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Logic tester circuit

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mcgill

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Hello to all, i am a journeyman electrician who is trying to get an electronics degree. my assignment is to build a logic tester from a kit and write a report on it. one of the requirements of the report is to justify the existance of every component in the project (as if to a potential client) to show no extra parts are included. i assembled the kit and it works according to the test procedures. while i understand most of what goes on in the design,i have a few questions:
the first thing after the probe is a .001 microfarad capacitor in parallel with a 100 kilo ohm resistor, how do i explain this?
also the probe sees a couple of diodes, one with its cathode facing Vcc and the other with its anode facing ground, i'm thinking these bleed off excess voltage to protect the rest of the circuit?
the kit is an elenco lp-525k
also it is not considered cheating if i make reference to your help in the bibliography
thanks for your help
 
The 100K resistor limits the current into the detector, but the detector has shunt capacitance which will take a long time to charge thru 100K. The .001 uF cap charges the input capacitance so that short pulses can be detected. You are correct, the diodes prevent the voltage from damaging the input circuit.
 
thanks for the responses
in parallel with the hi-lo detector is the pulse detector where the signal runs thru what looks to me like a class B amp and down the circuit from that are a couple of diodes in parallel that both facing the same way ,one diode ties into the base of an NPN transistor and the collector of a PNP transistor. the other diode ties into the emitter of the NPN. now where i'm confused is that if that is a class B amp, why have both diodes facing the same way? (cathodes facing the transistors)
thanks again
 
mcgill said:
thanks for the responses
in parallel with the hi-lo detector is the pulse detector where the signal runs thru what looks to me like a class B amp and down the circuit from that are a couple of diodes in parallel that both facing the same way ,one diode ties into the base of an NPN transistor and the collector of a PNP transistor. the other diode ties into the emitter of the NPN. now where i'm confused is that if that is a class B amp, why have both diodes facing the same way? (cathodes facing the transistors)
thanks again
Can you post a schematic?
 
Q4 and Q3 form a latch. If you turn on Q4, it turns on Q3 which turns on Q4 and Wham! the circuit is latched until Q4 is turned off. Since there are two diode drops in series with D3, D3 will never turn on but perhaps the fast rise of the pulse comes thru the capacitance of D3 and turns Q4 on. C3 won't sustain a very long pulse but it must be long enuf that D4 pulls the emitter of Q4 up to turn it off. The output is at Q3 emitter and is a very narrow pulse. C4 is quite large, so the pulse amplitude will be small. It does not look like a reliable circuit.
 
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