Forums :: NEW! Web Resources :: Features :: Photo Gallery :: Vintage Radio Shows :: Archives
Support This Site: Contributors :: Advertise


It is currently Jul Sat 31, 2021 4:02 pm


All times are UTC [ DST ]





Post New Topic Post Reply  [ 11 posts ] 
Author Message
 Post subject: Flyback Driver Questions
PostPosted: Jun Tue 08, 2021 2:28 am 
Member
User avatar

Joined: Jan Thu 01, 1970 1:00 am
Posts: 4765
Location: Texas
I've reverse engineered a flyback driver. So I could compare its circuit with a standard one, I connected the two schematics. The one that I reverse engineered is Circuit A that produces inverting output. The 16 KHz 12 Volt pulse generator drives the gate of an IRF840 FET that powers the flyback primary.

The better flyback drivers use a totem pole or push-pull pair of transistors to drive the gate rather than connect the pulse generator (555 timer) directly to the gate. The transistor pair produces faster and cleaner switching.

All these drivers I've seen on the net are of the non-inverting arrangement as in circuit B. The circuit I reverse engineered is inverting and I want to know why. Non inverting output is 12 V 90% duty cycle. When inverted it becomes 10% duty cycle. Why would the designer go with the inverting topology?


Attachments:
Image8.jpg
Image8.jpg [ 98.1 KiB | Viewed 866 times ]
Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Tue 08, 2021 3:50 am 
Member
User avatar

Joined: May Wed 18, 2011 2:40 am
Posts: 5571
Location: Littleton, MA
Macrohenry wrote:
Why would the designer go with the inverting topology?

Minimize the RMS coil current to keep the flyback primary winding from burning up?

_________________
Steve Byan https://www.byan-roper.org/steve/steve-at-play/


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Tue 08, 2021 1:56 pm 
Member

Joined: Jan Thu 01, 1970 1:00 am
Posts: 5204
Location: Rochester NY USA
The 555 timer can't produce a duty cycle less than 50%, so needs to be inverted. It's a simple switch-mode controller with minimal extra parts required.

_________________
My web page: https://bit.ly/2rxq4qx


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Fri 11, 2021 4:42 pm 
Member
User avatar

Joined: Oct Mon 03, 2011 5:33 am
Posts: 603
Location: Tucson, Arizona
You can't drive a power MOSFET like an IRF840 with a 555 timer directly. The 555 isn't capable of providing enough peak current to drive the gate capacitance fast enough, either high or low. There needs to be a low impedance output driver like the two complementary bipolar transistors in the circuits below, or something similar like a monolithic gate drive IC. For a simple off-line flyback, I'd use a UCC38C45 or similar single quadrant PWM controller, which has an internal totem pole output driver that can deliver 1A peaks, a current limit circuit, an error amplifier and reference. This controller, and similar ones in the series, have been used for decades for this kind of application. The T.I. web site has copious application notes on its use.

The peak primary current in the transformer is proportional to the output load and duty cycle, and inversely proportional to the switching frequency and magnetizing inductance. For a discontinuous conduction mode flyback, the primary current will look like a sawtooth during the D state and with dead time during the 1-D state. For a continuous conduction mode flyback, the primary current will look like a trapezoid (a square wave with a superimposed sawtooth) during the D state and with dead time during the 1-D state. Note that the equations for duty cycle D are different between a DCM flyback and a CCM flyback.


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Sun 20, 2021 11:27 am 
Member

Joined: Jan Thu 16, 2020 12:29 am
Posts: 1401
Macrohenry wrote:
Why would the designer go with the inverting topology?


I cannot see that your question has been adequately answered yet.

If you look at the other option B, you will see that the output stage is merely a complimentary emitter follower. An emitter follower cannot give full rail to rail output swing. As the base voltage approaches the power supply, or common voltage, there is the Vbe drop of the base-emitter junction. Also the ability of the device to pull the output up to the max power rail voltage, or to zero, is diminished, because it requires that the collector voltage is higher than the base voltage to maintain the transistor in an emitter follower mode. This is why in audio output stages where you have that complimentary follower arrangement, bootstrapping is often employed, with R-C coupling from the output to the base of the upper device, to help the upper device, where the output is derived from the emitter turn on better, at least dynamically. And the emitter follower, as an output device, in a switching application, can never be driven into saturation with a low C-E voltage drop.

