Many thanks, this is a very good explanation. I am using an OB2269CP and there is approximately a 9K internal PWM pull-up resistor to 6V. This makes the Emitter current of the Opto about 0.3mA in the center of its range, say 3V to FB pin. This reduces the LED current to about 0.6mA for which I had to scale up the LED resistors significantly. Was finally able get a faulty SMPS to work once I understood the principles (don't know how it ever worked before it went faulty because FB was always being pulled below 1V by over current to LED before the TL431 reached 2.5V on ref pin, and the chip goes into Eco mode!).
Very interesting. This isn't what I would have expected. I would not have anticipated the LED current to increase. I would have expected the light intensity of the LED in the opto to decrease over time which would reduce the current on the transistor side. I would have then designed to have a sufficiently high LED current so that the transistor current was sufficiently high.
I actually did test this out. I went out to a forward converted I had previously designed that has a very similar circuit consisting of an optocoupler and an LM431, and I replaced the optocoupler with one that had low CTR (about 20%), sure enough the current on the LED side increased to about 12mA. So, I then replaced the optocoupler again with one that has a CTR of about 300%. Sure enough the current decreased to less than 1mA. So, now im worried about biasing current of the LM431. With the very high optocoupler, we don't seem to have enough biasing current as required for the LM431.
Hi Robert It's a great video. I still have a question. Why Vka= 10 V at BOL? I confuse that what's the function of TL431. At the video 22, it just like programmed zener diode. In this circuit, we set Vout at 12V. However, Vka is 10V. Could you explain more? Thank you~
Because at BOL the ILED is only about 2mA and total current on Rup is 3mA, so Vka at TL431= .003*225-VLED =10.3V. At EOL the total current goes up to about 30mA and so the Voltage at Vka decreases significantly.
Great Video and explanation, thank you. I am a bit confused about your statement regarding the aging of the opto coupler. As the opto gets older, the CTR drops, I understand this - makes perfect sense. You say that the emitter current stays constant and therefor the Led current rises. As far as I understand it, the CTR drops because for the same Led current the LED emits less light over time, leading to less emitter current and therefor lower CTR. How can the Led current rise and the emitter current stay constant? I thought the Led current is dictated by the output voltage (12V in this case), the voltage divide and the TL431. The emitter current is dictated by the Led current and CTR.
Hi Pete, great question. The current and the voltage in the emitter of the opto remains constant because of the negative feedback from the loop. As the CTR decreases, the emitter voltage and current will drop slighty, the vout will decrease and then the error amp in the TL431 will see that the vout drop so the voltage at the cathode of the TL431 will drop/decrease and this in turn will increase the current in the LED and in turn increase and keeps both the emitter's voltage and current constant. If you send me your email, I can send you the average model that I used and you can work with the model and get a more intuitive fell on how the system works. My email is rbola35618@aol.com
Thank you Robert. Now I understand. I guess I did not close the loop in my mind when I was thinking of the aging effect. Great videos, I do learn a lot from them. Please keep it up. Waiting for the next installment.
So if the device has 2R and R internal resistors, as opposed to R and R, we can assume 4.1V on the COMP pin ? Also I checked the device UC1843, it has 2R and R, could you link me to a specific datasheet that says it has R and R ?
Hi Nabil. That is a great question. I tried to simplify the explanation by changing the first resistor from 2R to R. You are correct that when the error amp COMP will go to around 4.2V when the PWM is at full load. However, we usually set the CS line to about 80% of 1V or 0.8V. That way you have 20% overcurrent margin. That also means that the COMP will drop down to 80% of 4.2V to around 3.2V. This is where the 3.2V comes from. Great question! RB
I had been waiting for this video since long. I had requested twice for the video on this topic. It is very informative & very useful to me as many doubts are cleared especially explanation on how the opto emitter current & voltage remain constant even in aging. Teaching method is also very effective & explanation is very nice which I have not seen in any book or on web. Thanks for taking efforts for making the topic simple & teaching it to our level. Now I am eagerly waiting for next video on compensation part. Thanks once again.I have a question that my out-put voltage will be 320V DC then can I connect it to the opto LED or I have to connect from other low voltage (12v) on secondary side? Is there any effect on feedback?Thanks.
Thank you Satyavijay for the feedback. It's took a long time. Last year was a bad year for me and did not have much time for making videos. I lost my father on September 30, and my wife got diagnosed with cancer. So most of my spare time was taking care of my wife. My wife is doing much better now and has finished her chemo. Hopefully, I will have more time to make videos. Thank you for your support. Robert
Sorry to hear. My sincere sympathies to you and family. I hope you and your family will come out the bad time very soon. Please guide me how to connect LED of opto to 320V DC output ? Thanks.
Thanks Catalin. Required resistor will be of more wattage of about 10watts. Instead, if I connect it from 12v from secondary side (but voltage sensing from resistor divider from 320 VDC) will it do and is there any effect on feedback control?
