Mosfet / Gate drive problems

[Rixen]

Hi all, new to the forum.

Im studying electronics, only about 8 months in now and decided to pick up some projects at home, so I had something to do in my spare time.

I'm trying to make a small induction heater, powered by a bipolar PSU that I made earlier, going to control it with a PIC, but im having problems with this, when I turn on the power, MOSFET's get very hot and if I look on my scope the signal is very distorted, even with no load applied.

Im also not quite sure what is the function of Cboot, I cant figure it out from the datasheet, value added was something I found on an article on google.

Were supposed to learn about Mosfets in like a year, but im just using them as switches so it was probably fine, well, except it's not really working I suppose.

MOSFET's are IRLS3034's
Gate driver is: IRS2153D
RVCC on the schematic is 1.8omhs, BC547, D1,L1 are not connected yet, wanted to make sure I had a good signal first.

+IN is +32v
-IN is -32

Here is the schematic I made with inspiration from other peoples circuits

Any help you can give would be greatly appreciated!

English is not my native language, so I hope you can bear with me, im trying my very best

 

[duke37]

Welcome to the forum, I am sure someone can help. Your english is better than most.

The two fets are in series, either the top one or the bottom one should be switched on, never both. The output will swing from zero to the supply voltage maybe 20V.

The gate must be a few volts above the source to turn on the fet, this driver can turn on the lower fet with the 16V supply. The top fet needs a few volts above its source (maybe 20V) which is not available from the 16V supply thus pin8 of the IC is driven by the capacitor from the output and with suitable clipping will swing it above the fet source. You could look up details of voltage multipliers.

If both fets are on, they will generate a lot of heat. Perhaps
1. The IC is not turning one off perhaps due to a poor ground line. P4 looks to be floating.
2. One of the fets is leaky.
3. You are running too fast so there is not sufficient time for the fet to turn off. Is the frequency 10.1MHz?

 

[Rixen]

Thanks, much appreciated!

P4 is going to ground, on the datasheet they have it attached to the negative rail, I saw other circuits where it went to ground, like mine, im not quite sure which is correct.
The frequency is 10.1 KHz

So my option is to find a way to increase the voltage on the HS gate? that needs some googling I think

 

[hevans1944]

According to the IR Application Note, the line you are supplying with -32 V should be at common potential. This is likely causing the low-side MOSFET to be driven into conduction regardless of whether LO on pin 5 is high or low.

 

[Rixen]

There's a while to go..

1.2vPP, the frequency is way off, oh and it's not exactly the square wave

On the plus side, it looks like both mosfets are atleast doing something.

According to the IR Application Note, the line you are supplying with -32 V should be at common potential. This is likely causing the low-side MOSFET to be driven into conduction regardless of whether LO on pin 5 is high or low.

Im not sure what you mean by this, both +In and -In share the same ground, sorry if I misunderstand.

 

[hevans1944]

... Im not sure what you mean by this, both +In and -In share the same ground, sorry if I misunderstand.

The gate drive signal for the low-side MOSFET is a positive-going, square-wave pulse from pin 5 (LO) to pin 4 (COM). It is intended that this pulse be applied between the gate and the source terminals of an N-channel MOSFET to turn the MOSFET on when the pulse is positive, and to turn the MOSFET off when the pulse is zero.

Adding a negative voltage (-IN) to the source terminal has the effect of increasing the gate voltage, in a positive direction, by the same amount. This means the gate-to-source voltage for the low-side MOSFET varies from a minimum of +32 V when pin 5 is zero with respect to common, to a maximum of +47.4 V when the pulse amplitude is +15.4 V (the internal zener clamp voltage) on pin 5 with respect to common. At no time does the gate-to-source voltage go to zero, therefore the low-side MOSFET is always on. Solution: ditch the -32 V supply and connect -IN to COM.

The only place you should see a saw-tooth waveform is on pin 3 (CT) with respect to pin 4 (COM). The high-side and low-side gate drive pulses are non-overlapping positive pulses with respect to the high-side and low-side MOSFET source terminals. The capacitor labeled Cboot makes this happen for the high-side pulses. The application note mentions a value of 0.1 μF for this capacitor.

 

[Rixen]

Ok, lets see if I got it.

I need to lower the gate voltage to make sure it hits zero, since that negative source goes positive on the gate, making it always on, no matter what the IC does basically?

