Understanding Charge Pumps

Hi,

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

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. 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? 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

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

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.

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.

Reply to
MooseFET

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

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| switch_b

=A0 =A0 =A0 =A0FlyingCap

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|

Reply to
kaba.electronics

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0| switch_b

=A0 =A0 =A0 =A0 =A0FlyingCap

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0|

f

It is sometimes easier to look at the charge and not the current in these situations. Q =3D C*V. Follow the charge around the circuit. When you are done you can worry about the currents.

George

Reply to
ggherold

Presumably the output load is bypassed with more caps to ground.

During phase 1, when the flying cap is being charged, the load is discharging its own bypass caps, so droops below 10 volts some.

When the cap is switched, we have 10 volts as a series circuit Sa + Cap + Sb, dumped into the load which has drooped below 10 volts by some difference, call that Delta volts. The cap will try to dump a shot of charge into the load to pick it back up. With a finite load and a finite cap, it won't make it all the way back up to 10.

Obviously, the average current provided by the cap must equal the average load current. The exact waveform of the charging current depends on the circuit parasitics: switch resistances, any ESR in the cap, any stray inductance. In the ideal/impossible case, the current is an impulse, infinite in height, zero duration, area (charge) as needed, namely Q = Iload/F.

Usually the current will spike at I = Delta / (2 * Rsw) and decay exponentially, tau = 2 * Rsw * C, or something like that. It can get complicated.

John

Reply to
John Larkin

Right here, you have assumed your way towards not understanding. The capacitors and switches are nevre ideal. Think of them as having simple linear resistance. At the instant the switches close the flying capacitor has less than 5V on it. There will be a current spike at that time. This spike will transfer charge into the capacitor.

There is a voltage difference between 5V + Vcap and Vout. This difference and the resistance of the switches etc controls this current spike. As the current flows, the flying capacitor discharges slightly and the output one charges. The voltage on the flying capacitor when the switch opens again is what will be there on the next cycle's start.

Reply to
MooseFET

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