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Hi...
Kindly refer to the attached circuit. Please note that the pin-14 of the 74HCT14 is NOT connected to VCC (or +5V). When the input A is connected to a high logic (approximately +4.5V), the YELLOW led is turned on. If the LED is disconnected from the pin-14 of the 74HCT14, it's OFF. How can this LED be turned on without +5V supply to the VCC of the 74HCT14?
Thanks.
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Leakage current? Things are very different in the real world, than they are in theory & on paper. Don't forget that integrated circuits are just many, many transistors. If the IC is unpowered, and there is no bias for the transistors, who knows what could be happening!
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I'd suspect input clamping diodes are allowing the 4.5V to flow through to Vcc.
Normally reverse biased until the input goes over Vcc.
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I'd suspect input clamping diodes are allowing the 4.5V to flow through to Vcc.
Normally reverse biased until the input goes over Vcc.
Thanks. Do you have any idea to prevent this to happen? -
Only way to defeat that flow is a blocking diode on the Vcc input, only allowing the normal power into the chip, but pay attention to the minimum Vcc.
Careful use of a mosfet may also help to reduce this voltage drop, but pay attention to the substrate's parasitic diode influence. -
Only way to defeat that flow is a blocking diode on the Vcc input, only allowing the normal power into the chip, but pay attention to the minimum Vcc.
Careful use of a mosfet may also help to reduce this voltage drop, but pay attention to the substrate's parasitic diode influence.
Thanks a lot.
[1] Today I found a weird phenomena in my old DSP system, which uses a Texas Instruments TMS320C31 DSP Starter Kit (DSK).
[2] The DSK is connected to PC via parallel port for programming and debugging.
[3] A 12-bit digital output board (DOB) is interfaced to the DSP Starter Kit.
[4] Each digital output is buffered with a 74HCT14 and a LED is used as indicator.
[5] When a 12-bit data (let's 1100 0000 0000 0000) is sent to the DOB, the LEDs connected to bit-11 and bit-10 are ON.
[6] Then, I turn OFF the DSP system (i.e. the DSK and the DOB are OFF and the computer is still ON). When I turn ON the DSK and the DOB again, the LEDs connected to bit-11 and bit-10 are still ON.
[7] It seems that the system has a "non-volatile" memory which can remember the last digital output state.
[8] Step [3] to [5] are repeated. Similar to step [6], but this time I turn OFF the DSK, DOB, and the computer. Then, all are powered up again. The LEDs connected to bit-11 and bit-10 are OFF.
[9] I really don't know where does the "memory" come from?
[10] Step [3] to [5] are repeated. Similar to step [6], but this time, while keeping the computer alive, I turn OFF the DSK and the DOB. Then, I disconnect the parallel port cable from the DSK and reconnect it. Then, both DSK and DOB are powered up again. The LEDs connected to bit-11 and bit-10 are OFF.
[11] It looks to me that the "memory" is from the "parallel port".
[12] Why can this happen?
Thanks.
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Maybe the parallel port is supplying enough power to something on the DSK to maintain state, but isn't connected to the LEDs? You'd have to look at the schematic diagram to see why this might be.
In general, you shouldn't assume anything about the power on state. Use a power on reset circuit or similar. -
Only way to defeat that flow is a blocking diode on the Vcc input, only allowing the normal power into the chip, but pay attention to the minimum Vcc.
How would this help, no current has to flow through the Vcc pins if the clamping diodes are forward biased. As long as the current draw is small, it's possible to power many digital circuits through just the clamping diodes (with one input high, another input low).
Careful use of a mosfet may also help to reduce this voltage drop, but pay attention to the substrate's parasitic diode influence.
It's usually not recommended to provide voltages on any of the other pins when Vcc is not powered, exactly for this reason. Those clamping diodes are not designed to carry much current. -
Only way to defeat that flow is a blocking diode on the Vcc input, only allowing the normal power into the chip, but pay attention to the minimum Vcc.
