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H-Bridge Shorting

Suraj143

Active Member
I have attached the left half of my hbridge inverter driving stage.The problem is while the PIC initialise the H-bridge pins (~10uS), the 12V rail gets shorted. How to overcome this? I use PIC12F675 with normal GPIO pins.
 

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danadak

Well-Known Member
Most Helpful Member
You have no R's from T1 T5 base to ground. When PIC starts up
(check datasheet) it may leave bases floating and cause both
drivers to turn on, which drops supply to PIC so it goes nuts as
well. PIC starts up with outputs tristated ? Check datasheet.

A 10K to ground on both T1 and T5 bases might help. That also helps
prevent turning on T1 and T5 due to leakage.


Regards, Dana.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
As the two halves work the opposite ways, both will either be ON together, or OFF together, and it's the ON together which is your issue.

10K's on the bases as suggested may help, but I would suggest adding an extra inverter on one of the halves, another NPN (or small FET - like a 2N7000) before T1 and T5, as well as the 10K's. You will have to alter your programming accordingly.

Below is part of an experimental test board I've been playing with, which is both a boost and/or buck converter, PL1 and PL2 are three pin headers so you can select the extra inverter or not for each half. Notice I also choose 10K for the resistors from gate to Vss on the first FET's, along with 10K pull-ups on their drains. This allowed me to select inverting on either, both, or neither of the two halves.

Rather like your issue, I was concerned as to what happens as the PIC initially powers up.

Buck_Boost.png
 

Suraj143

Active Member
Ok many thanks.

Adding 10K on both T1 &T5 Bases to ground wouldn't help.It will still shorts the output.

Am I correct .!!
 

danadak

Well-Known Member
Most Helpful Member
"Normally" if this is a PWM driven bridge one sets up PWM dead band
settings to insure there is no overlap -

Here you can see the dead band is set to 10 clocks and observe that
the PH1 and PH2 phases do not overlap by that amount.

1651320701897.png




Regards, Dana.
 

Suraj143

Active Member
@ Dana
Thanks, Whats that software? Never used that before.I actually adding a push button to start the inverter.Until presses the inverter will be in OFF position.I think I have to add another transistor stage before the T1 & T5...!

What about replacing T1 &T5 with PNP ones and add Base to GND with 10Ks?
 
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ronsimpson

Well-Known Member
Most Helpful Member
T8 bottom FET turns on/off nicely. T4 top FET does not turn on well. It can only pull up to about 6 volts. The turn on voltage is not good. Look at Nigel's circuit in #3. Hi is using a N and also a P-MOSFET. The top driver is very different.
1651324289461.png
 

Nigel Goodwin

Super Moderator
Most Helpful Member
@ Dana
Thanks, Whats that software? Never used that before.I actually adding a push button to start the inverter.Until presses the inverter will be in OFF position.I think I have to add another transistor stage before the T1 & T5...!

What about replacing T1 &T5 with PNP ones and add Base to GND with 10Ks?
You're not making things any better - read my previous post - you need to change ONE half of the circuit, not both. As it stands, and also with your proposed modification of BOTH halves, there is a short period where BOTH transistors are turned ON, with a single inverter stage on one half one will be ON, and the other OFF.

The only problem is the VERY short time as the PIC initialises, once that has passed you have full control of the I/O pins and what they are doing - so the very first thing your program needs to do is turn both transistors OFF.

If you like, you could add the three pin headers as I did on mine, so you can then configure it in whatever way you wish.

You might also like to change your antique PIC (nice as it was in the past) for something rather more modern - like the 16F18313 - which is still 8 pin, but has far more facilities than the 675, including PWM hardware with dead-band etc. for exactly your purpose. The 12F1840 is another 8 pin device that includes dead-band PWM, and the 16F18313 and 12F1840 have become my 8 pin PIC's of choice.

Incidentally, one version of my partial schematic above had a socket to connect to any PIC board, the other had an 8 pin DIL socket on board instead, which I used a 12F1840 in. The 12F1840 version also had a serial port connection to an FTDI232 board to allow serial communication from a PC - so you could adjust it's settings from a PC, which it then stored in data EEPROM.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
T8 bottom FET turns on/off nicely. T4 top FET does not turn on well. It can only pull up to about 6 volts. The turn on voltage is not good. Look at Nigel's circuit in #3. Hi is using a N and also a P-MOSFET. The top driver is very different.

