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Odd 74HCT1G125 behaviour
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Odd 74HCT1G125 behaviour
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$begingroup$
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
8718
$endgroup$
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
8718
$endgroup$
$begingroup$
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
1 hour 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
8718
$endgroup$
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
8718
asked 2 hours ago
MattyT2017MattyT2017
8718
asked 2 hours ago
MattyT2017MattyT2017
8718
asked 2 hours ago
MattyT2017MattyT2017
8718
asked 2 hours ago
MattyT2017MattyT2017
8718
8718
$begingroup$
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
85.6k397197
$endgroup$
add a comment |
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$begingroup$
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
85.6k397197
$endgroup$
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
85.6k397197
$endgroup$
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
85.6k397197
$endgroup$
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
85.6k397197
answered 2 hours ago
The PhotonThe Photon
85.6k397197
answered 2 hours ago
The PhotonThe Photon
85.6k397197
answered 2 hours ago
The PhotonThe Photon
85.6k397197
85.6k397197
add a comment |
add a comment |
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$begingroup$
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
1 hour ago