I think your voltage issue is because one polarity the current flows across the normal source drain path. and the other polarity it crosses the body diode.

Yes, I also think the problem are the diodes... but why's that? Why they are lowering the voltage so much? Is there any way to avoid that issue? I'd like to have at least equal voltage in both directions.

Also, I think Irving is leaving the coil drive circuit floating instead of the signal circuit.

I see I uploaded wrong schematics in my last post above! I'll update them in a second... something is missing

 

I have updated the schematics on my previous post:

https://forum.allaboutcircuits.com/...er-to-hundreds-of-magnets.171045/post-1536557

The MOSFET gate trigger driven by the opt-coupler is actually floating and connected to the DC source. But the MOSFET source grounded issue remains. I think I should not worry about that if I can find a way to have the same voltage for both polarities. Thoughts on that are very welcome!

Thank you again.

 

To lower the voltage loss across the diodes you could switch to Schottky diodes.

But, that wont help with the asymmetry problem.

 

Yes, it has to be completely floating, else there will be a short on one polarity or another - you must remove ground 3.

Is that an N-channel MOSFET? the symbol is a p_channel JFET... The drain goes to the diode bridge common anodes, the source to the common cathodes.

What MOSFET are you actually testing with? And do you have specification of solenoid coils? I've used 40ohm based on your original post of 20v @ 0.5A.

Your solenoid volts will (and must) be 2 diode drops below the supply volts, if its not its because you're shorting somewhere and its not working as you think it is. You could, as suggested, use Schottky diodes, which would give you 0.8v back, but easier/cheaper to raise supply volts to 24v.

I'm not convinced that powering the gate drive from H-Bridge +ve supply is going to work reliably. On one polarity you don't have enough gate drive and on the other you're over-driving it.

 

Yes, it has to be completely floating, else there will be a short on one polarity or another - you must remove ground 3.

Great, I'll forget about that and leave that circuit completely floating.

Is that an N-channel MOSFET? the symbol is a p_channel JFET... The drain goes to the diode bridge common anodes, the source to the common cathodes.

My apologies for the incorrect symbol, that's actually an opt-coupler. My used program didn't have the right symbol for that, so I sued a combination of a LED and a JFET.

What MOSFET are you actually testing with? And do you have specification of solenoid coils? I've used 40ohm based on your original post of 20v @ 0.5A.

The MOSFET I am using to drive the solenoid is a N-channel IRFZ44N. About the solenoid (electromagnets) I use, yes, your specifications are correct. But when I have measured the voltage I didn't connect it, I just plugged the meter in place of the solenoid to measure the voltage, that's how I found that voltage difference according to the given polarity.

Your solenoid volts will (and must) be 2 diode drops below the supply volts, if its not its because you're shorting somewhere and its not working as you think it is. You could, as suggested, use Schottky diodes, which would give you 0.8v back, but easier/cheaper to raise supply volts to 24v.

I'm not convinced that powering the gate drive from H-Bridge +ve supply is going to work reliably. On one polarity you don't have enough gate drive and on the other you're over-driving it.

As I wrote above, I just measured the voltage at the solenoid connection which wasn't connected. I can certainly raise the voltage if I want to, but how to address the different voltage according to the polarity? Is that due to your mentioned "stress" on the MOSFET? If so, how would you suggest powering the gate other than from the H-Bridge +v supply? Can I control that maybe with a zener diode as you suggested in your previous schematics? Just thinking out loud!

Thank you again very much

View attachment 213566

 

I just plugged the meter in place of the solenoid to measure the voltage, that's how I found that voltage difference according to the given polarity.

That gives no useful measurement as you're not drawing any current. That's the open circuit voltage which is meaningless. Voltage is a measure of the difference across a component due to the current flowing through it. You must measure across the actual solenoid to get true information.

Secondly you're over thinking the problem. The solenoid will have a resistance of 40ohm +/- probably with a manufacturing tolerance of 5% or more. So I'd expect the voltage across it, at 0.5A, to be between 19 and 21v anyway.