In a nutshell, if you derive the output from an output stage, from the emitters, the base drive voltage has the B-E drive voltage subtracted from it, this is what results in near unity voltage gain for the emitter follower. One way to fix this is to drive the transistor's B-E junction from the output of a transformer with an isolated windings, so the emitter voltage does not subtract from the drive voltage. (Look up a split driver transformer to see how this is done with audio amplifiers, I hope I am not boring you with the details)

However, if the output is taken from the collectors (inverting topology), and sufficient B-E drive is achieved for each output transistor, the output device can be driven into saturation, then nearly the full power supply output voltage is available to drive the load, in addition, the output stage has voltage gain (which the complimentary emitter follower does not). So, obviously, for a switiching application, like some flyback supply (not analog audio) it is a distinct advantage in energy efficiency, to drive the output device into saturation, which happens in circuit A. So the transistors will run cooler and need less heat-sinking, if any.

So if the output of this circuit A, is used to drive the gates of power mosfets, it will be better than circuit B.

But don't forget, if you are using that collector output configuration, to put at least a 10R to 47R resister in series with your mosfet gates to create a mild LPF at the gate inputs and limit the peak current on the switching transients and help protect the driver transistors and improve RF immunity. Power mosfets often have a substantial gate input capacitance.


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Sun 20, 2021 6:59 pm 
Member
User avatar

Joined: Jan Thu 01, 1970 1:00 am
Posts: 4765
Location: Texas
I love this kind of knowledge, thanks. It makes sense,thanks.

I have more appreciation for the designer. Maybe they knew exactly what they were doing. This makes me wonder why rhe emitter follower configuration is more commonly found in flyback driver circuits on the web.


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Mon 21, 2021 3:47 am 
Member

Joined: Jan Thu 01, 1970 1:00 am
Posts: 2725
Location: Saskatoon
The driver may not need to switch rail to rail. It only has to go above the MOSFET gate's turn-on voltage and below the turn-off voltage, and source or sink plenty of current in the process, in order to switch the MOSFET as fast as possible.


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Mon 21, 2021 4:25 am 
Member
User avatar

Joined: Mar Sun 22, 2020 5:56 am
Posts: 1893
Location: Arvada, CO, 80004
I enjoy reading this. It helps me learn.

_________________
Electronics are filled with smoke. It’s my job to put the smoke back in when they fail.
Cheers,
Jay


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Mon 21, 2021 12:16 pm 
Member

Joined: Jan Thu 16, 2020 12:29 am
Posts: 1401
Macrohenry wrote:
I love this kind of knowledge, thanks. It makes sense,thanks.

I have more appreciation for the designer. Maybe they knew exactly what they were doing. This makes me wonder why rhe emitter follower configuration is more commonly found in flyback driver circuits on the web.


For circuits with relatively low power levels, the thermal dissipation may be so low that the extra dissipation from an emitter follower design is neither here nor there. It becomes more important in power output stages where some Amps of collector current is flowing and it is desirable to have the transistors in saturated switching mode. In this mode, the B-E voltage drop, typically around 0.75V is greater than the C-E voltage drop, typically around 0.3V, which cannot happen with the emitter follower configuration.

That C-E vs B-E voltage relationship is one definition of saturation. There are others, since most transistors, even power transistors have a DC current gain of at least 15 to 20, normally much more, it is safe to assume, that the transistor will be in a saturated switching mode if its base current is at least 1/10 of its collector current.

So lets say you are switching some load, like a motor, relay or other power load and the collector current is some value I, you need a base current of I/10 to ensure saturation for practically any transistor you choose.

And, don't forget that if you are switching inductive loads, when the transistor switches off, the rapidly collapsing magnetic field results in a high voltage spike, because the inductor opposes a change in current, and the magnetic field energy ultimately exchanges for the potential energy of the electric field of the self capacitances (a resonant circuit), and those capacitances being fairly low, the voltage spike on account of the stored magnetic energy can be 10 times or more the supply voltage and exceed the transistor's voltage rating, so it pays to have a snubber diode across the inductive load.

