The H-Bridge module shown has a total of 20 pins (with 2 extra for thermistors), and is required to facilitate low frequency switching of polarity (~4 Hz only) in an array of electromagnets to enable a rotating magnetic field. I am untrained in electronics and having simply acquired this module on spec, now realise that, although I can see the basic circuitry of the H-Bridge, it appears unnecessarily sophisticated for the simple task at hand.

For example, there are 4 DC [--] input pins and 4 DC [+] output pins, and I was hoping that someone here could explain why any more than one of each is necessary. At the same time, there are 2 sets of pins labeled 'ph' to which a motor or electromagnet may be connected, so it seems that this H-Bridge is designed to allow the simultaneous operation of two circuits. However, there are only 4 gate pins (and ?4 source pins), so a fairly complicated MCU would be necessary to orchestrate such a circuit.

So, presuming that the basic H-Bridge involves 2 pairs of IGBTs, [T14 -ph2-ph1-T11] and [T12-ph1-ph2-T13] requiring only single DC-2 [--] and DC-1 [--] in-pins as ground respectively, is it sufficient to use two of the four DC [+] out-pins, and, for the simple switching of polarity required, to open and close these pairs by controlling the voltage across their respective G-S pins (~5V)? The module is rated for 75 A/650 V which is more than sufficient for the application using 12 V car battery to the electromagnets through this switch.

(By the way, it is late at night where I am, so if I'm a little tardy in replying to any answers or suggestions, please bear with me. My apologies if more clarity or information is required).

 

The images of two files are upside down, please correct them and re-upload again.

Edit :
Another method is to use the rotation function of pdf to rotates the direction of image.

 

The images of two files are upside down, please correct them and re-upload again.

Sorry--that's how I received them from my 'secretary'; it will take some time, a day or two probably, to rectify. Thanks for your patience.

 

For example, there are 4 DC [--] input pins and 4 DC [+] output pins

You put them in parallel to reduce any pin resistive voltage drop at high load currents.

there are 2 sets of pins labeled 'ph' to which a motor or electromagnet may be connected

Same as above.

is it sufficient to use two of the four DC [+] out-pins

Depends upon the current.
Anything over 5A, I would use all four pins.

for the simple switching of polarity required, to open and close these pairs by controlling the voltage across their respective G-S pins (~5V)?

Yes.
But note that the IGBT's need a gate-emitter voltage that is 15V which means the top IGBT's must have a gate voltage that is 15V above the supply voltage to fully turn on (see below where they measure the saturation voltage).
For this you can use a MOSFET H-bridge driver circuit with a built-in high-side supply.

 

You put them in parallel to reduce any pin resistive voltage drop at high load currents.
Same as above.
Depends upon the current.
Anything over 5A, I would use all four pins.
Yes.
But note that the IGBT's need a gate-emitter voltage that is 15V which means the top IGBT's must have a gate voltage that is 15V above the supply voltage to fully turn on (see below where they measure the saturation voltage).
For this you can use a MOSFET H-bridge driver circuit with a built-in high-side supply.

View attachment 181163

So, these pins are behaving as bottle-necks? In this case, if I have two terminals to a 12 V battery, [+] and [--], do you mean by putting these pins in parallel that I should use 4 wires from those terminals to the 4 respective in/out pins? Similarly 2 wires to the electromagnet terminals?

Thanks for reminding me about the need for [12 + 5]V. I had considered using an Arduino set-up; you're saying that it might be simpler to use the MOFSET driver?

[Also, the expected current will be 40-50 A, but only for short periods].

 

Connect Pins 5 , 6, 15 & 16 together and then run a large wire to the +12 Volt Battery
All four (+) pins wired in parallel can carry the full 75+ amps current.

Connect Pins 3 , 4, 17 & 18 together and then run a large wire to the -12 Volt Battery
All four (-) pins wired in parallel can carry the full 75+ amps current.

You need to research "gate driver for high side h-bridge"
T12 & T14 are on the "high Side" of the h-bridge.
They require a special Gate Driver.

Connect Pins 9 & 10, Output Phase 1, together and then run a large wire to the electromagnet
Connect Pins 11 & 12, Output Phase 2, together and then run a large wire to the electromagnet

Your notes imply ... S11 - S14 are not required
Use S11 - S14 for your Gate Drive circuity

Your notes imply ... two (2) Full H-Bridges "inside".
Nope, this is single Full H-Bridge package

 

Like mvas said "gate driver" such as https://www.ti.com/product/UCC23513/technicaldocuments I did not check the appropriateness for your application.

S and G allow access to V(GS) without the voltage drop associated with the high current lines. The part above is an isolated gate driver.

V(gs-threshold) is where the MOSFET starts to conduct, and it's a place to start the selection process. When it's low enough, the part may ne labeled a LOGIC Gate/LOGIC FET. You need to look at your operating point to see what V(gs) is required. V(gs) has a maximum voltage too.

You have 4 inputs to deal with FWD, REV, Brake and Coast. (Talking motors here).