Just to be sure of the solution here, put the -In (-32v? / -15.4v?) to COM and Low Side MOSFET source to ground?

I switched the CBOOT capacitor to a 100nF one.

Thanks for the help and for taking the time to explain these things, I really appreciate it

 

[Rixen]

Just want to add, I can do some measurements / screenshot the waveforms / pins, if it is required.

 

[hevans1944]

Ok, lets see if I got it.

I need to lower the gate voltage to make sure it hits zero, since that negative source goes positive on the gate, making it always on, no matter what the IC does basically? ...

What you need to do is connect the source of the low-side MOSFET to pin 4 (COM) instead of to a negative voltage referenced to that same pin 4 (COM).

... Just to be sure of the solution here, put the -In (-32v? / -15.4v?) to COM and Low Side MOSFET source to ground? ...

No. The "ground" goes away and is replaced by the -32 V power supply rail. That will "float" the IRS2153D on the -32 v rail, and it should work provided the Vcc supply common is capable of floating at -32 V. In other words, you need an isolated power supply for the Vcc supply and all those "ground" connections will be floating on and connected to the -32 V rail. This will cause problems with the transistor switch you are using to control the IRS2153D from an MCU. NPN transistors don't operate well with their collectors at a negative potential with respect to their emitters. So, depending on how fast you need to turn the IRS2153D on and off, perhaps an optical isolator could be substituted for the BC547 transistor. The diode half of an optical isolator is easy to drive with an MCU output.

... I switched the CBOOT capacitor to a 100nF one.

Please let us know how that worked out.

...Thanks for the help and for taking the time to explain these things, I really appreciate it

It appears you are trying to obtain bi-polar (positive and negative) outputs at the connection between the drain of the low-side MOSFET and the source of the high-side MOSFET. To accomplish this with proper voltage pulses on the MOSFET gate-to-source terminals, requires this circuit to "float" the IRS2153D and its Vcc supply on the negative power supply rail. The output will then swing between the IRS2153D "common," or -32 V, and the positive rail, or +32 V.

The other side of the induction coil was connected to the -32 V rail in your original circuit, but you should now connect it to the "common" of the ±32 V power supply to take advantage of the full ±32 V swing of the output.

There is an alternative to floating the Vcc power supply: float the ±32 V supply instead. You would need to isolate the ±32 V "ground" because it is not necessary for driving the induction coil. The important thing is the source terminal of the low-side MOSFET must be connected to pin 4 (COM) on the IRS2153D. So, your choice, either isolate the ±32 V supply so its negative rail can be tied to the Vcc power supply common, or isolate the Vcc power supply so its common can be tied to the negative power supply rail. The result will be the same, but the "grounds" will be different.

Please keep us posted on how this all works out. Good luck in your studies! It looks like you have a lot of nice "toys" to play with! Yes, electronics is supposed to be fun. I have been totally amazed that someone has always wanted to pay me real money to have fun for the past fifty-something years. Now, if I could just convince Uncle Sam and the Social Security Administration to continue that tradition...

Hop

 

[Rixen]

Alright

What you need to do is connect the source of the low-side MOSFET to pin 4 (COM) instead of to a negative voltage referenced to that same pin 4 (COM).

Ok, so this is what I think I have to do when I read this, but it doesn't make a whole lot of sense to me, is it right?

No. The "ground" goes away and is replaced by the -32 V power supply rail. That will "float" the IRS2153D on the -32 v rail, and it should work provided the Vcc supply common is capable of floating at -32 V. In other words, you need an isolated power supply for the Vcc supply and all those "ground" connections will be floating on and connected to the -32 V rail. This will cause problems with the transistor switch you are using to control the IRS2153D from an MCU. NPN transistors don't operate well with their collectors at a negative potential with respect to their emitters. So, depending on how fast you need to turn the IRS2153D on and off, perhaps an optical isolator could be substituted for the BC547 transistor. The diode half of an optical isolator is easy to drive with an MCU output.

Ground is gone and replaced with the -32v rail, i've added the LM317 which im using to supply the VCC of the IC, this is the updated diagram, and pictures of the breadboard, I turned this on and RVCC instantly burned out, but I think I might just have fried the IC tinkering so much, got some spares though. as for the MCU control, yeah it looks like that NPN has to go, damn.

