, because when all columns are given 25 mA, it would result in a total current of 200 mA. The 74HC595N can only handle 70 mA.
As I understand, the 74HC595 controls the rows one by one. However, if all columns of a row are getting a current of 25 mA, will there be 200 mA through a single source pin of the 74HC595? Or should there be max. 70 / 8 = 8.8 mA per LED? The TPIC can handle 150 mA per pin, so that more than enough.
Can the 74HC595 handle 8 LEDs at 20 mA in the above circuit?
led led-matrix 74hc595
led led-matrix 74hc595
asked 2 hours ago
Michel KeijzersMichel Keijzers
5,75992562
asked 2 hours ago
Michel KeijzersMichel Keijzers
5,75992562
asked 2 hours ago
Michel KeijzersMichel Keijzers
5,75992562
asked 2 hours ago
Michel KeijzersMichel Keijzers
5,75992562
asked 2 hours ago
Michel KeijzersMichel Keijzers
5,75992562
5,75992562
1
Bit of a duplicate of this question?
– Finbarr
1 hour ago
@Finbarr not a duplicate, but a nice bit of extra information, thanks for the link.
– Michel Keijzers
1 hour ago
add a comment |
1
Bit of a duplicate of this question?
– Finbarr
1 hour ago
@Finbarr not a duplicate, but a nice bit of extra information, thanks for the link.
– Michel Keijzers
1 hour ago
1
1
Bit of a duplicate of this question?
– Finbarr
1 hour ago
Bit of a duplicate of this question?
– Finbarr
1 hour ago
@Finbarr not a duplicate, but a nice bit of extra information, thanks for the link.
– Michel Keijzers
1 hour ago
@Finbarr not a duplicate, but a nice bit of extra information, thanks for the link.
– Michel Keijzers
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
All the LED current flows through the Vcc pin of the 74HC595 (and the GND pin of the TIPxxx). The absolute maximum rated current through Vcc (or GND) of a typical 74HC595 is indeed 70mA.
The absolute maximum peak current per LED is thus 8.75mA, or an average current of about 1mA per LED (1/8 duty cycle per LED).
In practice you should stay WELL away from the absolute maximum value.
To put it explicitly, this is a hobbyist-level circuit, designed by someone who doesn't care or doesn't know much about reliability (assuming they actually recommended anything like the currents you stated). Using the 74HC595 to drive a high-side driver array or prebiased PNP transistor duals would be much better. They're designed as logic shift registers, not as load drivers.
Using such drivers you could also get a much higher brightness. An average current of 10mA per LED requires a total current of 640mA, obviously, which means that the source drivers need to handle 80mA each (with all potentially on at once) and the sink drivers need to hand 640mA each (with each one seeing a 1/8 duty cycle).
answered 2 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
Thanks, so it would help to use a TPIC for sourcing too (than I can use 150 / 8 = 18.75 mA per LED, in one row) ?
– Michel Keijzers
2 hours ago
1
The TPIC has open-drain DMOS transistors so it wouldn't work. You need a high-side (source) driver.
– Spehro Pefhany
2 hours ago
Ah great, than I could use a 74HC595 to drive multiple 2N7000 transistors for sourcing and a TPIC595B for sinking. I think the TPIC can do not 640 mA, but I will scan through the rows anyway.
– Michel Keijzers
2 hours ago
1
You need p-channel transistors, eg. AO3401 and to invert the logic (low = on). The TPIC can handle about 250mA/output with only one on (see figure 10).
– Spehro Pefhany
1 hour ago
Thanks again for that tip ... however I will search for some throughole similar P channel transistor since I never used SMD so far.
– Michel Keijzers
1 hour ago
|
show 3 more comments
In this example, the 74HC595 sources say 8-10mA per output as determined by R1-R8. One output of the TPICx595 is turned on a time to sink 64-80mA of current if all 8 LEDs are on. That's it. The 640mA number is not correct.
Each column is turned on for 2-3mS, then it is turned off, data for the next column is turned on, and the next column drive is turned. Repeat for all 8 columns.
That's the basis of multiplexing - cycle thru 8 columns quickly and fool the eye into thinking all 64 LEDs can be on at once, when in reality only 8 are ever turned on.
answered 21 mins ago
CrossRoadsCrossRoads
1,3008
Than I will check if 8-10 mA is good enough (probably it is), and use a 74HC595 and TPICB595 (as you proposed earlier). At least I now have a better understanding. I will do first some tests before I ask more (for most experts here) trivial questions.
