transformer driving mosfet

[Ubergeek63]

Isn't it a simple case of using a DC-blocking capacitor in series with the primary?? Seems both affordable and practical...

no, it is a matter of ratios. at 50% duty cycle the core is completely reset and operates at zero FLUX bias, anything else puts bias on the core. that is why you find reset windings on forward converters.

the area above zero and below need to cancel out.

 

[dougy83]

the area above zero and below need to cancel out.

And you're saying that this doesn't happen once you've blocked the DC-component of the drive??

I've attached some pictures in case you're interested; one is the dc-coupled gate drive transformer (GDT), while the other is the ac-coupled GDT. The DC one can be seen to have an increasing inductor current, while the other does not. The trace marked V(n002) is the voltage across C2.

 

[Ubergeek63]

And you're saying that this doesn't happen once you've blocked the DC-component of the drive??

I've attached some pictures in case you're interested; one is the dc-coupled gate drive transformer (GDT), while the other is the ac-coupled GDT. The DC one can be seen to have an increasing inductor current, while the other does not. The trace marked V(n002) is the voltage across C2.

not the current the CORE FLUX. for a full discussion and solution see the attatched

 

[shortbus=]

Quote dougy83 "Why would the secondary be collapsing."

When the mosfet is no longer conducting, when the Micro shuts it off.

 

[tcmtech]

Typically in transformer based gate drive circuits the primary is driven off of a dedicated driver Ic and the secondary is then snubbed by a pair of zener diodes placed back to back in series to limit any possible switching spike to a level below the gates maximum voltage rating.

Thats all there really is to it. the rest is just turns ratio, duty cycle and frequency that determines what type of transformer is used.

 

[dougy83]

Quote dougy83 "Why would the secondary be collapsing."

When the mosfet is no longer conducting, when the Micro shuts it off.

The gate drive transformer is not driven by a single-ended driver - more likely by a totem pole, so the primary always has a low impedance source/driver. Therefore the secondary won't just 'collapse'.

 

[dougy83]

not the current the CORE FLUX. for a full discussion and solution see the attatched

What? As far as I'm aware, the flux in an area is the (area * mag. field) and the mag field is the (current * some independant integral of the length of the conductor). This means the flux is proportional to the field, the field is proportional to the current, and therefore the current and flux are proportional too. So "not the current the CORE FLUX" makes no sense to me.

That document shows says the volt-seconds must cancel out on each cycle or there will be bias. Measuring (by eye) the volt-seconds of the AC-coupled signal (shown in the pictures) across the primary: positive: 7.35V*4us = 29.4u, negative: -2.66V*11us=29.3u. Allowing for reading errors, it's balanced.

 

[Ubergeek63]

What? As far as I'm aware, the flux in an area is the (area * mag. field) and the mag field is the (current * some independant integral of the length of the conductor). This means the flux is proportional to the field, the field is proportional to the current, and therefore the current and flux are proportional too. So "not the current the CORE FLUX" makes no sense to me.

That document shows says the volt-seconds must cancel out on each cycle or there will be bias. Measuring (by eye) the volt-seconds of the AC-coupled signal (shown in the pictures) across the primary: positive: 7.35V*4us = 29.4u, negative: -2.66V*11us=29.3u. Allowing for reading errors, it's balanced.

had you bothered reading it AND understanding it you would have seen that if you supply a 90% duty cycle to a 10V FET gate you would need a 100V reset pulse in order to reset the core. Are you suggesting he put a +/-100V supply in his system to get a 10-90% duty cycle?

 

[dougy83]

As far as I can see, that's for a DC coupled driver only.

Are you suggesting he put a +/-100V supply in his system to get a 10-90% duty cycle?

Not sure how you arrived at that conclusion.

 

[Ubergeek63]

As far as I can see, that's for a DC coupled driver only.Not sure how you arrived at that conclusion.

simple, if you need +10V to saturate the FET at 90% duty cycle to keep the volt seconds equal you need -90V at 10% duty cycle 10V*90%=90V*10% there fore you would need a -100V supply to reset the core unless you plan on ordering a custom transformer 1:1:10 using the 10T winding as a reset winding. but that still only gets you to 10% to 90% duty cycle. many PWMs run 1% to 99% or more.

