Is it normal that if i supply a Nand FZH101 with others pins disconnect and i mesure the 2 input (of all 4 gate of the ic) i've about 7,5 volt and the output is near 0 volt?
In a common cd4011 the 2 input are at 0 volt(as shoul be) and the output are
15Vcc. Any help is welcome. Thanks Regards, max
NAND: both inputs high = output low , any input low = output high
Looks ok.
See the circuit on the last page. Both inputs disconnected: the 10k resistor is pulling the Z-diode in the middle up. If you short any of the two inputs to GND the outputlevel should change.
MfG JRD
"maxi" schrieb im Newsbeitrag news:TSI0i.4060$ snipped-for-privacy@twister1.libero.it...
Yes
No. On CD4011 if inputs are floating they may do bad things to your chip.
-- Manfred Winterhoff, reply-to invalid, use mawin at gmx dot net homepage: http://www.geocities.com/mwinterhoff/ de.sci.electronics FAQ: http://dse-faq.elektronik-kompendium.de/ Read 'Art of Electronics' Horowitz/Hill before you ask. Lese 'Hohe Schule der Elektronik 1+2' bevor du fragst.
Hi,
thanks very much!!!
So from the schematic that you linked i coul use a FZH111 as a FZH101 but not the inverse?
Now if a disconnected input of FZH101 result as a High level input(as PDF says) if i use the FZH with a common CD4011(no prroblem for the minor current and pin out) and i leave a input disconnected the logic function will go centerly wrong? And if i simulate on a 4011 a input disconnected (as was on fzh101)as connected to 15 Vcc(high) the logic function will be correct?
Many thanks
The FZH111 has the N signal bonded out Pin 1. The FZH101 hasn´t. So you have to look if your board has the pin unconnected, then it will work.
Unused inputs of the CD40xx should be connected to GND or 12V. Leaving them open was possible on the FZHxxx and TTL-logic: they have internal pullup-circuits.
An external pullup resistor to 15V would fix it. Easy to patch on DIL-IC with SIL resistor-arrays. Just don´t solder on the pins to the outputs.
Another problem at the CD40xx inputs is: the other FZHxxx on the board may not provide the required CMOS hi/lo-level. A CD40xx schmitt-trigger and resistors may (just) work.
MfG JRD
So i could use a cd4093 with a pullup resistor of 10K to 15 Vcc on both input of the 4 nand...........for simulate a FZH101???
Regards,
MAx
With a 15V supply the input-lowlevel for the CD4093 is about 6,3V and the highlevel about 9,0V. For the FZ101 output is 1,0V and 14,0V. So on paper it would match very well.
But for a safe, simple replacement one should look on the board what actully is driving the input ( it may not be a FZxxx-output, or that output may drive several gates ) and what loading the output has ( it may not be another FZxxx-input ).
The answer from 34.4.2007 still holds:
So there are ( more complex ) alternatives to the CD4093 if required. The advantage of bipolar circuits is not only that they drive more current at the output. But they are more EMI-resistant. And slower, which may be an advantage here.
MfG JRD
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