It appears you are trying to obtain bi-polar (positive and negative) outputs at the connection between the drain of the low-side MOSFET and the source of the high-side MOSFET. To accomplish this with proper voltage pulses on the MOSFET gate-to-source terminals, requires this circuit to "float" the IRS2153D and its Vcc supply on the negative power supply rail. The output will then swing between the IRS2153D "common," or -32 V, and the positive rail, or +32 V.

This is exactly what im trying to do and what I think I have to do, to get that good looking square wave and the induction that im after.

The other side of the induction coil was connected to the -32 V rail in your original circuit, but you should now connect it to the "common" of the ±32 V power supply to take advantage of the full ±32 V swing of the output.

I think this is something I dont understand, where is the common exactly? in my head I want it to be ground, or perhaps im just calling it ground, I could upload the diagram for my PSU if we need it, but I dont think that's it

Please keep us posted on how this all works out. Good luck in your studies! It looks like you have a lot of nice "toys" to play with! Yes, electronics is supposed to be fun. I have been totally amazed that someone has always wanted to pay me real money to have fun for the past fifty-something years. Now, if I could just convince Uncle Sam and the Social Security Administration to continue that tradition...

Will do, and thanks!

Yeah I have a few toys, my friends, old colleagues and well.. most people that I know of, prefer to spend their money on expensive clothes, getting drunk and usually their bank accounts are in the red after a week or two every month, I never drink so, while they do that, my own account is building up nicely, so I can occasionally afford nice things

Edit1: I was going to upload a picture of the breadboard, but the site censures it for some reason.

 

[Alec_t]

I was going to upload a picture of the breadboard, but the site censures it for some reason.

File too big? Keep it below 500k (take pity on folk with limited/poor broadband ). Wrong format? Use PNG or GIF or JPG.

 

[Rixen]

File too big? Keep it below 500k (take pity on folk with limited/poor broadband ). Wrong format? Use PNG or GIF or JPG.

Hmm, dont think it's too big, nor is it the wrong format.. it's the exact same as every other picture I've posted so far.

Edit: turns out it was some problem with tinypic.com, uploaded on another site, heres the picture of the breadboard:

 

[hevans1944]

So what is the common rail of the +16 V power supply connected to? As I mentioned in an earlier post, this power supply must be "floated" on the -32 V rail, meaning the +16 v power supply common must be connected to the -32 V rail. Clearly this cannot be done if the +16 V power supply shares a common connection with the ±32 V power supply because that would short out the -32 V rail!

The IRS2153D needs to have a voltage applied between Vcc and COM that is provided from a power supply whose output is current-limited and less than the maximum Vclamp internal zener voltage, or 16.8 V. See Note 1 on page 2 of this Application Note:

This IC contains a zener clamp structure between the chip VCC and COM which has a nominal breakdown voltage of 15.4 V. Please note that this supply pin should not be driven by a DC, low impedance power source greater than the VCLAMP specified in the Electrical Characteristics section.

If your +16 V power supply still has its common connected to the common of the ±32 V power supply, you are applying 48 V between pin 1 (Vcc) (through the 120 Ω current-limiting resistor, Rvcc) and pin 4 (COM) which is now connected to -32 V. The internal 15.4 V zener diode is connected between pin 1 and pin 4 and the current through the zener diode should be limited to 5 mA by appropriate choice of Rvcc. So, for a moment after turning the power on, Rvcc conducts a current of 272 mA [(48V - 15.4V) / 120 Ω] instead of 5 mA. And, for a moment, it dissipates 8.86 watts before going up in smoke. The zener and the internal electronics of the IRS2153D melt down waaaay before you notice that Rvcc is a crispy critter. I think you have NOT floated the +16 V power supply.

... something I dont understand, where is the common exactly? in my head I want it to be ground, or perhaps im just calling it ground

"Common" is a node in a circuit to which voltage measurements are referenced. It does not necessarily mean common is connected to "earth" or power-line ground, although this is general practice.

You appear to be using two power supplies: a 15 VDC power supply providing +15 V with respect to its common; and dual 32 VDC supplies providing both +32 V and -32 V with respect to its common. It appears that you have connected the commons of these two power supplies together and perhaps also to power-line "ground" or "earth" ground. So far, so good. Nothing wrong with that except it won't work with this circuit.

