Odd 74HCT1G125 behaviourOdd number frequency dividerWhen scoping the CLK and DATA lines of a PS/2 keyboard it...
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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
$begingroup$
@crj11 - It's generally strongly frowned upon to depend on the implementation details of an IC's output drivers. If you're going to do this, use an external schottky clamp diode too.
$endgroup$
– Connor Wolf
4 mins 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
$begingroup$
@crj11 - It's generally strongly frowned upon to depend on the implementation details of an IC's output drivers. If you're going to do this, use an external schottky clamp diode too.
$endgroup$
– Connor Wolf
4 mins 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
$begingroup$
@crj11 - It's generally strongly frowned upon to depend on the implementation details of an IC's output drivers. If you're going to do this, use an external schottky clamp diode too.
$endgroup$
– Connor Wolf
4 mins 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
$begingroup$
@crj11 - It's generally strongly frowned upon to depend on the implementation details of an IC's output drivers. If you're going to do this, use an external schottky clamp diode too.
$endgroup$
– Connor Wolf
4 mins ago
$begingroup$
@crj11 - It's generally strongly frowned upon to depend on the implementation details of an IC's output drivers. If you're going to do this, use an external schottky clamp diode too.
$endgroup$
– Connor Wolf
4 mins 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
85.6k398198
$endgroup$
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
85.6k398198
$endgroup$
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
85.6k398198
$endgroup$
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
85.6k398198
$endgroup$
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
85.6k398198
answered 2 hours ago
The PhotonThe Photon
85.6k398198
answered 2 hours ago
The PhotonThe Photon
85.6k398198
answered 2 hours ago
The PhotonThe Photon
85.6k398198
85.6k398198
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
2 hours ago
$begingroup$
@crj11 - It's generally strongly frowned upon to depend on the implementation details of an IC's output drivers. If you're going to do this, use an external schottky clamp diode too.
$endgroup$
– Connor Wolf
4 mins ago