– Michel Keijzers
19 mins ago
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%2f416035%2fwhy-does-a-74hc595n-work-in-led-matrices%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
All the LED current flows through the Vcc pin of the 74HC595 (and the GND pin of the TIPxxx). The absolute maximum rated current through Vcc (or GND) of a typical 74HC595 is indeed 70mA.
The absolute maximum peak current per LED is thus 8.75mA, or an average current of about 1mA per LED (1/8 duty cycle per LED).
In practice you should stay WELL away from the absolute maximum value.
To put it explicitly, this is a hobbyist-level circuit, designed by someone who doesn't care or doesn't know much about reliability (assuming they actually recommended anything like the currents you stated). Using the 74HC595 to drive a high-side driver array or prebiased PNP transistor duals would be much better. They're designed as logic shift registers, not as load drivers.
Using such drivers you could also get a much higher brightness. An average current of 10mA per LED requires a total current of 640mA, obviously, which means that the source drivers need to handle 80mA each (with all potentially on at once) and the sink drivers need to hand 640mA each (with each one seeing a 1/8 duty cycle).
answered 2 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
Thanks, so it would help to use a TPIC for sourcing too (than I can use 150 / 8 = 18.75 mA per LED, in one row) ?
– Michel Keijzers
2 hours ago
1
The TPIC has open-drain DMOS transistors so it wouldn't work. You need a high-side (source) driver.
– Spehro Pefhany
2 hours ago
Ah great, than I could use a 74HC595 to drive multiple 2N7000 transistors for sourcing and a TPIC595B for sinking. I think the TPIC can do not 640 mA, but I will scan through the rows anyway.
– Michel Keijzers
2 hours ago
1
You need p-channel transistors, eg. AO3401 and to invert the logic (low = on). The TPIC can handle about 250mA/output with only one on (see figure 10).
– Spehro Pefhany
1 hour ago
Thanks again for that tip ... however I will search for some throughole similar P channel transistor since I never used SMD so far.
– Michel Keijzers
1 hour ago
|
show 3 more comments
All the LED current flows through the Vcc pin of the 74HC595 (and the GND pin of the TIPxxx). The absolute maximum rated current through Vcc (or GND) of a typical 74HC595 is indeed 70mA.
The absolute maximum peak current per LED is thus 8.75mA, or an average current of about 1mA per LED (1/8 duty cycle per LED).
In practice you should stay WELL away from the absolute maximum value.
To put it explicitly, this is a hobbyist-level circuit, designed by someone who doesn't care or doesn't know much about reliability (assuming they actually recommended anything like the currents you stated). Using the 74HC595 to drive a high-side driver array or prebiased PNP transistor duals would be much better. They're designed as logic shift registers, not as load drivers.
Using such drivers you could also get a much higher brightness. An average current of 10mA per LED requires a total current of 640mA, obviously, which means that the source drivers need to handle 80mA each (with all potentially on at once) and the sink drivers need to hand 640mA each (with each one seeing a 1/8 duty cycle).
answered 2 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
Thanks, so it would help to use a TPIC for sourcing too (than I can use 150 / 8 = 18.75 mA per LED, in one row) ?
– Michel Keijzers
2 hours ago
1
The TPIC has open-drain DMOS transistors so it wouldn't work. You need a high-side (source) driver.
– Spehro Pefhany
2 hours ago
Ah great, than I could use a 74HC595 to drive multiple 2N7000 transistors for sourcing and a TPIC595B for sinking. I think the TPIC can do not 640 mA, but I will scan through the rows anyway.
– Michel Keijzers
2 hours ago
1
You need p-channel transistors, eg. AO3401 and to invert the logic (low = on). The TPIC can handle about 250mA/output with only one on (see figure 10).
– Spehro Pefhany
1 hour ago
Thanks again for that tip ... however I will search for some throughole similar P channel transistor since I never used SMD so far.
– Michel Keijzers
1 hour ago
|
show 3 more comments
All the LED current flows through the Vcc pin of the 74HC595 (and the GND pin of the TIPxxx). The absolute maximum rated current through Vcc (or GND) of a typical 74HC595 is indeed 70mA.
The absolute maximum peak current per LED is thus 8.75mA, or an average current of about 1mA per LED (1/8 duty cycle per LED).
In practice you should stay WELL away from the absolute maximum value.
To put it explicitly, this is a hobbyist-level circuit, designed by someone who doesn't care or doesn't know much about reliability (assuming they actually recommended anything like the currents you stated). Using the 74HC595 to drive a high-side driver array or prebiased PNP transistor duals would be much better. They're designed as logic shift registers, not as load drivers.
