why use a resistor across an inverter?

I saw a circuit where a 74AC04 invertor had a 1Mohm resistor across it.
I've looked this concept up in my electronics books, but haven't had any
Does anyone know why?
thanks in advance
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Yep. It basically turns the inverter into a high-gain "linear" amplifier that will respond to small signals (and, of course, large ones!) at its input. Typically used with other components to make an oscillator.
How it works is fairly simple - the resistor applies feedback so that the output will settle at around half the supply voltage and so will the input. This region of operation, at neither "1" or "0", isn't normal for a digital circuit and isn't covered by the specification. So actually using it as a linear, analogue, amplifier isn't a good idea - but it works well in oscillators and threshold detectors.
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=?UTF-8?B?UGFsaW5kcuKYu21l?= wrote in news: snipped-for-privacy@corp.supernews.com:
This still confuses me a bit.
How is this considered high gain if the output is half what it should be (2.5 volts basing the chip to be exactly 5 volts)?
How do you konw the output will be "around" half?
Sounds like the output will either be 0 or 2.5 volts instead of 0 or 5 volts.
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Most linear amplifiers have the output set to around half the supply voltage in its quiescent state (ie no input signal). The inverter plus resistor acts like one of these amplifiers. The resistor added to the inverter makes the inverter respond to small signals. Normally an inverter is only used with large signals, ie 0 1 transitions on the input.
Imagine the inverter with no input for a moment. Let's say the input is pulled "high" by internal leakage. The output will be "low".
Now add the resistor. The "low" on the output will now put a "low" on the input - so the inverter will start to change to produce a "high". As the output rises, so will the input. As the input voltage reaches about half the supply rail voltage, the output voltage starts to fall until it too is about half the supply voltage. At which point, a stable state is reached.
The quiescent state of the output is going to be about half the supply voltage because, if it was much higher, it would take the input to a clear "high" state and if it was much lower, it would take the input to a clear "low" state - in both cases the changed input would produce a corresponding changed output. It is only in the undefined area around the half supply point that the inverter can stabilise.
Nope, the quiescent state will be around 2.5 volts. Applying an ac signal to the input pin will produce a corresponding ac output signal, going /either side/ of the 2.5v.
Can I suggest that you look at how a standard linear amplifier works? These too normally set the output to half the supply voltage in the quiescent state (although this dc component will typically be removed by a coupling capacitor). The gain is nothing to do with this quiescent state.
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It may be just to load the ouput during "no load" conditons that might make the inverter unstable. Some inverters do not work without any load and just fly up and down in an erratic way so the output votlage cannot be measured without a load.
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The Real Chris

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