difference between transformer, charge pump, opamp

difference between charge pump & opamp

Since any of charge pump or an op-amp can be used to double the input voltage...

What is the difference in the principle of the two and which should be used in which scenario?

Hi,

a charge pump is a device to generate a voltage as power supply.

An OPAMP is an amplifier, meant to amplify analog signals. An OPAMP needs a power supply.
The supply rails need to be wider than the expected output voltage. <--> An OPAMP can not generate an output voltage that is outside of the supply rails.

Klaus

Sorry, I can't understand:
How is it that a charge-pump can boost voltage without an external power supply? That way we would be able to drive any amount of current through the load by boosting voltage (I=V/R)?

A charge pump has a DC-powered oscillator that produces an AC voltage. Then it charges a capacitor to near the peak voltage of the AC. Then on the next AC cycle it charges another capacitor and adds its voltage to the voltage of the first capacitor.
A charge pump can also convert a positive voltage to a negative voltage using its oscillator, some diode rectifiers and some capacitors.

Most charge pumps have a small maximum output current. The ICL7660 voltage converter IC can double a fairly low voltage or convert positive to negative with a maximum output current of only 10mA to 20mA. Maxim make an IC with a little more output current.

Hi,

How is it that a charge-pump can boost voltage without an external power supply?

Click to expand...

For sure it needs input power. It may double it. So if you want 12V output you need 6V input.

Maybe my description was not clear: The charge pump is made to supply other circuitry with power.
In opposite it is not meant to amplify signals, like from a microphone.

Klaus

Hi,

KlausST said:

For sure it needs input power. It may double it. So if you want 12V output you need 6V input.

Click to expand...

It can't double input power right (input power = output power)? But it can double the voltage. What I'm confused is does that mean we can generate any amount of current (that would also mean any amount of power) from a given input voltage? Because if we connect a load across the multiplied voltage it would generate current proportional to the multiplied voltage, right?

Maybe my description was not clear: The charge pump is made to supply other circuitry with power.
In opposite it is not meant to amplify signals, like from a microphone.

Click to expand...

I understood this part.

Thanks.

Hi,

It can't double input power right (input power = output power)? But it can double the voltage. What I'm confused is does that mean we can generate any amount of current (that would also mean any amount of power) from a given input voltage? Because if we connect a load across the multiplied voltage it would generate current proportional to the multiplied voltage, right?

Click to expand...

It can't increase power....

If a device has 100% efficiency (just imaginary. In almost all cases it is less than 100%, this is also true for voltage doublers) then:
* Input power = output power
* if you increase voltage, then it will decrease current

Example with 10V input voltage, 1000 Ohms load:
* connect both and you will see 10mA current and 0.1W of power
* now use a voltage doubler
* connect the resistor at doubler output, now 20V
* you will see 20mA current through the resistor and 0.4W of power.

* here begins your understanding problem..
* you have to see the resistor as one system and the doubler as an independent system.
* the current in the resistor increases because of the increased voltage, it is just Ohm's law and independent of what power supply you use.
* Before you had 0.1W now you have 0.4W. The voltage doubler has 10V input voltage at 0.4W that means 40mA input current.

Now see the voltage doubler:
Input: 10V, 40mA, 400mW
Output: 20V, 20mA, 400mW
You see the voltage is doubled, but the current is half, so the power stays the same..

I hope this helps

Klaus

Yes, it makes sense if you look at the doubler and load as independent systems... Thanks.