Using such drivers you could also get a much higher brightness. An average current of 10mA per LED requires a total current of 640mA, obviously, which means that the source drivers need to handle 80mA each (with all potentially on at once) and the sink drivers need to hand 640mA each (with each one seeing a 1/8 duty cycle).
answered 2 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
All the LED current flows through the Vcc pin of the 74HC595 (and the GND pin of the TIPxxx). The absolute maximum rated current through Vcc (or GND) of a typical 74HC595 is indeed 70mA.
The absolute maximum peak current per LED is thus 8.75mA, or an average current of about 1mA per LED (1/8 duty cycle per LED).
In practice you should stay WELL away from the absolute maximum value.
To put it explicitly, this is a hobbyist-level circuit, designed by someone who doesn't care or doesn't know much about reliability (assuming they actually recommended anything like the currents you stated). Using the 74HC595 to drive a high-side driver array or prebiased PNP transistor duals would be much better. They're designed as logic shift registers, not as load drivers.
Using such drivers you could also get a much higher brightness. An average current of 10mA per LED requires a total current of 640mA, obviously, which means that the source drivers need to handle 80mA each (with all potentially on at once) and the sink drivers need to hand 640mA each (with each one seeing a 1/8 duty cycle).
answered 2 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
edited 2 hours ago
answered 2 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
answered 2 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
answered 2 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
204k4150408
Thanks, so it would help to use a TPIC for sourcing too (than I can use 150 / 8 = 18.75 mA per LED, in one row) ?
– Michel Keijzers
2 hours ago
1
The TPIC has open-drain DMOS transistors so it wouldn't work. You need a high-side (source) driver.
– Spehro Pefhany
2 hours ago
Ah great, than I could use a 74HC595 to drive multiple 2N7000 transistors for sourcing and a TPIC595B for sinking. I think the TPIC can do not 640 mA, but I will scan through the rows anyway.
– Michel Keijzers
2 hours ago
1
You need p-channel transistors, eg. AO3401 and to invert the logic (low = on). The TPIC can handle about 250mA/output with only one on (see figure 10).
– Spehro Pefhany
1 hour ago
Thanks again for that tip ... however I will search for some throughole similar P channel transistor since I never used SMD so far.
– Michel Keijzers
1 hour ago
|
show 3 more comments
Thanks, so it would help to use a TPIC for sourcing too (than I can use 150 / 8 = 18.75 mA per LED, in one row) ?
– Michel Keijzers
2 hours ago
1
The TPIC has open-drain DMOS transistors so it wouldn't work. You need a high-side (source) driver.
– Spehro Pefhany
2 hours ago
Ah great, than I could use a 74HC595 to drive multiple 2N7000 transistors for sourcing and a TPIC595B for sinking. I think the TPIC can do not 640 mA, but I will scan through the rows anyway.
– Michel Keijzers
2 hours ago
1
You need p-channel transistors, eg. AO3401 and to invert the logic (low = on). The TPIC can handle about 250mA/output with only one on (see figure 10).
– Spehro Pefhany
1 hour ago
Thanks again for that tip ... however I will search for some throughole similar P channel transistor since I never used SMD so far.
– Michel Keijzers
1 hour ago
Thanks, so it would help to use a TPIC for sourcing too (than I can use 150 / 8 = 18.75 mA per LED, in one row) ?
– Michel Keijzers
2 hours ago
Thanks, so it would help to use a TPIC for sourcing too (than I can use 150 / 8 = 18.75 mA per LED, in one row) ?
– Michel Keijzers
2 hours ago
1
1
The TPIC has open-drain DMOS transistors so it wouldn't work. You need a high-side (source) driver.
– Spehro Pefhany
2 hours ago
The TPIC has open-drain DMOS transistors so it wouldn't work. You need a high-side (source) driver.
– Spehro Pefhany
2 hours ago
Ah great, than I could use a 74HC595 to drive multiple 2N7000 transistors for sourcing and a TPIC595B for sinking. I think the TPIC can do not 640 mA, but I will scan through the rows anyway.
– Michel Keijzers
2 hours ago
Ah great, than I could use a 74HC595 to drive multiple 2N7000 transistors for sourcing and a TPIC595B for sinking. I think the TPIC can do not 640 mA, but I will scan through the rows anyway.
– Michel Keijzers
2 hours ago
1
1
You need p-channel transistors, eg. AO3401 and to invert the logic (low = on). The TPIC can handle about 250mA/output with only one on (see figure 10).
– Spehro Pefhany
1 hour ago
You need p-channel transistors, eg. AO3401 and to invert the logic (low = on). The TPIC can handle about 250mA/output with only one on (see figure 10).
– Spehro Pefhany
1 hour ago
Thanks again for that tip ... however I will search for some throughole similar P channel transistor since I never used SMD so far.