In some applications of electromagnets (slip clutches used for tension control), it is desirable sometimes to turn the electromagnet fully off, so you reverse the electrical field to kill the residual magnetism.

Your MOSFETS will have drops too.

Analog/digital builds generally require keeping analog ground, digital ground, earth ground, high current ground all separated and if required, tied together at a single point.

 

Connect Pins 5 , 6, 15 & 16 together and then run a large wire to the +12 Volt Battery
All four (+) pins wired in parallel can carry the full 75+ amps current.

Connect Pins 3 , 4, 17 & 18 together and then run a large wire to the -12 Volt Battery
All four (-) pins wired in parallel can carry the full 75+ amps current.

You need to research "gate driver for high side h-bridge"
T12 & T14 are on the "high Side" of the h-bridge.
They require a special Gate Driver.

Connect Pins 9 & 10, Output Phase 1, together and then run a large wire to the electromagnet
Connect Pins 11 & 12, Output Phase 2, together and then run a large wire to the electromagnet

Your notes imply ... S11 - S14 are not required
Use S11 - S14 for your Gate Drive circuity

Your notes imply ... two (2) Full H-Bridges "inside".
Nope, this is single Full H-Bridge package

Thank you for this great information--exactly the sort of instruction I need. I now know where to start, and what to learn. I found a site at 'modular circuits' which seems about my level.

 

Like mvas said "gate driver" such as https://www.ti.com/product/UCC23513/technicaldocuments I did not check the appropriateness for your application.

S and G allow access to V(GS) without the voltage drop associated with the high current lines. The part above is an isolated gate driver.

V(gs-threshold) is where the MOSFET starts to conduct, and it's a place to start the selection process. When it's low enough, the part may ne labeled a LOGIC Gate/LOGIC FET. You need to look at your operating point to see what V(gs) is required. V(gs) has a maximum voltage too.

You have 4 inputs to deal with FWD, REV, Brake and Coast. (Talking motors here).

In some applications of electromagnets (slip clutches used for tension control), it is desirable sometimes to turn the electromagnet fully off, so you reverse the electrical field to kill the residual magnetism.

Your MOSFETS will have drops too.

Analog/digital builds generally require keeping analog ground, digital ground, earth ground, high current ground all separated and if required, tied together at a single point.

Thanks for the excellent info.; I'll check out the driver you suggest too, probably quite a lot later after a fair bit of study. I'm not a big fan of the internet, but forums like this are certainly an admirable feature of it. Thanks again.

 

I'll check out the driver you suggest too

I didn't actually "suggest" it except for being a "place to start".

 

I didn't actually "suggest" it except for being a "place to start".

My mistake--should have said 'mentioned'. I appreciate it though, especially your explanation for the 4 pins. Thanks again.

 

I'm still quite confused about certain points; to begin with, what are the S pins for? If these are for 'Source=Collector', then how is the Gate-Emitter voltage controlled; or because these are IGBTs (?with N-P-N-P+), is that control possible using Gate-Source voltage? As you see, I can't quite understand how attaching just one wire to the gate pin functions unless there is a circuit somewhere from the driver; to ground or Vcc I suppose.

Also, when you say that the gate-emitter voltage needs to be 15 V above the supply voltage [12V], how is this possible when the maximum range for that is +/--20V? Is this a matter of calling the supply voltage --[12V], and adding +3V to +5V at the gate, or vice versa, and how is this to be done? As shown in figure 2 in the data sheet for the H-Bridge, the implication seems to be that the G-E voltage should be in the range 8-17V, better towards the higher end around 15V. (Gate-Emitter threshold ~4V).

The following is an excerpt from 'modular circuits' article on H-Bridge Drivers; which I can follow generally. Any comments?


High-side P-MOS drivers
So far we’ve only talked about driving N-MOS devices and driving them on the low-side. Let’s consider now the high-side drivers, first for P-MOS devices:



This configuration present some complications: P-MOS transistors are open (non-conducting) when their gate is at close to the same potential as their source, and closed (conducting) when the gate is at a significantly lower potential, -5…-15V lower. This means that in order to completely turn off a high-side P-FET we’ll have to drive it’s gate as high as it’s source, which is connected to the power supply. To turn the FET on, we have to lower the gate voltage by 5…15V below Vbat.

All the drive circuits we’ve discussed before can be used for high-side P-FETs with the following change: you have to power the driver stage from the same voltage as the bridge is operating on, that is Vbat. That way, the high-level output voltage will be Vbat, which will turn the P-FET off properly, and the low-level output voltage will be 0, that is almost always enough to turn the FET on. (You might have problems with extremely low Vbat voltages, where you would have to drive the gate to a negative voltage to turn the FET on properly. This however is a rare enough case to ignore simply because high-current H-bridges usually operate at higher voltages and low-voltage H-bridges have low-enough currents that a small logic-level FET can be used in them that can be turned on by -Vbat.)

******

 

Also, when you say that the gate-emitter voltage needs to be 15 V above the supply voltage [12V], how is this possible when the maximum range for that is +/--20V? Is this a matter of calling the supply voltage --[12V], and adding +3V to +5V at the gate, or vice versa, and how is this to be done?

