Odd 74HCT1G125 behaviourOdd number frequency dividerWhen scoping the CLK and DATA lines of a PS/2 keyboard it...
Can we harness gravitational potential energy?
Why is Agricola named as such?
Which communication protocol is used in AdLib sound card?
Am I a Rude Number?
What is a good reason for every spaceship to carry a weapon on board?
How much mayhem could I cause as a sentient fish?
What does it mean for a caliber to be flat shooting?
In Linux what happens if 1000 files in a directory are moved to another location while another 300 files were added to the source directory?
Why avoid shared user accounts?
Eww, those bytes are gross
How can I get my players to come to the game session after agreeing to a date?
What incentives do banks have to gather up loans into pools (backed by Ginnie Mae)and selling them?
If I delete my router's history can my ISP still provide it to my parents?
How can I play a serial killer in a party of good PCs?
Making him into a bully (how to show mild violence)
Is a new Boolean field better than a null reference when a value can be meaningfully absent?
Graph with overlapping labels
What sets the resolution of an analog resistive sensor?
Cookies - Should the toggles be on?
Why do cars have plastic shrouds over the engine?
Why zero tolerance on nudity in space?
Why is working on the same position for more than 15 years not a red flag?
Why is it that Bernie Sanders is always called a "socialist"?
False written accusations not made public - is there law to cover this?
Odd 74HCT1G125 behaviour
Odd number frequency dividerWhen scoping the CLK and DATA lines of a PS/2 keyboard it looks… rather oddWhat is the intended usage of “WE” or “PE/OE” pins on 4517 shift registerOdd Parity Output as Input to Second CircuitOP AMPs : ADC Buffer on voltage divider plus protectionOpen collector outputs with logic gates causing infinite loop scenarioVGA controller using FIFO memory, discrete ICs and Arduino Uno/Mega?Is XOR equal to XNOR when odd number of inputs are considered?Source/Sink on 74AHCT2G125DC not driving WS2712bsUnderstanding CMOS Circuit Behaviour with Resistive Loads using Thevenin theorem
$begingroup$
Ok,
So using a 74HCT1G125 buffer as an '5V output driver', but, in some cases it has a 12v, low impedance source presented at its output pin (but in these cases its enable is false, i.e the output should be Hi Z). When I do this, the 5v supply (that the buffer uses also) is pushed up to around 8v. Surely if its Hi-Z its an open circuit, how is the external 12v making its way and effecting the 5V?
Have checked and indeed the device is being 'told' to be in Hi-Z. I then put 12v onto its Output pin, and somehow, the VCC gets pushed up.
Thanks
digital-logic buffer
asked 2 hours ago
MattyT2017MattyT2017
9218
$endgroup$
add a comment |
$begingroup$
Ok,
So using a 74HCT1G125 buffer as an '5V output driver', but, in some cases it has a 12v, low impedance source presented at its output pin (but in these cases its enable is false, i.e the output should be Hi Z). When I do this, the 5v supply (that the buffer uses also) is pushed up to around 8v. Surely if its Hi-Z its an open circuit, how is the external 12v making its way and effecting the 5V?
Have checked and indeed the device is being 'told' to be in Hi-Z. I then put 12v onto its Output pin, and somehow, the VCC gets pushed up.
Thanks
digital-logic buffer
asked 2 hours ago
MattyT2017MattyT2017
9218
$endgroup$
$begingroup$
Assuming that your circuit still works with the resulting high gate driver resistance, you can fix this problem by putting a series resistor on the gate output. Output protection diodes can typically withstand several milliamps without causing problems. So if you size the series resistor for say 1ma at 12V - 5V => ~7K ohms, that might fix your problem.
$endgroup$
– crj11
2 hours ago
add a comment |
$begingroup$
Ok,
So using a 74HCT1G125 buffer as an '5V output driver', but, in some cases it has a 12v, low impedance source presented at its output pin (but in these cases its enable is false, i.e the output should be Hi Z). When I do this, the 5v supply (that the buffer uses also) is pushed up to around 8v. Surely if its Hi-Z its an open circuit, how is the external 12v making its way and effecting the 5V?
Have checked and indeed the device is being 'told' to be in Hi-Z. I then put 12v onto its Output pin, and somehow, the VCC gets pushed up.
Thanks
digital-logic buffer
asked 2 hours ago
MattyT2017MattyT2017
9218
$endgroup$
Ok,
So using a 74HCT1G125 buffer as an '5V output driver', but, in some cases it has a 12v, low impedance source presented at its output pin (but in these cases its enable is false, i.e the output should be Hi Z). When I do this, the 5v supply (that the buffer uses also) is pushed up to around 8v. Surely if its Hi-Z its an open circuit, how is the external 12v making its way and effecting the 5V?
Have checked and indeed the device is being 'told' to be in Hi-Z. I then put 12v onto its Output pin, and somehow, the VCC gets pushed up.
