Reply-To: "Kevin Aylward"
From: "Kevin Aylward"
References: <firstname.lastname@example.org> <email@example.com>
Subject: Re: Help analysing a CMOS ring oscillator
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Fri, 10 Jan 2003 08:01:30 -0000
NNTP-Posting-Date: Fri, 10 Jan 2003 08:01:35 GMT
Jeff Verive wrote:
> "Kevin Aylward" wrote in message
>> Jeff Verive wrote:
>>> To answer your question, CLoad is a sum of internal capacitance
>>> (including capacitance due to the device package), external probe
>>> capacitance, and external capacitance due to attached devices.
>> This is not completely correct, it its a bit more complicated than
>> this, hence my post. CLoad is also due to the miller capacitance of
>> the following stage. However, the miller gain is itself a result of
>> the load capacitance on its output. This goes on all around the
>> loop. There is not an easy way to model this, imo.
> Multiply the load capacitance by the gain for the given stage, if you
> want to get a closer approximation.
Ho hum.... That's the bloody point. You don't know the gain. The gain
depends on the load capacitance of the next stage (gm.Xload), the
capacitance of the next stage depends on the gain of the following
stage, the gain of that stage depends on the gain of the 1st stage,
hence you have to solve solve some simultaneous equations to get the
gain/phase of the loop.
But remember that the use of
> inverting feedback lowers overall gain and hence decreases Miller
This statement makes no sense.
> Stable ring oscillators often include gain limiting
> stages for this purpose (since open loop gain is highly temperature
> and voltage dependent).
Open loop gain is not really important in itself. What matters is how
stable the phase is at temperature/volts. However, the effective phase
shift is supply voltage dependant, so you do want to stabilise it.
> Anyway, you were looking for a simplified lumped Cload.
I wasn't, the original poster was, and I was explaining that it is not
trivial to come up with the effective gain in a loop like this. You cant
easily calculate the miller capacitance, because you don't know the
gain, because it depends on knowing what the capacitance is in the first
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