Join two bridges by a common ground 1

• May 6, 2016

• #2

ScottyUK

Electrical

May 21, 2003

12,915

DE

Connecting the grounds will not work. Hint: think about how each DC rail connects to the AC line when the diode pairs in each bridge conduct. It's OK for one half-cycle, but not for the other.

Can you use an opto as a level shifter to drive the transistor? If would likely be easier to drive a FET rather than a BJT unless the switching frequency is very high.

 

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• May 6, 2016

• Thread starter
• #3

LucasBS

Mechanical

Jun 4, 2010

79

BR

Thank you
I'll use an optocoupler I might have around

This high voltage transistor was chosen based on availability (stole it from CFL bulbs)
They're switching oriented

I was planning on work at 100kHz. Says the datasheet "Transition Frequency" >= 4.0MHz
I'm not sure if this information is what I think it is.

Would you kindly help me finding this information on the datasheet ?
There's a link below


Also, I fail to see the dynamic here. I thought the second half-cycle was equal to the first. On both cases the current flowing on the same directions
If you could elucidate this part. I'd also be grateful


And finally,
Is there another alternative instead of an optocoupler ?
No problem in redesigning everything from scratch
I've got four power sources to do, and one opto. Also, this opto does 60V, not 127 (the transistor is supposed to buck)

http://www.rectron.com/data_sheets/cd13002.pdf

 

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• Jul 15, 2016

• #4

OperaHouse

Electrical

Jun 15, 2003

1,379

US

What is the end result of what you are trying to do with this transistor?

 

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• Nov 18, 2016

• Thread starter
• #5

LucasBS

Mechanical

Jun 4, 2010

79

BR

Since the thread is still up...

Let's leave the transistor and the logic of the power supply behind
This part I'll do the empirical way

Because I think I really need the lesson on why I cannot join the goddam bridges
I fail to see a reason

Both inputs are phased... Both grounds are transmitting in the same direction on any part of the cycle...
I cannot see the problem in theory (only ¡plaac! and smoke in practice)

 

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• Nov 18, 2016

1
• #6

ScottyUK

Electrical

May 21, 2003

12,915

DE

Consider what happens when the X1-1 is positive with respect to X1-2.

The D4 cathode and D10 cathodes will be 0.7V below their respective grounds. If you link the two grounds then the drop across C3 will be zero and a lot of current will flow through D1, D6, and D10. There's a good chance that D6 will get hot. And then explode. It might get saved by D1 or D10 exploding first.

Try it and see. Include a fuse somewhere.

 

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• Nov 20, 2016

• Thread starter
• #7

LucasBS

Mechanical

Jun 4, 2010

79

BR

Thanks

Now I understand
That was the missing explanation


And my carbonized testing board confirmed twice it doesn't work
Forgot the fuses


Thank you again

 

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• Nov 21, 2016

• #8

ScottyUK

Electrical

May 21, 2003

12,915

DE

Pleased you can see why it doesn't work. Fuses are definitely the way forward!

 

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