There is nothing "magic" about connecting a common to (earth) ground. It is often not even necessary to do so. However, in the circuit you are trying to build, there are two commons to consider. Well, three, actually if you consider the COM on pin 4.

The voltage you use to power the electronics inside the IRS2153D is applied between pin 1 (Vcc) and pin 4 (COM). The positive-going MOSFET gate pulses this integrated circuit provides on pin 5 (LO) are referenced to pin 4 (COM), but the positive-going MOSFET gate pulses this integrated circuit provides on pin 7 (HO) are referenced to pin 6 (VS). The reason for this is the gate pulses must be provided between the gate and the source terminal of each of the two MOSFETs. However, only the low-side MOSFET has its source connected to pin 4 (COM). The high-side MOSFET has its source connected to pin 6 (VS) and also to the drain of the low-side MOSFET. Therefore the gate pulses for the high-side MOSFET are measured between pin 7 (HO) and pin 6 (VS), so you could consider pin 6 (VS) to be the "common" for pin 7 (HO) for pulse-measurement purposes. Be very careful trying to measure this pulse with your o'scope because it's "common" lead is probably connected to power-line ground. You could "float" the 'scope with an isolation transformer to make the measurement, but that is another can of worms.

In the original application note for this circuit, the "common" connection on pin 4 (COM) is made to a -AC rectified line and the Vcc connection on pin 1 (Vcc) is made to a +AC rectified line through Rvcc. The load, L and RL, is connected between the -AC rail and the common connection at the drain of the low-side MOSFET and the source of the high-side MOSFET. THERE IS NOT A GROUND ANYWHERE IN SIGHT! The integrated circuit IRS2153D "floats" between the +AC and -AC rails. The voltage across the load varies from zero (when the low-side MOSFET is conducting) to 64 V (when the high-side MOSFET is conducting).

So now along comes @Rixen who needs to "control" the IRS2153D with an MCU, and what better way to do this than to insert an NPN transistor between pin 3 (CT) and pin 4 (COM)? So far, so good. But wait! The transistor base and emitter must be referenced to the MCU "ground" for this to work! So let's tie all those "grounds" together!
Woops! That won't work because it shorts out the -32 V power supply rail which is now also connected to pin 4 (COM).

There are several solutions to this dilemma, but perhaps the simplest is to "float" the DC power supply providing power to the IRS2153D. You could do this with a battery (for test purposes) if for some reason your 15 V PSU has a hard-wired "earth" ground connection for its "common" lead. This won't solve the problem of how to control the IRS2153D from pin 3 (CT) though.

You could "float" the ±32 V power supply by disconnecting and insulating its "common" connection, although that may not be possible either if this power supply's "common" also has a hard-wired "earth" ground connection. However, if you could float the common connection on the ±32 V power supply, then you could connect pin 4 (COM) to an "earth" ground (which means the -32 V rail would also be connected to "earth" ground) and then connect the NPN transistor per your original circuit diagram. This is the method I would use, even if I had to construct the power supply from scratch using a transformer with a bridge rectifier attached to its 32 VAC secondary. After all, you don't need much voltage regulation for an induction heating coil.

On second thought, I would use an optical isolator in place of the transistor to turn the IRS2153D on and off. The thought of "accidentally" applying 32 V (or more) to the MCU is pretty frightening.

Hop

 

[Rixen]

So what is the common rail of the +16 V power supply connected to

Here are all the diagrams for the PSU I made so far, I wasnt planning on keeping that 16v psu for the "final product".

And.. once again I want to thank you for your time, I hope im not being too retarded, im trying hard to make this thing work

1:Mains IN

2OS

3:NEG

4: 5v/12v

5. MOSFET Driver IC PSU

If your +16 V power supply still has its common connected to the common of the ±32 V power supply, you are applying 48 V between pin 1 (Vcc) (through the 120 Ω current-limiting resistor, Rvcc) and pin 4 (COM) which is now connected to -32 V. The internal 15.4 V zener diode is connected between pin 1 and pin 4 and the current through the zener diode should be limited to 5 mA by appropriate choice of Rvcc. So, for a moment after turning the power on, Rvcc conducts a current of 272 mA [(48V - 15.4V) / 120 Ω] instead of 5 mA. And, for a moment, it dissipates 8.86 watts before going up in smoke. The zener and the internal electronics of the IRS2153D melt down waaaay before you notice that Rvcc is a crispy critter. I think you have NOT floated the +16 V power supply.

