High Current AC Voltage Divider?

[johnywhy]

Is a resistor divider suitable for a low-voltage, high-current AC voltage divider?

I have an 8V p-p AC sine wave, powering a load that's drawing 1.5A. I need to divide that AC voltage down by half, without distortion, with good precision, so the load will draw 3A @ 4V p-p. And i need it to stay cool.

Is there a simple passive way to do it?

I assume i could use a step-down trafo, but that's bulkier than i would like.

How about a capacitive divider?


Or, a capacitive/resistive divider?


Inductive divider?

[geggi1]

What you are describing is a transformer.
With a voltage divider you will never get a higher current out of the circuit than in. the loss will also be at least as high as the load.

Voltage dividers is normally only used in low current designs to match impedanze or reduse the voltage for mesuring or similars.

[johnywhy]

Isn't a capacitive power supply designed to deliver high current?

A capacitive power supply or capacitive dropper is a type of power supply that uses the capacitive reactance of a capacitor to reduce higher AC mains voltage to a lower DC voltage. A relatively large mains-voltage capacitor is required and its capacitance must increase with the output current, which leads to a higher-cost and bulky capacitor.


https://en.wikipedia.org/wiki/Capacitive_power_supply

[Gyro]

Isn't a capacitive power supply designed to deliver high current?

A capacitive power supply or capacitive dropper is a type of power supply that uses the capacitive reactance of a capacitor to reduce higher AC mains voltage to a lower DC voltage. A relatively large mains-voltage capacitor is required and its capacitance must increase with the output current, which leads to a higher-cost and bulky capacitor.


https://en.wikipedia.org/wiki/Capacitive_power_supply

That power supply delivers a mere 10mA and uses a single series capacitor, not a capacitive divider.. It also offers no safety isolation from the mains, so isn't suitable for anything that can be touched.

A capacitive divider (two capacitors) is no more suitable for providing any significant current than a resistive voltage divider. They are used (often in combination with a resistive divider) for measurement purposes only.

It sounds as if a proper wall wart PSU is what you need Edit: or bite the bullet and use that transformer - it's the only way you're going to achieve any current multiplication in any case].

[Andy Chee]

Is a resistor divider suitable for a low-voltage, high-current AC voltage divider?

I have an 8V p-p AC sine wave, powering a load that's drawing 1.5A. I need to divide that AC voltage down by half, without distortion, with good precision, so the load will draw 3A @ 4V p-p. And i need it to stay cool.

Is there a simple passive way to do it?

I assume i could use a step-down trafo, but that's bulkier than i would like.

What frequency is your sine wave?  If it is >20kHz the transformer can be a relatively compact ferrite transformer.

Otherwise, no, there is no passive way to do it.

Your distortion and precision specification suggests some unusual application.  Care to enlighten us?  Showing us a bigger picture may allow us to proffer alternate solutions.

[johnywhy]

That power supply delivers a mere 10mA and uses a single series capacitor, not a capacitive divider.. It also offers no safety isolation from the mains, so isn't suitable for anything that can be touched.

I'm not working with mains. I want to step down a 1.5A, 10V, 400 kHz AC signal.

Your distortion and precision specification suggests some unusual application.  Care to enlighten us?  Showing us a bigger picture may allow us to proffer alternate solutions.

What i can share is that i need to convert a bipolar 3A, 5V p-p, 400 kHz AC signal to a unipolar, positive-only 3A, 5V p-p, 400 kHz signal.

My first plan was to use a Villard doubler to convert the biplar wave to unipolar. The video:

https://youtu.be/ep3D_LC2UzU

My concern was that the V doubler will halve the current and double the voltage, from 3A, 5V p-p to 1.5A, 10V p-p. That's why i posted this question looking for a divider to restore the original.

But i think now this was a misconception. It looks like the p-p doesn't change, the input and output of the Villard will be the same current, same V p-p. In which case i don't need this divider.

[Andy Chee]

What i can share is that i need to convert a bipolar 3A, 5V p-p, 400 kHz AC signal to a unipolar, positive-only 3A, 5V p-p, 400 kHz signal.

At 400kHz, a transformer can be made using this core:

https://www.digikey.com.au/en/products/detail/epcos-tdk-electronics/B65803P0000Y038/3914034

Is that transformer still too "bulky" for your application?

[johnywhy]

At 400kHz, a transformer can be made using this core:
https://www.digikey.com.au/en/products/detail/epcos-tdk-electronics/B65803P0000Y038/3914034
Is that transformer still too "bulky" for your application?

Not to bulky! Thx for that. Why did you select RM type?
But, as explained in my last comment, i might not need to step down after all.

[Andy Chee]

No reason for RM core, any core with the same volume (or more, within reason) should work at 400kHz @ your current.

[johnywhy]

No reason for RM core, any core with the same volume (or more, within reason) should work at 400kHz @ your current.

Thx! Why did you choose T38 material?

[Andy Chee]

No reason for RM core, any core with the same volume (or more, within reason) should work at 400kHz @ your current.

Thx! Why did you choose T38 material?

Actually now that you mention it, that material may not be appropriate.

In general, high frequency AC calls for ferrite material. 

However your application of a unipolar output, may potentially impose a net DC through the secondary winding.  If this is the case, you will need something like a powdered iron material instead or larger core size or both.

Need your schematic to make an informed decision on that.  If you use a full wave diode bridge on the output, then ferrite remains viable.  If you've got anything else hanging off it, then perhaps not.