You're making the assumption that the gate is referenced to ground in that situation, but in fact it is referenced to the source voltage. When the mosfet turns on (in a high side situation) the source comes up to the drain voltage(ignoring the small drop from the mosfet, the Rds) So if you have 12V on the gate and 12V on the source You need to add the extra voltage to keep the gate on and making the mosfet conduct. Since your source is now at 12V and not ground or 0, adding another 12V to the gate doesn't make gate to source 24V but only 12V.

As to where the extra voltage comes from, when using a gate driver chip, Google for the term "bootstrap circuit".

 

I talked about an "isolated gate driver", so the "S" is the source pin and "G" is the gate pin. Vgs is relative to the source pin. lets add "form an isolated source" in this discussion.

 

I talked about an "isolated gate driver", so the "S" is the source pin and "G" is the gate pin. Vgs is relative to the source pin. lets add "form an isolated source" in this discussion.

Yes, when reading back over your answer and others here after studying a bit, things get a lot clearer. What was confusing me for a while, apart from the point that Shortbus clarified, is that BJTs regulate the C-E current through base-emitter voltage, while MOSFETs and evidently IGBTs apply gate-source voltages; and the source is analogous to the collector, right. The H-Bridge I'm using has IGBTs, and, given the position of the S pins, the MOSFET part, if it can be called that, must therefore be NMOS.

I've found a good article on isolated gate drivers at 'AnalogDialog'; and am also trying to understand bootstrapping in that context, of the appropriate gate-driver set-up for this H-Bridge. I haven't got to the part about ''forming an isolated source'' yet, but am looking forward to it with the usual mixture of trepidation, alright terror, and curiosity afflicting the technically unlearned and fundamentally lazy compelled to study something as goddam complicated as electronics.

A couple of things: In your opinion, from scant information, does it look like I'll also need a low-side gate driver? I'm thinking yea. And if these IGBTs are rated for 650V, Vce or Vds, and reversing through the diodes, then controlling 12V should be easy, right.

 

You're making the assumption that the gate is referenced to ground in that situation, but in fact it is referenced to the source voltage. When the mosfet turns on (in a high side situation) the source comes up to the drain voltage(ignoring the small drop from the mosfet, the Rds) So if you have 12V on the gate and 12V on the source You need to add the extra voltage to keep the gate on and making the mosfet conduct. Since your source is now at 12V and not ground or 0, adding another 12V to the gate doesn't make gate to source 24V but only 12V.

As to where the extra voltage comes from, when using a gate driver chip, Google for the term "bootstrap circuit".

Thanks for that--very helpful. I'm reading about bootstrapping, and can follow the argument generally--although it will take a lot more study before I can remember it well enough to think about productively. Is it correct to say that it is basically a way of using the same power supply, 12V in this case, for the gate driver and the H-Bridge; kind of juggling it using capacitors and a diode? If it is, it would make better sense to me.

 

FET's are voltage controlled devices that take very little current. Operation much like a tube or valve. It has very high input impeadance.

Like all devices, they have parasitics which makes them non-ideal. You have gate capacitance to overcome and you have leakage current to provide a path for. This you always see a low value resistor in series with the gate and a high value resistor from Vgs (leakage path);

Transistor have a "miller capacitance" . there is also a technique called a "speed-up capacitor". it's inherently a current controlled device with a Vbe drop that is about 0.7V and varies about -10 mV/dec C.

Don't know if this https://www.silabs.com/documents/public/application-notes/AN486.pdf could help understand the "bootstrap cirtcuit"

There are two fundamental things that I learned in school and it makes some stuff fall in place.

1. The voltage across a capacitor cannot change instantaneously.
2. The current through and inductor cannot change instantaneously.

it makes sense using the fundamental equations.

3. Parasitics matter.

Wierd examples: There is a Tektronics instrument where something like three 1K resistors are used in series. One 3K resistor won't work.
The inductive parasitics are higher for the 3K resistor. There are instances where SMT resisters are mounted on their side to minimize parasitics.

One closer to home is that waving a wire generates current. I had to deal with that one. The earth has a magnetic field right. A moving wire in a magnetic field generates a current, right? It was a few pA, but nevertheless a current. That was the realm in which I was measuring.

Circuits I worked on, you could not tough them with your bare hands and all flux must be removed. Components mounted on Teflon standoffs because of leakage currents.

In my set-up, I could put down a piece of paper and two probes and measure it's resistance.

When I changed a light bulb every 6 months, I had to "suit up" a little. Face shield, some body protection, no touching with fingers, remove foot traffic from the room etc. The bulb was pressurized to 10 atmospheres, so about 150 PSI. The bulb had a polarity, but nothing to prevent you from connecting it the wrong way. A fingerprint while operating on might cause an explosion.

Something like 22 V 42 A to operate, 40 kV to start. It was a nominal 1000 W arc lamp.

When learning, these nuances are left out, thus circuits get created without LED current limiting resistor. Transistors don;t get resistors in their bases. You don't get the extra two FET resistors.