If I add a resistor going from COM to -32v, say 6.8k at 5mA etc, would that not cause it to "float"? otherwise im not entirely sure what that term means

I tried.. just to restore some of my sanity, removing the MOSFET's for now, just to see if I could atleast make that IC do something, here's the results of that.

PIN2:

PIN3:

PIN5:

PIN6:

PIN7:

PIN8:

That's atleast positive I suppose.

If I ever get this thing working, do you think it would be possible to replace RT with a digipot? something like MCP414x series.

 

[hevans1944]

Let's back up to the beginning and your original circuit. As I suspected, all of your power rails share a common "ground" connection, i.e., the center-tap of the power transformer. This will not work for reasons already discussed!

The original problem you discovered with this circuit was the MOSFETs both got very hot. The reason for this was the low-side MOSFET source was tied to the -32 V rail. This effectively biased the gate of this MOSFET positive by 32 V and the positive pulses coming out of pin 5 (LO) just added to that. The result was the low-side MOSFET was turned on all the time. The high-side MOSFET also has positive-going pulses applied to its gate whenever the low-side MOSFET is turned off. Problem is, the low-side MOSFET never turns off! This places the -32 V rail on the source terminal of the high-side MOSFET too, effectively biasing the high-side MOSFET gate to +32 V (with respect its source terminal) plus whatever amplitude the positive pulse is. The result was the high-side MOSFET was also turned on all the time. With both MOSFETs turned on simultaneously, this creates a "dead short" across the +32 V to -32 V rails. No wonder the MOSFETs got hot!

The only thing that might have saved the MOSFETs from self-destruction is the Rds(on) value is specified to be only 1.7 mΩ with 10 V gate-to-source drive and 195 A drain current! That's 64 watts dissipation. Rugged little beastie. Your little power supply transformer apparently is not capable of providing enough current to destroy the MOSFET when the MOSFET is fully conducting. But that doesn't mean you are home free. The maximum gate-to-source voltage is specified to be only ±20 V before you punch through the oxide layer and destroy the gate. You are hammering it with 32 V! You may get away with that a few times, but it is not good engineering practice. The positive pulses applied by the IRS2153D to the MOSFET gates with respect to their source terminals should be the same amplitude as the zener voltage, about 15 V.

So, how can you fix this problem? Well, first the low-side MOSFET source must be connected to pin 4 (COM). This is true no matter how many other bells and whistles and circuit modifications you make from now on. For now, DO NOT CONNECT -32 V rail to ANYTHING! The original circuit should now work with a peak-to-peak output of 32 V at the connections between the low-side MOSFET drain, the high-side MOSFET source, and the load. The other side of the load connects to the low-side MOSFET source, which is now back to being at ground potential, per your original circuit with the -32 V rail not being used.

But, wait, Hop! I want to drive the load with bi-polar ±32 V! I want to turn the IRS2153D on and off with a transistor controlled by my MCU! I want to vary the frequency with a digital potentiometer replacing the 6.8 kΩ timing resistor! I want my ground, too! Well, you can have most of those things, but the -32 V rail has to be at ground potential. The only way I know to do that is to float the ±32 V power supply so connecting the -32 V rail to common doesn't short out the rail.

If you want bi-polar power applied to the load, one side of the load would connect to the "common" of the ±32 V supply. The transformer, full-wave bridge rectifier, filter capacitors, and ±32 V voltage regulator circuits must not share any connections with the rest of your circuit, and most especially the "common" for this power supply must not connect to the "common" of the IRS2153D circuitry. The only other thing that shares the transformer center-tap "common" is your induction heater load. Since the -32V rail is connected to the "common" of the rest of the circuit, the center-tap on the transformer will be 32 V positive with respect to that common. You would need to make sure the transformer insulation can handle that. When the low-side MOSFET is "on" that will apply -32 V to the load with respect to the transformer center tap. When the high-side MOSFET is "on" that will apply +32 V to the load with respect to the transformer center tap. You must not "ground" the center-tap of the transformer because that would short out the -32 V rail. This means you either need two transformers, or a transformer with two secondary windings: one center-tapped winding for the ±32 V supply, and another lower-voltage winding for the low-voltage power supply.

