Hello,
Just a wireless power transfer with a couple of 0,50 mm wires.
First having square wave from 555 and flipping it with LM358N this 16 V. 800 mA. max. power.

On the transferred site, again a LM358N OpAmp to create a square wave and boosting the voltage. The result is approximately 310 mA with 9 V. on the green 4.21mH coil.

The purpose is to use the green 4.21 mH coil as a magnet, it becomes an electro-magnet but the power is not enough.
Actually there are 12 of these coils in series.
Any advice ?
(All the coils are made of 0,50 mm. coppers. Inductance values are estimated.)

 

The LM358 is probably not a good choice for this circuit. It has a fairly low GBW of 1 MHz and a slew rate of 0.3 v/uS. which is too slow for a switching circuit like this.
Regards,
Keith

 

How much power are we talking about transferring?
Over what distance?

 

The LM358 is probably not a good choice for this circuit. It has a fairly low GBW of 1 MHz and a slew rate of 0.3 v/uS. which is too slow for a switching circuit like this.
Regards,
Keith

Thank you very much.

 

How much power are we talking about transferring?
Over what distance?

As much as possible. Maximum 5 mm. distance.

 

As much as possible. Maximum 5 mm. distance.

That is not a useful answer. Here is why. You cannot get more power out of a transfer scheme than you put in. This means the efficiency of any transfer scheme will be less than 100% Now, "as much as possible". is limited by how much you are willing to put in. So let us start with how much you are willing to put in and what efficiency you are aiming for. Can we do that?

 

Where does the LM358 on the "transferred" side get it power from?
What are the length parameters on the JFETs for?
What frequency is the squarewave from the 555?
Why is the amplifier in the lower circuit operating open loop?

This circuit looks like a chimera.

 

LM358 on the receiver gets its power from the coil. And by creating a square wave, boosting it to 310 mA with 9 V.
The source have 16 V. max 800 mA power.
MOSFETS are IRFZ44N. https://pdf1.alldatasheet.com/datasheet-pdf/view/620717/NJSEMI/IRFZ44N.html
The 555 's frequency is about 600 Mhz.

 

Where does the LM358 on the "transferred" side get it power from?
What are the length parameters on the JFETs for?
What frequency is the squarewave from the 555?
Why is the amplifier in the lower circuit operating open loop?

This circuit looks like a chimera.

LM358 on the receiver gets its power from the coil. And by creating a square wave, boosting it to 310 mA with 9 V.
The source have 16 V. max 800 mA power.
MOSFETS are IRFZ44N. https://pdf1.alldatasheet.com/datasheet-pdf/view/620717/NJSEMI/IRFZ44N.html
The 555 's frequency is about 600 Mhz.

 

LM358 on the receiver gets its power from the coil. And by creating a square wave, boosting it to 310 mA with 9 V.
The source have 16 V. max 800 mA power.
MOSFETS are IRFZ44N. https://pdf1.alldatasheet.com/datasheet-pdf/view/620717/NJSEMI/IRFZ44N.html
The 555 's frequency is about 600 Mhz.

That cannot be correct. A 555 cannot work at 600 MHz., and neither can an LM358. It is beginning to look like you don't have a clue about what you are doing.

 

The LM358 is probably not a good choice for this circuit. It has a fairly low GBW of 1 MHz and a slew rate of 0.3 v/uS. which is too slow for a switching circuit like this.
Regards,
Keith

How about a 5532 OpAmp ? http://www.ti.com/lit/ds/symlink/ne5532.pdf?&ts=1588918444948

 

As much as possible. Maximum 5 mm. distance.

I want to build a boat.

Q: How big do you want the boat to be?
A: As big as possible.

Can you understand how useless and impractical is such an answer?

 

I want to build a boat.

Q: How big do you want the boat to be?
A: As big as possible.

Can you understand how useless and impractical is such an answer?

Yes Mr Chips. Sure.
I am sorry about the impraticalness of my answer.
I would like to say; I want to get as much as possible power on the transferred side of my circuits.
I use on Tx side 16 V. max 800 mA. and with this combination of circuits, I get approximately 9V. with 310 mA. on my Rx side.
I try to increase this power of my Rx side.
Excuse me for my English. I am not a native Englishman, neither an electric, nor an electronic engineer.
Just an Industrial Designer who try to learn some electronics.
Thank you for your interest and time.

 

That cannot be correct. A 555 cannot work at 600 MHz., and neither can an LM358. It is beginning to look like you don't have a clue about what you are doing.

Yes sorry. This is approximately 600 KHz. (KiloHertz) sorry.

 

Yes Mr Chips. Sure.
I am sorry about the impraticalness of my answer.
I would like to say; I want to get as much as possible power on the transferred side of my circuits.
I use on Tx side 16 V. max 800 mA. and with this combination of circuits, I get approximately 9V. with 310 mA. on my Rx side.
I try to increase this power of my Rx side.
Excuse me for my English. I am not a native Englishman, neither an electric, nor an electronic engineer.
Just an Industrial Designer who try to learn some electronics.
Thank you for your interest and time.

Tx side 16 V. max 800 mA = 12.8W
RX side 9V. with 310 mA = 2.79W
Efficiency = 22%
Is this reasonable efficiency? How would you increase the power at the receiver?
You either have to increase the Tx power or the efficiency.

 

"How would you increase the power at the receiver?"
This is my question is about.
If you have any idea. Please do not hesitate to share it with me.
To increase the power in Tx side is increasing the heat of the Mosfets at an undesired level.
Thank you.

 

Your MOSFETs are getting hot because they are staying too long in the linear region.
You need to drive the MOSFETs harder and faster to minimize the effect of the capacitance at the gate of the MOSFET.

 

Your MOSFETs are getting hot because they are staying too long in the linear region.
You need to drive the MOSFETs harder and faster to minimize the effect of the capacitance at the gate of the MOSFET.

Thank you.

 

[QUOTE="ouzcoskun, The 555 's frequency is about 600 Mhz [/QUOTE]

https://www.ti.com/lit/ds/symlink/lm555.pdf?&ts=1589807336993 page No-6. Rise time 100 ns, fall time 100 ns. Further You may calculate Yourself. If not believe to calculator, in the same document look into page 11 graph fig-16.

 

Thank you.