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In this article, http://www.edn.com/design/test-and-measurement/4391393/Single-hex-inverter-IC-makes-four-test-gadgets,

http://m.eet.com/media/1168037/single-hex-inverter-ic-makes-four-test-gadgets-fig1.jpg
A circuit is used to detect a user degined low logic level and high logic level, by lighting a red & green led. It uses an "inverter" that to me works just like a mosfet gate. Top inverter looks like a mosfet with red led at source, bottom inverter looks like mosfet with green led at drain.

Regardless, they do not function as inverter at all (inverting logic 0 & 1 ). Logic inverter takes input of 0 or 1 (and has both supply & GND connections) . But in this case, looks like the gate to the "inverter" must be > threshold for it to work.

The use of inverter symbol confuses me. Why the need for it ?

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2 Answers 2

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Assuming that the inverters are CMOS ones, the circuit is equivalent to:

schematic

simulate this circuit – Schematic created using CircuitLab

While in the schematics you provided it seems like the circuits are interrelated, in fact they are not. Two completely separate blocks having a common input.

The basic idea behind this circuit is this:

  • When M2 is ON (input voltage is LOW), the current can flow through the red LED
  • When M3 is ON (input voltage is HIGH), the current can flow through the green LED

M2 and M3 are functioning as pass-transistors in this case.

The function of M2 and M4 are less important - they pull the voltages down/up when the LEDs are not required to be ON. I believe that the circuit would function without them as well. However, the presence of the full inverter allows for a very steep ON/OFF characteristic for LEDs - there is no wide transition region between ON/OFF states.

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  • \$\begingroup\$ Actually, you mean When M2 is closed & M3 is closed. Am I correct ? Also in either case, M1 & M4 are not needed as they are open at all times. The point is why use 2 inverters with 4 MOSFET when just 2 MOSFET would do the job ? \$\endgroup\$ Aug 27, 2013 at 7:29
  • \$\begingroup\$ @user1502776, when M2 and M3 are ON :) I hate this open/closed terminology - don't know why did I use it here. I edited the answer with the reasons for using an inverter rather than a single transistor \$\endgroup\$
    – Vasiliy
    Aug 27, 2013 at 7:39
  • \$\begingroup\$ That is exactly what I would like to know. Thank you ! \$\endgroup\$ Aug 27, 2013 at 7:51
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This circuit is a very simple logic analyzer. You probe something, and the circuit will tell you if the probed point is a logical HI or LO by lighting one of the two LEDs. Because an inverter chip has its own threshold before it output an inverted signal, it can also be used as a simple analog to digital converter. The surrounding resistors are used to specify the exact threshold you want to use as a trigger.

Another function of this type of circuit is to ensure one of the LEDs only lights when there is an input at the probe.

If the input is Logic HI, current flows probed point to ground, causing the bottom NOT gate to output a LO. Current flows from the source through the green LED with the bottom NOT gate serving as a sink. The top NOT gate outputs a LO, so the red LED is off.

If the input is Logic LO, current flows from the source to the probed point, with the top NOT gate inverting the signal to output a HI. The top NOT gate then serves as a current source for the red LED. The bottom NOT gate outputs a HI, so the green LED is off.

If there is no input, the top NOT gate is pulled HI with a LO output, and the bottom NOT gate is pulled LO with a HI output, so neither LED will light.

I built a very simple digital design assistant a while ago with a similar circuit, although mine has less functionality. I used a single inverter to do essentially the same thing.

Inverter Circuit

If the input is HI, one LED will light. If the input is LO, the other LED will light. The inverter serves as either the source or the sink, depending upon the input. If there is no input, neither LED will light. The difference in this circuit from the one you showed is that mine does not have the resistors to set the exact desired threshold for triggering - it relies on the chip's internal threshold to function.

Typically, this type of switching would be performed by complimentary mosfet transistors, as noted in the original circuit description. The shown circuit just shows how to perform a similar function with a commonly acquired logic gate.

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