Floating is a simple concept: there must be no electrical continuity between the "floating" circuit and earth ground. Imagine the circuit placed in a balloon floating high above the Earth. Easier said than done because of leakage resistance and leakage capacitance, but usually a transformer provides enough isolation between the power-line ground and the secondary winding(s) to allow circuits attached to the secondary winding(s) to be floated up to several kilohertz and several hundred volts before the parasitic capacitance between primary and secondary windings becomes a significant factor.

When using your Rigol 'scope to measure the high-side MOSFET gate pulses, you can "float" the scope with an isolation transformer or (if so equipped) operate it from internal batteries. This would allow you to connect the probe common to pin 6 (VS) and the probe tip to pin 7 (HO) to measure the gate pulse applied to the high-side MOSFET. Bear in mind that this "floats" the entire 'scope, so whatever potential (with respect to "ground") the common lead of the probe is at will also be on the chassis and controls of the 'scope. Make sure there is no way the 'scope chassis can come in contact with a real "earth ground" while you are floating the 'scope to take a measurement. And, also, if you "float" one probe common you have floated both probes' commons. The two probe commons are connected together inside the 'scope, and probably connected also to the power-line (green wire) ground inside the scope.

Some daring souls will just use a "cheater" plug to disable that green-wire ground, relying on the 'scope's internal transformer for isolation from the power-line ground. But who knows what's inside a 'scope these days? An external isolation transformer is a safer method. If it comes equipped with a 3-prong grounding socket, with the green-wire grounding wire connected to the green-wire in the power cord, you may have to use a "cheater" plug to defeat the power-line ground connection.

Your 'scope traces look good. The little "glitches" at the square-wave transitions are there because the gate drive signals do not overlap. You can see this if you trigger the 'scope using the rising edge of the low-side gate signal on one channel and display the high-side gate signal (with AC coupling to remove the DC offset) simultaneously on the other channel.

Is it all clear now? Are you ready to re-connect the MOSFETs?

Hop

 

[Rixen]

The only thing that might have saved the MOSFETs from self-destruction is the Rds(on) value is specified to be only 1.7 mΩ with 10 V gate-to-source drive and 195 A drain current! That's 64 watts dissipation. Rugged little beastie. Your little power supply transformer apparently is not capable of providing enough current to destroy the MOSFET when the MOSFET is fully conducting. But that doesn't mean you are home free. The maximum gate-to-source voltage is specified to be only ±20 V before you punch through the oxide layer and destroy the gate. You are hammering it with 32 V! You may get away with that a few times, but it is not good engineering practice. The positive pulses applied by the IRS2153D to the MOSFET gates with respect to their source terminals should be the same amplitude as the zener voltage, about 15 V

Well, it looks like the MOSFET's are still alive, but they did manage to melt part of my breadboard, or well, the bridge rectifier IC melted my breadboard, i've got plenty of those FETS to test with, but i'll try look for some replacements in the mean time.

So, how can you fix this problem? Well, first the low-side MOSFET source must be connected to pin 4 (COM). This is true no matter how many other bells and whistles and circuit modifications you make from now on. For now, DO NOT CONNECT -32 V rail to ANYTHING! The original circuit should now work with a peak-to-peak output of 32 V at the connections between the low-side MOSFET drain, the high-side MOSFET source, and the load. The other side of the load connects to the low-side MOSFET source, which is now back to being at ground potential, per your original circuit with the -32 V rail not being used.

Okay, I went back to the original circuit, removed the -32v rail completely, took the low side MOSFET source and connected it straight to COM, there was nothing on the output, so I thought, hmm, well there's no smoke or heat either, I then measured the pins, and there was 32v DC on every pin of the IC, thought the MOSFET's might have actually died and probably destroyed that IC, replaced it all but with the same result.

Just for absolute crystal clarity, CVCC and CT are now going to the "ground" line as per the original circuit? (COM and CT no longer connected ofcourse) well I tried both putting them to "ground" and leaving them floating, with no difference, out of desperation I suppose.

But, wait, Hop! I want to drive the load with bi-polar ±32 V! I want to turn the IRS2153D on and off with a transistor controlled by my MCU! I want to vary the frequency with a digital potentiometer replacing the 6.8 kΩ timing resistor! I want my ground, too! Well, you can have most of those things, but the -32 V rail has to be at ground potential. The only way I know to do that is to float the ±32 V power supply so connecting the -32 V rail to common doesn't short out the rail.

