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backfeed voltage from Logic IC chips

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lewisP

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Back-feed means when a Logic IC chip back-feeds voltage or current from it's input to a previous stages output. What causes an IC chip to back-feed from its input pin outputting voltage and current. How can an electronic technician troubleshooting and track down a back-feed problem or an IC chip that is back-feeding?
 

SunnySkyguy

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All transistors have a Miller Capacitance which causes feedback of input signal and also the inverse back-feed from load generated transients that result from either poor load impedance matching on a transmission line (small) or inductive kickback from switched currents (large)

FF's and unbuffered registers are notorious, which is why any long trace register signals should be buffered and/or terminated.
 

lewisP

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Why do Flip flops and Registers cause back-feeding voltages from there inputs, because of the switching voltage?

I found a bad multiplexer that was back-feeding voltage from its input pin to a previous stage output logic gate which was causing the logic gates output to be stuck high because of the back-feeding voltage from the multiplexers input. When a Logic IC chip internally goes back it outputs a voltage from it's input which is called back-feeding. Back-feeding can cause a previous output pin to be stuck high. Have you technicians seen or heard of these logic back-feeding problems?
 

Gorgon

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You may have an input source (backfeed) voltage to an output if the input's overvoltage protection is damaged/ shorted. This happens normally if you have an ESD damage to the input, or have put in an overvoltage to the input.
 

lewisP

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The input over-voltage protections from the datasheets says it gets damaged by a negative voltage more -1.4 volts. The input protection diodes internally inside the logic chip can get damaged when more then negative -1.4 volts. This causes a back-feeding voltage outputting from the IC chips input. They call the input transistor the steering circuit, if the input transistor that has a dual gate gets shorted the +5 VCC goes through the logic IC chip and goes out the input pin causing a back-feeding voltage. I'm not sure why the input protection diodes on the dual base or dual gate input transistor can only handle -1.4 volts because they get damaged.
 

SunnySkyguy

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I agree. If the current exceeds the rail protection clamp diodes (Vth+If*Rs) , then internal SCR latchup may damage low ESR CMOS from inductive transients or overvoltage from ESD, followed by shoothru DC thermal faults.

Normally Miller capacitance is small, and previous stage has a low enough ESR to absorb the transient. Since Cfb*ESR (or..CissRdsOn) is relative constant (for same lithography/Vds rating) this is only a real problem on unbuffered registers to inductive loads or "defective or ESD damaged" parts where ESD has closed the junction gap reducing ESR and increasing Ciss causing excessive leakage and "back-feed".

A Huntron tracker is an effective sweeping current source for isolating these faults on external pins, while ICT vector tests verify existence of any internal faults. Floating inputs can also be a cause , prey to ESD even without contact. (Technical correct term is EOS, or electrical over stress.) Even LEDs without ESD zener protection are commonly accused of being bad parts , when the user is unaware of EOS risks in their assembly and handling process. I know this from 1st hand experience.(s).

Registers and FF's are edge sensitive, so a 10ns bus glitch can toggle the state with backfeed.
 

lewisP

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I use a huntron 2700 to measure CMOS resistive shorts with the huntron black common probe on either reference to ground or +VCC and probe each IC pin with the red huntron probe. I use a Fluke meter also to measure CMOS resistive shorts on each pin of an IC logic chip. I use the huntron 2700 setting range is 3vac at 60hz with a source impedance at 100ohms. If you apply a DC voltage in any pin of a CMOS chip when the circuit board is turned off it will damage the CMOS chip. The huntron applys an AC waveform at 3 vac at 60hz.
 

SunnySkyguy

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i recall protection diodes are limited to 10mA dc , in your case 2Vppac would be (2-.25)/100 * ~1/3 duty factor so <10mA so safe.

duty factor average conduction of offset sine, rms to dc ...estimated...

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I recommended a square wave current source at low freq, but a sine works well to. Huntron might show a bit of liasing in the VI curve shown at 60Hz, but 1000 Hz would be more sensitive to high input or gate capacitance to phase shift. mA chosen should not exceed diode rating. E.g. 10v+ 1k -_-_-
 

lewisP

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Some datasheets say -1.4 volts peak , but you're saying -2 volts peak. I have seen flip flops circuits output a negative logic overshoot and undershoot voltage. Any Logic chip switches will cause some negative voltage on the output that is overshoot or undershoot. Using Tri-state buffer/line drivers solve these issues from damaging CMOS input pins?

A resistive short is a partial short internally inside the logic IC chip. Do you use an ohm meter to test for resistive shorts on each IC chip of the suspected chip you think is bad?

Early 70's Op amp IC would latch up and say stuck in a high state mostly which was a defect in design which is caused i think by the miller effect.

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I recommended a square wave current source at low freq, but a sine works well to. Huntron might show a bit of liasing in the VI curve shown at 60Hz, but 1000 Hz would be more sensitive to high input or gate capacitance to phase shift. mA chosen should not exceed diode rating. E.g. 10v+ 1k -_-_-

Why would you want a square-waveform in the VI curve or XY mode of an oscilloscope, the VI curves will be distorted, bending and curving the signatures which would think you had a bad IC chip

60Hz, but 1000 Hz would be more sensitive to high input or gate capacitance to phase shift.

Use the 1000hz would show CMOS resistive shorts better? huntron has 200hz and 2000hz settings not 1000hz

The Source impedance of the huntron you can choose from 10ohms, 100ohm 1K ohms, 10K ohm, 100kohm

I always use the 100ohms impedance setting.
 

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Why would you want a square-waveform in the VI curve or XY mode of an oscilloscope, the VI curves will be distorted, bending and curving the signatures which would think you had a bad IC chip

.


A square wave current source or a a high voltage high R resistance voltage source such as 10V,1K gives a more linear impedance response and if capacitance variations are significant, this is far more sensitive. Obvious cycle rates can be reduced if too sensitive.

A sine wave will give a better looking picture because the cosine will be smooth even intensity and tend to show more zener like resistive properties than reactive. ( capacitive ) which are better clues of ESD wounds. Of course a 1kHz sine wave is suitable as well.
 

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