You can get some nice little 8 pin drivers, IR2112? (or something like that) - I haven't got the details here, only at work, but there are a few slightly different versions. They provide dead-band in hardware, plus boot-strapping for driving the top FET to cure the low drive issue.
 

danadak

Well-Known Member
Most Helpful Member
The chip I was referring to is a PSOC 5LP, it has that capability onchip as well as these
onchip -

1651330894569.png


The compiler and IDE (PSOC Creator) are free, very useful board with debug is $ 15.

1651331072363.png


Just as an aside your chip only has CCP on it, there are PICs with ECCP which also have a
PWM with dead band capability.



Some ref material to compute the Gate R for the MOSFETS -

https://toshiba.semicon-storage.com/info/docget.jsp?did=59460

https://www.kynix.com/Blog/Selection-of-Drive-Resistor-MOSFET-Gate-Drive.html


How to design the gate resistor value?

This is the datasheet of the driver IC I am working on (LM5112). Following is the application diagram of the module. Basically this is the GATE driver circuit for the MOSFET with PDM signal as the
electronics.stackexchange.com
electronics.stackexchange.com

https://www.infineon.com/dgdl/Infin...N.pdf?fileId=5546d462518ffd8501523ee694b74f18

Regards, Dana.
 
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Suraj143

Active Member
You're not making things any better - read my previous post - you need to change ONE half of the circuit, not both. As it stands, and also with your proposed modification of BOTH halves, there is a short period where BOTH transistors are turned ON, with a single inverter stage on one half one will be ON, and the other OFF.
Thanks.Now I got it.Added another transistor to the lower side driver.

Or just use a PMosfet in high side without adding an inverterting transistor.
 

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Suraj143

Active Member
T8 bottom FET turns on/off nicely. T4 top FET does not turn on well. It can only pull up to about 6 volts. The turn on voltage is not good. Look at Nigel's circuit in #3. Hi is using a N and also a P-MOSFET. The top driver is very different.
Now I'm a bit confused.I cannot understand this "It can pull up to about 6 volts". Can you explain this?

I'm refering the EGS002 module design.It uses 4 N-channel FETs in their H-bridge design.The High side Gate driving via IR2110S (But inside it also has a FET totempole arrangment)
 

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eTech

Well-Known Member
Why not just use a 74HC132 NAND gate to drive the BJT totem pole?
One input of the NAND to VCC (5+) and the other to the PIC output. It would invert the output (could change this is software) but it would solve the tri-state boot issue. And those other BJT's wouldn't be needed.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Now I'm a bit confused.I cannot understand this "It can pull up to about 6 volts". Can you explain this?

You're using the two devices in completely different modes - the bottom FET is common source, so is a digital switch - either ON or OFF. But your top FET is common drain, so works as an analogue amplifier, and will get very hot. I've not checked the spec. of the devices, but presumably the previously mentioned 6V is what it takes to turn the FET fully ON? - so with the gate at 12V the source will still be at 6V, because it has to be 6V lower than the gate.

The bottom transistor feeds 12V to the gate, and the source is 12V lower, so it's switched ON nice and fully.

It's exactly the same with both FET's and bi-polar, for switching you want the load to be in the drain/collector, not the source/emitter.

I'm refering the EGS002 module design.It uses 4 N-channel FETs in their H-bridge design.The High side Gate driving via IR2110S (But inside it also has a FET totempole arrangment)
The IR2110 (Like the IR2112 I mentioned previously) has boot-strapping, so allows the top FET to turn ON fully, by creating more than 12V to feed the top FET gate. It's fully explained in the datasheet and application notes.

So either use a boot-strapped driver IC, or change the top FET to P-type.
 

Suraj143

Active Member
You're using the two devices in completely different modes - the bottom FET is common source, so is a digital switch - either ON or OFF. But your top FET is common drain, so works as an analogue amplifier, and will get very hot. I've not checked the spec. of the devices, but presumably the previously mentioned 6V is what it takes to turn the FET fully ON? - so with the gate at 12V the source will still be at 6V, because it has to be 6V lower than the gate.

The bottom transistor feeds 12V to the gate, and the source is 12V lower, so it's switched ON nice and fully.

It's exactly the same with both FET's and bi-polar, for switching you want the load to be in the drain/collector, not the source/emitter.


The IR2110 (Like the IR2112 I mentioned previously) has boot-strapping, so allows the top FET to turn ON fully, by creating more than 12V to feed the top FET gate. It's fully explained in the datasheet and application notes.

So either use a boot-strapped driver IC, or change the top FET to P-type.
Many thanks for the clear explanation.All solved.
Can I add a boot strapping network to my circuit as well?Because the FET driver chips may get late to arrive.
 

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