Yeah I think i'll forget all the extras and niceties and bi-polar PSU atleast for now anyway, my focus has got to be on the thing actually working.

Floating is a simple concept: there must be no electrical continuity between the "floating" circuit and earth ground. Imagine the circuit placed in a balloon floating high above the Earth. Easier said than done because of leakage resistance and leakage capacitance, but usually a transformer provides enough isolation between the power-line ground and the secondary winding(s) to allow circuits attached to the secondary winding(s) to be floated up to several kilohertz and several hundred volts before the parasitic capacitance between primary and secondary windings becomes a significant factor.

This makes sense, but damn, it looks like I will eventually need a new transformer.

Is it all clear now? Are you ready to re-connect the MOSFETs?

Damn I hate to bring all those bad news

 

[hevans1944]

Okay, I went back to the original circuit, removed the -32v rail completely, took the low side MOSFET source and connected it straight to COM, there was nothing on the output, so I thought, hmm, well there's no smoke or heat either, I then measured the pins, and there was 32v DC on every pin of the IC, thought the MOSFET's might have actually died and probably destroyed that IC, replaced it all but with the same result.

Just for absolute crystal clarity, CVCC and CT are now going to the "ground" line as per the original circuit? (COM and CT no longer connected ofcourse) well I tried both putting them to "ground" and leaving them floating, with no difference, out of desperation I suppose.

There should not be +32 V on ANY of the IC pins. The only place +32 V is applied is to the drain of the high-side MOSFET.

Pin 1 (Vcc)connects to the junction of Rvcc=120 Ω and the 22 μF capacitor, pin 1 (Vcc) being powered with +16 V from the LM317 through current-limiting resistor Rvcc.

Pin 4 (COM) goes to your power supply common (the center-tap of your power transformer). Call this common connection "ground" and it should also be connected to power-line ground.

Pin 3 (CT) is connected to a timing capacitor Ct=10 nF and the other end of Ct is connected to pin 4 (COM).

Make sure Rt=6.8 kΩ is connected between pin 2 (RT) and pin 3 (CT).

Make sure Cboot=100 nF is connected between pin 8 (VB) and pin 6 (VS).

Connect the two MOSFETs in series, the high-side MOSFET source connecting to the low-side MOSFET drain.

The drain of the high-side MOSFET connects to +32 V.

The source of the low-side MOSFET connects to common.

The gate of the high-side MOSFET connects to pin 7 (HO).

The gate of the low-side MOSFET connects to pin 5 (LO).

After verifying everything is wired correctly, you should see a square wave on the drain of the low-side MOSFET with an amplitude that varies from +32 V to zero. You may need to place some sort of resistive and/or inductive load across between the drain and the source of the low-side MOSFET.

 

[Rixen]

It worked.. it worked.. it worked!

Holy crap what a relief, thank you so much, I cant get my arms down right now!

But what the hell! I even tried this connection earlier! with no load however, but I did try it!

Channel 1 is after a small inductor and resistor that I added

Channel 2 is the high MOSFET side source, low MOSFET side drain, and VS point.

 

[hevans1944]

Note the ringing on the waveform, which occurs during the transition period when both MOSFETs are off. I would have expected this ringing to be damped by the body diode in the low-side MOSFET, but maybe you need an external diode across the inductor and perhaps one across the drain-to-source of the low-side MOSFET too. With a large enough inductance and current, the ringing amplitude could reach amplitudes large enough to damage the MOSFETs.

Looks like you are well on the way to heating something up with your induction coil. Invest in a larger transformer for the ±32 V supply, maybe increase the voltage too. Check the datasheets for the MOSFETs to see how high you can go. The IRS2153D will handle up to 600 V according to its data sheet. Be careful! The higher the power you are switching, the more likely that errors will have spectacular (perhaps lethal) results. Good luck in your studies of power engineering... it is a fascinating and rewarding endeavor.

Hop

 

[Rixen]

Yep, i'll start looking for a transformer and since the majority of the current power supply is rated for higher voltages than this, I dont expect any problem raising the voltages too!

Any suggestions for the diodes? the 1N's may be too slow for this, perhaps schottky's?

And maybe I should consider, atleast a used isolation transformer while messing around with this, I got the same feeling, this may eventually become abit dangerous..