Magnet field strength is current related, not voltage related. Voltage represents the 'pressure'to put the required current through a component. Your solenoids are rated at 0.5A @ 20v, it's the current that matters, and that's the current that gives their rated magnetic field strength. The 32v rating is saying that's the voltage which will drive the maximum current (0.8A) through it and above which it will overheat & burn out.

The difference in field strength between 19 and 21v is not huge. If that's a problem then you need to be driving them to give a much higher operating current, say 0.6A (26+ volts) or finding solenoids with a greater magnetic field density at 0.5A.

 

If so, how would you suggest powering the gate other than from the H-Bridge +v supply? Can I control that maybe with a zener diode as you suggested in your previous schematics?

In my previous schematic in post #94 I showed an independent isolated nominal 10v gate supply from a simple dc-dc converter (oscillator, transformer, diode, capacitor combo) That's the only safe way to get an isolated and fixed voltage gate supply.

To do it your way, you'd need a 32v auxiliary supply, or a 10v isolated supply connected to the 22v rail, and then regulated down to 10v at the actual gate - so you might as well do it my way, either with the discrete component solution from my diagram, or a separate commercial isolated dc-dc supply for each switch.

 

The IRFZ44N MOSFET needs a gate voltage of 10v to be fully switched on. Otherwise it's fine for this.

Regarding the symbol, I wasn't referring to the optocoupler but to the symbol you used for the MOSFET.

 

That gives no useful measurement as you're not drawing any current. That's the open circuit voltage which is meaningless. Voltage is a measure of the difference across a component due to the current flowing through it. You must measure across the actual solenoid to get true information.

Secondly you're over thinking the problem. The solenoid will have a resistance of 40ohm +/- probably with a manufacturing tolerance of 5% or more. So I'd expect the voltage across it, at 0.5A, to be between 19 and 21v anyway.

Magnet field strength is current related, not voltage related. Voltage represents the 'pressure'to put the required current through a component. Your solenoids are rated at 0.5A @ 20v, it's the current that matters, and that's the current that gives their rated magnetic field strength. The 32v rating is saying that's the voltage which will drive the maximum current (0.8A) through it and above which it will overheat & burn out.

The difference in field strength between 19 and 21v is not huge. If that's a problem then you need to be driving them to give a much higher operating current, say 0.6A (26+ volts) or finding solenoids with a greater magnetic field density at 0.5A.

You are right, I am overthinking and I'd need to measure at the solenoid. Let's forget about this, as you said, this is not an issue! Thank you for helping me understand.

 

In my previous schematic in post #94 I showed an independent isolated nominal 10v gate supply from a simple dc-dc converter (oscillator, transformer, diode, capacitor combo) That's the only safe way to get an isolated and fixed voltage gate supply.

To do it your way, you'd need a 32v auxiliary supply, or a 10v isolated supply connected to the 22v rail, and then regulated down to 10v at the actual gate - so you might as well do it my way, either with the discrete component solution from my diagram, or a separate commercial isolated dc-dc supply for each switch.

Yes, I'll see how to improve my schematics based on your suggestions

 

The IRFZ44N MOSFET needs a gate voltage of 10v to be fully switched on. Otherwise it's fine for this.

Regarding the symbol, I wasn't referring to the optocoupler but to the symbol you used for the MOSFET.

From these specifications I read a range of 2-4v for the "Gate Threshold Voltage":

https://www.infineon.com/dgdl/irfz44npbf.pdf?fileId=5546d462533600a40153563b3a9f220d

Am I reading wrong?

Thanks again!

 

Hello,

What will be the gate voltage?
if it is only 5 Volts from a microcontroller, have a look at the IRLZ44, wich has a gate threshhold of 1 - 2 Volts.