 

[shortbus=]

It still goes back to what I said in post #5. It's much easier, faster, cheaper, less board space and fewer parts to use one of the dedicated gate driver chips!

 

[MrAl]

Hello,


A driver transformer for a MOSFET should have certain properties such as:
Minimum leakage inductance by using 1:1 turns ratio, a gapless toroid core, and bifilar windings (trifilar for push-pull).
Minimum number of turns to reduce capacitance.
Cores with small diameters and narrow hysteresis loops to reduce magnetizing currents.
Low eddy losses and high flux capacities at the operating frequency.
Driver transformer example: Magnetics Inc. core 52485-1D, winding two windings of 32 turns each, two in hand for a bifilar winding, using one as primary and one as secondary of course. For push pull, three windings of 32 turns each three in hand for a trifilar (one input and two outputs).

Opto coupler designs are even simpler if you dont work with transformers or inductors too often. Remember darlingtons are just two transistors, and a transistor can be driven fast or slow depending on how it is used. A transistor that is driven into saturation is slow but one that is not is fast. A darlington is made of two transistors therefore if it is driven into saturation it is slow but if not it can be fast. It's not the darlington configuration that slows it down it is the way it is driven.
Using a logic opto coupler might be a good idea too, along with a driver transistor or two.
The power supply for the opto coupler drive circuit can be easily made from a transformer, rectifiers, filter caps, and even a small voltage regulator...it isnt that expensive really especially if you only need one. If you need four (sometimes used with bridges) then you use a transformer with four windings, four bridge rectifiers, four filter caps, four small three terminal voltage regulators. The windings are isolated so they help to make up isolated power supplies that can be ground referenced to the source(s) of the MOSFET(s) to be switched.

 

[Ubergeek63]

It still goes back to what I said in post #5. It's much easier, faster, cheaper, less board space and fewer parts to use one of the dedicated gate driver chips!

high side drive chip:

https://www.electro-tech-online.com/custompdfs/2010/11/FAN7361.pdf

 

[crutschow]

Isn't it a simple case of using a DC-blocking capacitor in series with the primary?? Seems both affordable and practical...

A capacitor will indeed block any DC through the transformer primary, but that means the average DC output is also zero. Thus the plus and minus peak value of the output waveform will depend upon the duty cycle. For example, it would be equal plus and minus volts for a 50% duty-cycle pulse. This is, of course, not desired if you want a pulse that goes between zero and some positive voltage to drive a N-MOSFET gate.

One way around that would be to also capacitor-couple the output into a DC restore (clamp) circuit. This consists of a series capacitor and a diode (anode to ground). This will clamp the pulse negative peak at about -0.7V, independent of duty-cycle.

The reason to use a transformer instead of an opto-coupler for a high-side driver, is that the transformer can transfer power across the isolation to give a high voltage drive to the gate of the high-side driver MOSFET greater than the drain voltage. An opto-isolator can not do this without an additional source of voltage on the output.

 

[Ubergeek63]

The reason to use a transformer instead of an opto-coupler for a high-side driver, is that the transformer can transfer power across the isolation to give a high voltage drive to the gate of the high-side driver MOSFET greater than the drain voltage. An opto-isolator can not do this without an additional source of voltage on the output.

actually most circuits use boot strapping to get the rail... it costs next to nothing.

 

[crutschow]

actually most circuits use boot strapping to get the rail... it costs next to nothing.

If it's built into the driver chip, then certainly it's simple. If you have to generate it externally for an opto-isolator output, that makes it a little more complicated and costly.

 

[Ubergeek63]

If it's built into the driver chip, then certainly it's simple. If you have to generate it externally for an opto-isolator output, that makes it a little more complicated and costly.

it is always supported and never built in on HV high side switchers. you MIGHT see the diode inside but mostly not. remember boot strap circuits are just 3 cheap parts: a resistor, a diode, and a capacitor.