This voltage spike is not really "Back emf" which you will see it commonly called all over the internet and occasionally in some textbooks. The term back emf was originally coined to describe the the reactive emf generated across an inductor's terminals when voltage was applied to the inductor's terminals (not disconnected when the field collapses, but watch people fight over this detail) and this emf opposes the applied voltage, hence the equation for the inductor V = - Ldi/dt, the minus sign showing the fact that the back emf opposes the applied voltage. The reason for the argument though, is that the inductor will always obey its equation, however you analyse its activity, whether you connect or disconnect it from some voltage source or apply alternating current to it. So when the inductor's current is attempted to be reduced, by reducing the applied voltage, the magnetic field reduces and attempts to maintain the flow of current. Or, if the applied voltage increases the inductor opposes this with the back emf, opposite to the applied voltage change. Disconnecting the inductor from the supply voltage source (or switching off the transistor driving it), is just an extreme case and the inductor attempts to maintain the current flow, and the only way this can occur is if the self capcitances are charged.


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Mon 21, 2021 5:31 pm 
Member

Joined: May Tue 30, 2006 4:46 pm
Posts: 10764
Location: Santa Rosa, CA
Quote:
The driver may not need to switch rail to rail. It only has to go above the MOSFET gate's turn-on voltage and below the turn-off voltage, and source or sink plenty of current in the process, in order to switch the MOSFET as fast as possible.


Power MOS Rds depends on gate drive voltage. Driving just above threshold will result in higher Rds and more heat dissipation. One needs to have enough current to charge the gate input capacitance quickly, and also a current sink to discharge the gate capacitance. Commercial gate drive ICs are designed to do these things.

https://www.vishay.com/docs/91070/sihf840.pdf

Vgs(th) is 4.0 Volts, but the Rds is specified at 10V drive. https://www.vishay.com/docs/91070/sihf840.pdf

Rich


Top
 Profile  
 
 Post subject: Re: Flyback Driver Questions
PostPosted: Jun Mon 21, 2021 10:17 pm 
Member

Joined: Jan Thu 16, 2020 12:29 am
Posts: 1401
Rich, W3HWJ wrote:
Quote:
The driver may not need to switch rail to rail. It only has to go above the MOSFET gate's turn-on voltage and below the turn-off voltage, and source or sink plenty of current in the process, in order to switch the MOSFET as fast as possible.


Power MOS Rds depends on gate drive voltage. Driving just above threshold will result in higher Rds and more heat dissipation. One needs to have enough current to charge the gate input capacitance quickly, and also a current sink to discharge the gate capacitance. Commercial gate drive ICs are designed to do these things.

https://www.vishay.com/docs/91070/sihf840.pdf

Vgs(th) is 4.0 Volts, but the Rds is specified at 10V drive. https://www.vishay.com/docs/91070/sihf840.pdf

Rich


This is one of the attractions of Logic level mosfets with a threshold voltage around 2.5V and by 5V they are fully enhanced. For many common garden standard mosfets, take the IRF540 for example, they are available with logic level inputs, and the part number changes to IRL540.

While power mosfet's, on the face of it, can appear more desirable because their Rds- on resistance can be extremely low, so for any drain current, their drain-source voltage drop is lower than a BJT which is stuck typically around 300 to 400mV or 1.2V for a Darlington, the large input capacitance of power mosfets is a very big disadvantage. And since, capacitance opposes a change in voltage, the higher the operating frequency, the more demands are placed on the mosfet's driver, and then it has to be a power stage in itself. This is why many mosfet drivers have been developed. Power mosfets are wonderful for DC switches where you are not relentlessly trying to charge and discharge the gate capacitance at a high frequency.

It is interesting that if these days somebody sees some new design done with BJT's in it, they first thing they seem to say is " I could have built a better one with mosfets" which often might not be the case because the designer stuck with BJT's for good reasons.

The power BJT with the lowest C-E voltage drop at saturation, is the obsolete Germainium type, when it is often under 200mV, a 2N174 being a good example, or the 2N1100 or the European ADZ12. This made these transistors pretty well ideal in DC/DC converters of the 1960's era converting 12V for portable tube equipment up to a few hundred volts.The circuits were self oscillating and super simple. The standard configuration was known as the Royer Oscillator, which is well described in the ARRL handbook.


Top
 Profile  
 
Post New Topic Post Reply  [ 11 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users and 6 guests



Search for:
Jump to:  


































Privacy Policy :: Powered by phpBB