Bertus

 

From these specifications I read a range of 2-4v for the "Gate Threshold Voltage":

https://www.infineon.com/dgdl/irfz44npbf.pdf?fileId=5546d462533600a40153563b3a9f220d

Am I reading wrong?

Thanks again!

Yes and no...

Gate threshold voltage is the lowest voltage that barely turns the MOSFET on. It is not in itself a useful number. If you look at how its specified you'll see it says that's at a drain current of 250uA. Vth or VGS(th) is best thought of as the threshold below which the MOSFET is actually off. If youlook at the curves for Vds v Id you'll see they give curves for Vgs from 4.5v to 15v. The performance of the MOSFET, particularly its Rds(on) resistance, is not specified/guaranteed below Vgs of 4.5v and across temperature and voltage tolerances, if you are designing for reliability, you should be looking at double that.

If you need a MOSFET to do any heavy lifting (i.e a few hundred mA upwards) you need to give it sufficient Vgs to get the on-resistance down. If you're driving it from a microcontroller you need a logicFET which has a guaranteed Rds(on) at a Vgs of 2 - 3v. The IRLZ44N (as opposed to the IRFZ44N) parts, as suggested by @bertus, have been selected during manufacture to work at 2.5v. Even so, you'll only handle something like 3A max. There are better logicFETs (eg IRF7456PbF , 10A @ Vgs = 2.5v).

Anyway, for your needs the IRFZ44N is fine, you just need to drive it properly...

 

Yes and no...

Gate threshold voltage is the lowest voltage that barely turns the MOSFET on. It is not in itself a useful number. If you look at how its specified you'll see it says that's at a drain current of 250uA. Vth or VGS(th) is best thought of as the threshold below which the MOSFET is actually off. If youlook at the curves for Vds v Id you'll see they give curves for Vgs from 4.5v to 15v. The performance of the MOSFET, particularly its Rds(on) resistance, is not specified/guaranteed below Vgs of 4.5v and across temperature and voltage tolerances, if you are designing for reliability, you should be looking at double that.

If you need a MOSFET to do any heavy lifting (i.e a few hundred mA upwards) you need to give it sufficient Vgs to get the on-resistance down. If you're driving it from a microcontroller you need a logicFET which has a guaranteed Rds(on) at a Vgs of 2 - 3v. The IRLZ44N (as opposed to the IRFZ44N) parts, as suggested by @bertus, have been selected during manufacture to work at 2.5v. Even so, you'll only handle something like 3A max. There are better logicFETs (eg IRF7456PbF , 10A @ Vgs = 2.5v).

Anyway, for your needs the IRFZ44N is fine, you just need to drive it properly...

So.... will about 10v be better to trigger it?

 

It needs to be at least 4.5v but no more than 15v... I always aim for 10v unless its a high-speed switching requirement, then it gets more complicated...

 

It needs to be at least 4.5v but no more than 15v... I always aim for 10v unless its a high-speed switching requirement, then it gets more complicated...

I'll try 10v then. Now sure what you mean with "hi-speed"... in my application the max switching speed I'll need will be probably in the range of milliseconds, not faster than that. I'll test and report. Thanks!

 

Hi-speed is <100nS... not an issue for you. The gate of a MOSFET is a capacitor, so to get it from 0 to 10v it takes a certain amount of current since V(t) = I * t / C. From the specs the total gate charge is 63nColumbs @ 10v, and a Columb is 1A per sec. So, to a first approximation, if you wanted to switch it on fully in 63nS you'd need at least a 1A drive capability. At 6.3mS you'll need at least 10mA.

 

Hi-speed is <100nS... not an issue for you. The gate of a MOSFET is a capacitor, so to get it from 0 to 10v it takes a certain amount of current since V(t) = I * t / C. From the specs the total gate charge is 63nColumbs @ 10v, and a Columb is 1A per sec. So, to a first approximation, if you wanted to switch it on fully in 63nS you'd need at least a 1A drive capability. At 6.3mS you'll need at least 10mA.

Awesome information. Thank you so much!