Thanks
digital-logic buffer
digital-logic buffer
asked 2 hours ago
MattyT2017MattyT2017
9218
asked 2 hours ago
MattyT2017MattyT2017
9218
asked 2 hours ago
MattyT2017MattyT2017
9218
asked 2 hours ago
MattyT2017MattyT2017
9218
asked 2 hours ago
MattyT2017MattyT2017
9218
9218
$begingroup$
Assuming that your circuit still works with the resulting high gate driver resistance, you can fix this problem by putting a series resistor on the gate output. Output protection diodes can typically withstand several milliamps without causing problems. So if you size the series resistor for say 1ma at 12V - 5V => ~7K ohms, that might fix your problem.
$endgroup$
– crj11
2 hours ago
add a comment |
$begingroup$
Assuming that your circuit still works with the resulting high gate driver resistance, you can fix this problem by putting a series resistor on the gate output. Output protection diodes can typically withstand several milliamps without causing problems. So if you size the series resistor for say 1ma at 12V - 5V => ~7K ohms, that might fix your problem.
$endgroup$
– crj11
2 hours ago
$begingroup$
Assuming that your circuit still works with the resulting high gate driver resistance, you can fix this problem by putting a series resistor on the gate output. Output protection diodes can typically withstand several milliamps without causing problems. So if you size the series resistor for say 1ma at 12V - 5V => ~7K ohms, that might fix your problem.
$endgroup$
– crj11
2 hours ago
$begingroup$
Assuming that your circuit still works with the resulting high gate driver resistance, you can fix this problem by putting a series resistor on the gate output. Output protection diodes can typically withstand several milliamps without causing problems. So if you size the series resistor for say 1ma at 12V - 5V => ~7K ohms, that might fix your problem.
$endgroup$
– crj11
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You aren't allowed to put more than VCC+0.5V on the output pin, even if it's in high-Z state. If you do, you could damage the chip.
What's happening in your circuit is there's a diode connected from the output pin to VCC, oriented so that it will be reverse biased in normal operation (cathode connected to VCC). This diode is there to shunt current during ESD events.
When you connect 12 V to the output pin, you forward bias this diode and deliver current out of the VCC pin. Depending what else is connected to the same VCC net, you could damage those other parts, or high current through the ESD diode could damage the 1G125.
answered 2 hours ago
The PhotonThe Photon
85.6k397198
$endgroup$
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%2f424733%2fodd-74hct1g125-behaviour%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
$begingroup$
You aren't allowed to put more than VCC+0.5V on the output pin, even if it's in high-Z state. If you do, you could damage the chip.
What's happening in your circuit is there's a diode connected from the output pin to VCC, oriented so that it will be reverse biased in normal operation (cathode connected to VCC). This diode is there to shunt current during ESD events.
When you connect 12 V to the output pin, you forward bias this diode and deliver current out of the VCC pin. Depending what else is connected to the same VCC net, you could damage those other parts, or high current through the ESD diode could damage the 1G125.
answered 2 hours ago
The PhotonThe Photon
85.6k397198
$endgroup$
add a comment |
$begingroup$
You aren't allowed to put more than VCC+0.5V on the output pin, even if it's in high-Z state. If you do, you could damage the chip.
What's happening in your circuit is there's a diode connected from the output pin to VCC, oriented so that it will be reverse biased in normal operation (cathode connected to VCC). This diode is there to shunt current during ESD events.
When you connect 12 V to the output pin, you forward bias this diode and deliver current out of the VCC pin. Depending what else is connected to the same VCC net, you could damage those other parts, or high current through the ESD diode could damage the 1G125.
answered 2 hours ago
The PhotonThe Photon
85.6k397198
$endgroup$
add a comment |
$begingroup$
You aren't allowed to put more than VCC+0.5V on the output pin, even if it's in high-Z state. If you do, you could damage the chip.
What's happening in your circuit is there's a diode connected from the output pin to VCC, oriented so that it will be reverse biased in normal operation (cathode connected to VCC). This diode is there to shunt current during ESD events.
When you connect 12 V to the output pin, you forward bias this diode and deliver current out of the VCC pin. Depending what else is connected to the same VCC net, you could damage those other parts, or high current through the ESD diode could damage the 1G125.
answered 2 hours ago
The PhotonThe Photon
85.6k397198
$endgroup$
You aren't allowed to put more than VCC+0.5V on the output pin, even if it's in high-Z state. If you do, you could damage the chip.
What's happening in your circuit is there's a diode connected from the output pin to VCC, oriented so that it will be reverse biased in normal operation (cathode connected to VCC). This diode is there to shunt current during ESD events.
When you connect 12 V to the output pin, you forward bias this diode and deliver current out of the VCC pin. Depending what else is connected to the same VCC net, you could damage those other parts, or high current through the ESD diode could damage the 1G125.
answered 2 hours ago
The PhotonThe Photon
85.6k397198
answered 2 hours ago
The PhotonThe Photon
85.6k397198
answered 2 hours ago
The PhotonThe Photon
85.6k397198
answered 2 hours ago
The PhotonThe Photon
85.6k397198
85.6k397198
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%2f424733%2fodd-74hct1g125-behaviour%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Assuming that your circuit still works with the resulting high gate driver resistance, you can fix this problem by putting a series resistor on the gate output. Output protection diodes can typically withstand several milliamps without causing problems. So if you size the series resistor for say 1ma at 12V - 5V => ~7K ohms, that might fix your problem.
$endgroup$
– crj11
2 hours ago