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Home > Archive > Electrical Engineering > June 2006 > why use a resistor across an inverter?
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| Author |
why use a resistor across an inverter?
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| 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
luck.
Does anyone know why?
thanks in advance
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| Palindr☻me 2006-06-13, 5:21 am |
| Peter wrote:
> 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
> luck.
>
>
> Does anyone know why?
>
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.
--
Sue
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| =?UTF-8?B?UGFsaW5kcuKYu21l?= <me9@privacy.net> wrote in
news:128ssn2htr2l1be@corp.supernews.com:
> Peter wrote:
> 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.
>
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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| Palindr☻me 2006-06-13, 10:21 pm |
| Peter wrote:
> =?UTF-8?B?UGFsaW5kcuKYu21l?= <me9@privacy.net> wrote in
> news:128ssn2htr2l1be@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)?
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.
>
> How do you konw the output will be "around" half?
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.
>
> Sounds like the output will either be 0 or 2.5 volts instead of 0 or 5
> volts.
>
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.
--
Sue
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| The Real Chris 2006-06-14, 7:21 pm |
| Hello,
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.
Chris
"Palindr?me" <me9@privacy.net> wrote in message
news:128un1pitreac31@corp.supernews.com...
> Peter wrote:
>
> 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.
>
> --
> Sue
>
>
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