Olin wrote:
> > How is this? When you power up the circuit, your capacitor will be
> > discharged. Which means that the positive end will be at either 0V
> > (if the negative end is tied to ground) or at 5V (if it's tied to
> > 5V).
>
> Not from the charge pump's point of view.  Yes, if the circuit has been
> off a long time the voltage accross all the caps will be 0.  However,
> the +10V supply is brought to +5V directly by the +5V supply.  This
> happens as the +5V supply comes up, and is not done by the charge pump,
> and therefore doesn't slow it down getting to +10V.

To a first order approximation, you're right. However, second-order effects
come into play.

For one thing, if you return the +10V filter capacitor to ground, you are
charging it to twice the voltage, and the current to do this must come
through the charge pump, which basically looks like a fixed resistance.
This has two implications. First, although the RC time constant is the
same in either case, it takes a fixed amount of additional time to reach
a particular voltage compared to returning it to +5V. The second thing
is that you are forcing twice as much current through the charge pump.

If the charge pump switches and capacitors have zero resistance, you can
readily see that it takes exactly one extra clock cycle for the output
capacitor to get to a particular voltage. However, these components not
only have resistance, the switches can also become nonlinear, saturating
at a fixed amount of current. This extends the charging time even further.

As far as noise on the +5V supply line, I would point out that if the +10V
output capacitor is returned to +5V, the charge pump draws one pulse of
current from the +5V supply for every cycle. In this configuration, the
+5V supply also sees the +10V load current, both the DC and AC components.

On the other hand, if you return the capacitor to ground, you won't have
the +10V load current flowing through the +5V supply, but you'll see two
pulses of current into the charge pump for each cycle. I can't see that it
makes any appreciable difference one way or the other.

So overall, in the interest of getting faster startup and recovery from
transient loads by avoiding saturation effects in the switches, I think
it would be better to connect the bottom end of the +10V filter capacitor
to the +5V supply.

-- Dave Tweed

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