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optocoupler signal II

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hi,
What type of opto coupler are you using.? and where are the signal images being measured.??

Please post the circuit around the opto.
 
hi,
The scope images you posted, are they really in the 150Vdc range.??

Also the HCPL is a push/pull output , where is the Vee connection on pin 5.??
 
the Vcc is about 15V. i forgot, pin 5 is also connected to ground.
what do you mean by 150 vdc? :confused:

hi,
Look at this marked up image of your scope traces.

It doesn't help us to give a meaningful answer to your questions if you do not post the complete circuit section.

No component values shown.
 
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sorry. i have drawn the circuit for you.

i'm working on a buck converter. and now still at the PWM stage. i'm getting the outputs of both the optocoupler and they're like the square wave i posted.

i hope this picture is clear enough :)

View attachment 61440
 
You are killing us with too much information? lol

There are capacitors across pins 5 and 8 of the opto. What value? I think the top cap is too small!
Bottom opto: pin5 needs to connect to the bottom FET source.
I do not understand VCC.
Inverter from opto pin2 to opto pin2. What part? How are you limiting the LED current?

Dead time; how are you keeping both FETs from turning on at the same time?

I have 10 more questions for next time.
 
You are killing us with too much information? lol

There are capacitors across pins 5 and 8 of the opto. What value? I think the top cap is too small!
Bottom opto: pin5 needs to connect to the bottom FET source.
I do not understand VCC.
Inverter from opto pin2 to opto pin2. What part? How are you limiting the LED current?

Dead time; how are you keeping both FETs from turning on at the same time?

I have 10 more questions for next time.

sorry. i was focusing on uploading the picture fast that i forgot all the other information :p

the capacitors are both 0.1uF. is this too small?

the inverter is supposedly for the purpose for getting outputs from both the opto-coupler that is opposite (inverted) from each other. wouldn't this make sure they won't both turn on at the same time? sorry, i'm a bit new in this stuff. hope u can help me :)

i'm using a 1k ohm resistor from the output pin 9 of the TL494. i've posted the circuit before for the PWM part. but here it is again View attachment 61442

i'm supplying 15 Vdc to the Vcc part.

i hope that's enough information :p
 
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The top opto must live from the energy stored in the 0.1uf cap for one cycle. I don't have time to dig through the data sheet now. It takes energy to charge up the G-S capacitance in the FET. It takes energy to charge up the G-D cap to 150 volts. It takes more power to keep the opto receiver working for 1/11khz of time. All of this must come from the 0.1uf cap.

I hope you do not have VCC connected as shown. + goes to opto-top and - toes to opto-bottom.

I know we have talked about this before; the bottom opto pin 5 must connect to the bottom FET source.
 
The top opto must live from the energy stored in the 0.1uf cap for one cycle. I don't have time to dig through the data sheet now. It takes energy to charge up the G-S capacitance in the FET. It takes energy to charge up the G-D cap to 150 volts. It takes more power to keep the opto receiver working for 1/11khz of time. All of this must come from the 0.1uf cap.

I hope you do not have VCC connected as shown. + goes to opto-top and - toes to opto-bottom.

I know we have talked about this before; the bottom opto pin 5 must connect to the bottom FET source.

regarding the capacitors, do you mean to say that the value of caps for the top optocoupler should be larger?

omg! you're right about the Vcc. it's a silly mistake, i drew it all wrong. and the same goes to the pin 5. i need to be extra attentive and careful when drawing next time :p
 
I have not done the math on the top cap. I hope you understand that when the top FET is off, and the bottom FET is on, the cap is charged through a diode. When the top FET is on the opto is not being powered from VCC. Its power is coming from the 0.1uf cap only. When that cap runs down the opto will not work. While the FETS are turning on/off the gate current is 0.6 amps all from the 0.1uf cap.
 
Look at the IR2111 data below. It solves the shoot though problem. It removes the opto. If you are going to use it slow (11khz) then increase the capacitor on the top FET driver. May by 10X because I have never used it below 100khz.

Also see FAN7390 and L6388.

The IR2111 is a very simple part that solves many of our problems. There are others but make sure they have built in protection that will not allow both FETs from being on at the same time. I know you think you don't have that problem but you do.
 

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hi,
Look at this marked up image of your scope traces.

It doesn't help us to give a meaningful answer to your questions if you do not post the complete circuit section.

No component values shown.

omg. i forgot to change the probe's settings. so it viewed times 10 the real waveform. so it's suppose to be 15V only :)
 
I have not done the math on the top cap. I hope you understand that when the top FET is off, and the bottom FET is on, the cap is charged through a diode. When the top FET is on the opto is not being powered from VCC. Its power is coming from the 0.1uf cap only. When that cap runs down the opto will not work. While the FETS are turning on/off the gate current is 0.6 amps all from the 0.1uf cap.

i've changed both the caps to 1uF. and the waveform is as follow. will i have a problem with both switching of IGBTs later in the buck circuit?
View attachment 61531

and this is the complete circuit.
View attachment 61532

also, i'm wondering if the optocouplers and tl494 could share the same power supply. hmm.. :confused:
 
Why do you have one IGBT connected to TL494 pin 9 and the other connected to TL949 pin 10. The 494 was never designed to work this way.

The top IGBT is a "switch" the bottom IGBT is a "diode". The way you have it now the "diode" will only be on for a short time. You will do better to drop the bottom IGBT and just use a diode.

Why not use MOSFET and not the IGBT?
 
Why do you have one IGBT connected to TL494 pin 9 and the other connected to TL949 pin 10. The 494 was never designed to work this way.

The top IGBT is a "switch" the bottom IGBT is a "diode". The way you have it now the "diode" will only be on for a short time. You will do better to drop the bottom IGBT and just use a diode.

Why not use MOSFET and not the IGBT?

i tried that and tried taking from the same pin too. the resulting PWM seems to be the same.

this project is actually given and i need to carry on with it. but why is the MOSFET better for this application? :confused: the replacement for the IGBT i just found out to make it into a synchronous buck converter. but since u mention about the diode, i think i'll try the converter with the diode first rather than jumping straight away to the diode-less converter :D

thanks ronsimpson!
 
You are using the wrong part (494) if you want a synchronous diode. The timing on the bottom IGBT gate is wrong. Look at the TL494 data sheet to see how it should be used. There are many examples as to how to use the 494.
I usually use MOSFETs with low voltage. IGBT have too much loss at low voltage.
 
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