Re: Understanding Charge Pumps

On Sep 23, 9:57 am, kaba.electron...@xxxxxxxxx wrote:
Hi Mosfet,

The part you describe is the part I already understand. I guess the
problem is the messed up ascii art.

Let me try to describe the ckt:

Switch A - from 5V to C1
Switch B - from C1 to ground
Switch C - from 5V to C2
Switch D - from C2 to out
Flying cap is connected between C1 and C2

Out has a load capacitor and a load resistance - both connected to
ground

First part of the cycle B and C are on - A and D are off

Assume ideal swiches - this is the part of the cycle where the cap is
charged to 5V - this I understand

Now the next half of the cycle - A and D are on - This is the part
where the output cap goes from 5V to 10V and the other end of the cap
goes from 0 to 5V.

During this half of the cycle - WHAT DETERMINES THE CURRENT FLOWING
THROUGH SWITCH A? - is my question

You mention the current that is used to replace the charge lost when
there is load on the output. I understand this. This is the current
that flows through switch C into the cap and charges it while the
other end of the cap is connected to ground through switch B.

Hope the question is clearer 'mosfet'. I appreciate any further
insight you can provide.

Thanks
kaba

On Sep 23, 9:17 pm, MooseFET <kensm...@xxxxxxxxx> wrote:



On Sep 23, 8:55 pm, kaba.electron...@xxxxxxxxx wrote:

Hi,

I am having difficulty understanding charge pumps and flying
capacitors. Can someone please explain?

I can't see your Ascii art because I'm having trouble with fonts but I
will explain based on your text.

Okay with reference to the circuit below:

                            +5V
+5V

                    switch_a |                              | switch_b

                                           FlyingCap

OUT
                    switch_c |                              |
switch_d  ------------outputcap--------GND

                        GND

When switch B and C are on, the cap gets charged to 5V. This I get.

Don't forget that neither the switches nor the capacitor are perfect.
It takes a little time for the capacitor to get charged.  If you
switch it away from the 5V before it is fully charged, its voltage
will be lower than 5V.

Then switch A and D are on and B and C are off. This is where my
understanding breaks down.

- There will be a spike current in switch A charging the base of the
flying capacitor to 5V. How much current will flow? Is the quantity of
this current determined by the size of the flying cap, size of the
output cap, output load?

The difference in the voltage on the flying capacitor and the output
capacitor is what drives this current.  The resistances of the
switches and the capacitors matter too.  It is an RC time constant
like situation the current will have an exp(-t/RC) shape to it.

After the charge pump has been running for a while the capacitors are
nearly fully charged.  If you start with it unloaded and think about
it, you will see that other than stray effects the current spikes
disappear after a while.

Once all of the capacitors are charged up, think about attaching a
load.  This will slightly discharge the output capacitor during each
cycle leading to a spike again.  The size of the spike depends on how
much the capacitor got discharged on each cycle.

What is the relationship to each of these.

- The charge on the capacitor remains constant during the second half
of the cycle. So the current during the second half of the cycle must
flow to charge the parasitic cap from the other terminal of switch A
to ground. Does this mean that this current is independant of the
flying cap?

Thanks for listening. Any help is much appreciated

Cheers
kaba- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

It is sometimes easier to look at the charge and not the current in
these situations. Q = C*V. Follow the charge around the circuit.
When you are done you can worry about the currents.

George

.

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