– Michel Keijzers
1 hour ago
Thanks again for that tip ... however I will search for some throughole similar P channel transistor since I never used SMD so far.
– Michel Keijzers
1 hour ago
|
show 3 more comments
In this example, the 74HC595 sources say 8-10mA per output as determined by R1-R8. One output of the TPICx595 is turned on a time to sink 64-80mA of current if all 8 LEDs are on. That's it. The 640mA number is not correct.
Each column is turned on for 2-3mS, then it is turned off, data for the next column is turned on, and the next column drive is turned. Repeat for all 8 columns.
That's the basis of multiplexing - cycle thru 8 columns quickly and fool the eye into thinking all 64 LEDs can be on at once, when in reality only 8 are ever turned on.
answered 21 mins ago
CrossRoadsCrossRoads
1,3008
Than I will check if 8-10 mA is good enough (probably it is), and use a 74HC595 and TPICB595 (as you proposed earlier). At least I now have a better understanding. I will do first some tests before I ask more (for most experts here) trivial questions.
– Michel Keijzers
19 mins ago
add a comment |
In this example, the 74HC595 sources say 8-10mA per output as determined by R1-R8. One output of the TPICx595 is turned on a time to sink 64-80mA of current if all 8 LEDs are on. That's it. The 640mA number is not correct.
Each column is turned on for 2-3mS, then it is turned off, data for the next column is turned on, and the next column drive is turned. Repeat for all 8 columns.
That's the basis of multiplexing - cycle thru 8 columns quickly and fool the eye into thinking all 64 LEDs can be on at once, when in reality only 8 are ever turned on.
answered 21 mins ago
CrossRoadsCrossRoads
1,3008
Than I will check if 8-10 mA is good enough (probably it is), and use a 74HC595 and TPICB595 (as you proposed earlier). At least I now have a better understanding. I will do first some tests before I ask more (for most experts here) trivial questions.
– Michel Keijzers
19 mins ago
add a comment |
In this example, the 74HC595 sources say 8-10mA per output as determined by R1-R8. One output of the TPICx595 is turned on a time to sink 64-80mA of current if all 8 LEDs are on. That's it. The 640mA number is not correct.
Each column is turned on for 2-3mS, then it is turned off, data for the next column is turned on, and the next column drive is turned. Repeat for all 8 columns.
That's the basis of multiplexing - cycle thru 8 columns quickly and fool the eye into thinking all 64 LEDs can be on at once, when in reality only 8 are ever turned on.
answered 21 mins ago
CrossRoadsCrossRoads
1,3008
In this example, the 74HC595 sources say 8-10mA per output as determined by R1-R8. One output of the TPICx595 is turned on a time to sink 64-80mA of current if all 8 LEDs are on. That's it. The 640mA number is not correct.
Each column is turned on for 2-3mS, then it is turned off, data for the next column is turned on, and the next column drive is turned. Repeat for all 8 columns.
That's the basis of multiplexing - cycle thru 8 columns quickly and fool the eye into thinking all 64 LEDs can be on at once, when in reality only 8 are ever turned on.
answered 21 mins ago
CrossRoadsCrossRoads
1,3008
answered 21 mins ago
CrossRoadsCrossRoads
1,3008
answered 21 mins ago
CrossRoadsCrossRoads
1,3008
answered 21 mins ago
CrossRoadsCrossRoads
1,3008
1,3008
Than I will check if 8-10 mA is good enough (probably it is), and use a 74HC595 and TPICB595 (as you proposed earlier). At least I now have a better understanding. I will do first some tests before I ask more (for most experts here) trivial questions.
– Michel Keijzers
19 mins ago
add a comment |
Than I will check if 8-10 mA is good enough (probably it is), and use a 74HC595 and TPICB595 (as you proposed earlier). At least I now have a better understanding. I will do first some tests before I ask more (for most experts here) trivial questions.
– Michel Keijzers
19 mins ago
Than I will check if 8-10 mA is good enough (probably it is), and use a 74HC595 and TPICB595 (as you proposed earlier). At least I now have a better understanding. I will do first some tests before I ask more (for most experts here) trivial questions.
– Michel Keijzers
19 mins ago
Than I will check if 8-10 mA is good enough (probably it is), and use a 74HC595 and TPICB595 (as you proposed earlier). At least I now have a better understanding. I will do first some tests before I ask more (for most experts here) trivial questions.
– Michel Keijzers
19 mins ago
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f416035%2fwhy-does-a-74hc595n-work-in-led-matrices%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
1
Bit of a duplicate of this question?
– Finbarr
1 hour ago
@Finbarr not a duplicate, but a nice bit of extra information, thanks for the link.
– Michel Keijzers
1 hour ago