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Dear Team,
I am using SN74AVC8T245PWR in a circuit. Same circuit is attached for your reference.
In one SN74AVC8T245PWR (Say IC1), Voltage across PIN 6 (A4) & PIN 11 (GND) is measured as 1V instead of 3.3V (Note the 3.3V pull up in the circuit attached).
So to confirm, I measured the voltage across same points in some other healthy ICs ((Say IC2 & IC3)). In it the voltage is measured as 3.3V.
This made me think, First IC ((Say IC1)) may be defective. I mean, that IC (PIN 6) sinks some amount of current (greater than ) from 3.3V pull up supply and due to this voltage across PIN 6 & GND may be lesser. Request you to confirm the same.
Also Pls let me know will the impedance at PIN 6 be same during power ON & power OFF conditions.I don't think both will be same. Because while measuring at power OFF condition in Healthy ICs (IC2 & IC3), Impedance across PIN 6 & GND is 4.7KOHM. I don't think same would maintain during power ON also. It should be around 1M during power ON. Right?
Thanks in Advance. Awaiting your reply.
Regards,
Devarajan R
All inputs should have high impedance (≫ 1 MΩ) at all times, even when powered off. If an input sinks current, then the device is damaged.
You cannot realiably measure while the device is connected to the rest of the circuit; you probably measured the pull-up resistor in parallel to the pin.
Thanks for the reply. It's clear now.
As per your statement if the 4.7KOHM observed across across PIN 6 & PIN 11(GND) is Pull-up resistor value, then do you mean that VCCA is somehow connected to GND in power OFF condition. If possible please share internal block diagram of IC.
Second point is, In datasheet I found a statement as below.
"The VCC isolation feature ensures that if either VCC
input is at GND, both ports are in the high-impedance
state"
What does that mean? Can you please elaborate.
Thanks in advance
The IC never connects VCC to GND. I assume that the circuit does this.
The VCC isolation feature ensures that both supplies must be above 0 V before the device outputs anything.