How would this help, no current has to flow through the Vcc pins if the clamping diodes are forward biased. As long as the current draw is small, it's possible to power many digital circuits through just the clamping diodes (with one input high, another input low).
Careful use of a mosfet may also help to reduce this voltage drop, but pay attention to the substrate's parasitic diode influence.
It's usually not recommended to provide voltages on any of the other pins when Vcc is not powered, exactly for this reason. Those clamping diodes are not designed to carry much current.
It helps in the posted diagram as it will block any current reaching the LED's anode.
Yes the chip will still be powered, never said it wasn't. -
It's usually not recommended to provide voltages on any of the other pins when Vcc is not powered, exactly for this reason. Those clamping diodes are not designed to carry much current.
Attached figure shows the block diagram of my system. Each module is supplied by a separate PSU. However, their COMMON (GND) are tied together. The output interface of the digital output board (DOB) is buffered with 74HCT245, whilst the input interface of the "LED Driver" board is buffered with 74HCT14. I hope that when I power off the PSU to the "LED Driver" board, the onboard LEDs should all be OFF. However, they're not. May be I should use other IC to replace the 74HCT14. What do you reckon? Thanks.
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Galvanic isolation?
If you use optocouplers as the inputs, you won't be able to bleed power from the other PCB.
That however then requires the driver board to be able to drive the optocoupler's LED.
Catch 22!
Another way maybe to use open collector (drain) drivers to the LED driver board - they cannot source power.
Add pullups on your LED driver board inputs connected to its own supply.
FWIW, this is the method used by floppy disk drive interfaces. -
Galvanic isolation?
If you use optocouplers as the inputs, you won't be able to bleed power from the other PCB.
That however then requires the driver board to be able to drive the optocoupler's LED.
Thanks.
Do you think digital isolator (e.g. ADUM1510), which requires far less driving current, can do the job?
Pls refer to the schematic below. If VDD2/GND2 are not connected to any power source, while VDD1 is powered and VIA is connected to a HIGH logic, what will be output state (HIGH, LOW, FLOATING) of VOA? And, do you think the LED will in OFF or ON state?
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You get a similar thing with a MAX232 IC, which is a good thing to be aware of.
If you have a max232 which has no power on it's VCC pin but which is connected to a live rs232 cable ( so getting 12v on its rs232 pins ).
This 12V can feed back out VCC and partially power the max232 and any other IC's in parallel with the max232.
The ATMega in my car that controls my carpc has this issue. The atmega can be off (battery disconnected) and i can still talk to the atmega through the serial link due to the max232 and atmega running from the 12V on the serial line.
The connection is pretty unstable and only seems to function properly in one direction. If i send a serial frame to the car it understands the frame and replies but the reply is all corrupted on the way out the max232. -
I hate looking at circuits with ICs drawn as boxes, it makes them very hard to read and figure out what's going on. In future always draw circuits with logic gates. a box means nothing without the datasheet.
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Some LED takes very little current and remains lit, even when the electricity is turned off. Did you have each LED with a resistor in series with it at the anode end of the LED?
http://led.linear1.org/1led.wiz
http://led.linear1.org/why-do-i-need-a-resistor-with-an-led/
http://www.hobby-hour.com/electronics/ledcalc.php
In a darken room, sometimes we could see the dim LED lit from the the current left in the capacitor of the power supply, when just turned off. I think the problem is in the power supply distribution or the LED driver circuit IC's design.
Lawsen -
Some LED takes very little current and remains lit, even when the electricity is turned off. Did you have each LED with a resistor in series with it at the anode end of the LED?
Thanks. Yes... a resistor is connected in series with the LED. -
I can light a high efficiency green LED in my house by holding one of the leads and connecting the other leg to a grounded radiator. Connecting a diode in reverse parallel and a capacitor in series improves the efficiency by acting as a voltage doubler.