[SOLVED] Can a charge pump gain be 1.2 or 1.3 instead of double or triple?

[CHL]

Hello

Basic charge pump circuits provide integer gain values such as a voltage doubler.

However, is it possible to make the gain be 1.2 or 1.3? not 2

For example, if I use 1.4V battery, I want to make 1.8V output using a charge pump.

 

[chris.mourad]

Hi

You'd better use a boost dc-dc to generate your 1.8V based on 1.4V input.
Otherwise you end up with a complicated charge pump design based on different capacitor values

 

[BradtheRad]

A joule thief is suitable to boost a low supply voltage.

To answer the thread title...
Suppose you start with a simple charge pump. You want to increase voltage by an amount which you can adjust. Choose either the capacitor value, or the operating frequency, to create a duty cycle that gives desired output.

This simulation is an example.



Notice there is no resistive drop limiting current flow. The current pulses are several Amperes. The amplitude is limited by factors such as how much the transistors are turned on, and supply impedance.

 

[CHL]

Hi

You'd better use a boost dc-dc to generate your 1.8V based on 1.4V input.
Otherwise you end up with a complicated charge pump design based on different capacitor values

Can you explain the disadvantage more specifically?

I thought that if the load current is so small, such as lower than 5mA, even a simple charge pump is fine.

Thanks for your help

 

[FvM]

Review the design of fractional charge pumps like LM3350.

 

[CHL]

A joule thief is suitable to boost a low supply voltage.

To answer the thread title...
Suppose you start with a simple charge pump. You want to increase voltage by an amount which you can adjust. Choose either the capacitor value, or the operating frequency, to create a duty cycle that gives desired output.

This simulation is an example.



Notice there is no resistive drop limiting current flow. The current pulses are several Amperes. The amplitude is limited by factors such as how much the transistors are turned on, and supply impedance.

Can you explain more details about the simple circuit?

When the npn is on, C1 and C2 are charged respectively and when the pnp is on, C1 and C2 become a parallel connection.

Should the charge in the C1 flow to the C2 and R to make the boosted output voltage?

Thanks!

 

[mtwieg]

The short answer is no, at least not without a significant penalty in efficiency.

The long answer is that charge pumps can have %100 only when multiplying or dividing by integer ratios (in one stage). To get efficient fractional ratios, you need to cascade a multiplier and divider. This generally is not worth the effort, unless you need a really simple ratio like 3/2 or 2/3.

 

[BradtheRad]

Can you explain more details about the simple circuit?

When the npn is on, C1 and C2 are charged respectively and when the pnp is on, C1 and C2 become a parallel connection.

Should the charge in the C1 flow to the C2 and R to make the boosted output voltage?

Thanks!

C1 is the charge pump capacitor. It charges through one diode, and discharges through a different diode. This is the typical action defining a charge pump.

To drive this action, the half-bridge acts as a SPDT switch.

Operation is further explained at this link (MIC5011 design techniques):

http://micrel.com/_PDF/App-Notes/an-1.pdf

By simulations I found that by making C1 small, it shows greater swing of its DC charge. It drops to 3.77V (in my schematic). This conveys less current. The output voltage is less than double.

C2 can be seen as a smoothing capacitor, or accumulating capacitor. I found by simulations that this capacitor can also be positioned across the load. Then it is the source of power to the load.