Hi Giridhar, Sorry I missed you question. The Ve = 1.9V comes from the fact that the internal PWM has a comparator that switches at 1 Volts and has two diodes in series and I make an assumption for the purpose to try to make the explanation simple. So the output of the op amp has to output 3.1V at steady state once everything is in regulation. If you look at the op amp, the noninverting is connected to the a reference of 2.5V. Therefore, then inverting input should also equal to the 2.5V. That means that a current flows from the output to the inverting of (3.1V-2.5V)/100kohm =6uA. Current of 6uA cannot flow into the input of the inverting node so it has to flow thru the input resistor. Therefore you will have a drop of 6uA*100kohm = 0.6V which is subtracted from the inverting input of 2.5V - 0.6V = 1.9V. I hope this make sense. Regard, Robert Bolanos. Commenst can be sent to rbola35618@aol.com
Hi Satyavijay, I have travelling for my daughter volleyball tournaments. I am hoping to make the final video which explain everthing including the bandwidth of the opto coupler. I hope to do it this weekend on Saturday May 5. Robert
Hi Raed. I have the first edition. This the second editionwww.amazon.com/Switch-Mode-Power-Supplies-Second-Simulations/dp/0071823468/ref=la_B001IOH604_1_1?s=books&ie=UTF8&qid=1520820825&sr=1-1
Many thanks, this is a very good explanation. I am using an OB2269CP and there is approximately a 9K internal PWM pull-up resistor to 6V. This makes the Emitter current of the Opto about 0.3mA in the center of its range, say 3V to FB pin. This reduces the LED current to about 0.6mA for which I had to scale up the LED resistors significantly. Was finally able get a faulty SMPS to work once I understood the principles (don't know how it ever worked before it went faulty because FB was always being pulled below 1V by over current to LED before the TL431 reached 2.5V on ref pin, and the chip goes into Eco mode!).
Thanks for the video. How does 50% CTR map to 0.00154 (47:27) ?
Very interesting. This isn't what I would have expected. I would not have anticipated the LED current to increase. I would have expected the light intensity of the LED in the opto to decrease over time which would reduce the current on the transistor side. I would have then designed to have a sufficiently high LED current so that the transistor current was sufficiently high.
I actually did test this out. I went out to a forward converted I had previously designed that has a very similar circuit consisting of an optocoupler and an LM431, and I replaced the optocoupler with one that had low CTR (about 20%), sure enough the current on the LED side increased to about 12mA. So, I then replaced the optocoupler again with one that has a CTR of about 300%. Sure enough the current decreased to less than 1mA. So, now im worried about biasing current of the LM431. With the very high optocoupler, we don't seem to have enough biasing current as required for the LM431.
Thank you Robert.
Very useful. Could you tell me what the jack alexanyer's mode is?
Hi Robert
It's a great video. I still have a question. Why Vka= 10 V at BOL? I confuse that what's the function of TL431. At the video 22, it just like programmed zener diode. In this circuit, we set Vout at 12V. However, Vka is 10V. Could you explain more? Thank you~
Because at BOL the ILED is only about 2mA and total current on Rup is 3mA, so Vka at TL431= .003*225-VLED =10.3V. At EOL the total current goes up to about 30mA and so the Voltage at Vka decreases significantly.
Great Video and explanation, thank you. I am a bit confused about your statement regarding the aging of the opto coupler. As the opto gets older, the CTR drops, I understand this - makes perfect sense. You say that the emitter current stays constant and therefor the Led current rises. As far as I understand it, the CTR drops because for the same Led current the LED emits less light over time, leading to less emitter current and therefor lower CTR. How can the Led current rise and the emitter current stay constant? I thought the Led current is dictated by the output voltage (12V in this case), the voltage divide and the TL431. The emitter current is dictated by the Led current and CTR.
Hi Pete, great question. The current and the voltage in the emitter of the opto remains constant because of the negative feedback from the loop. As the CTR decreases, the emitter voltage and current will drop slighty, the vout will decrease and then the error amp in the TL431 will see that the vout drop so the voltage at the cathode of the TL431 will drop/decrease and this in turn will increase the current in the LED and in turn increase and keeps both the emitter's voltage and current constant. If you send me your email, I can send you the average model that I used and you can work with the model and get a more intuitive fell on how the system works. My email is rbola35618@aol.com
Thank you Robert. Now I understand. I guess I did not close the loop in my mind when I was thinking of the aging effect.
Great videos, I do learn a lot from them. Please keep it up. Waiting for the next installment.
forgot to mention, I sent you an Email
Many thanks, this is a veri good!!! Hello is possible send me the material Excel!. Thanks
So if the device has 2R and R internal resistors, as opposed to R and R, we can assume 4.1V on the COMP pin ? Also I checked the device UC1843, it has 2R and R, could you link me to a specific datasheet that says it has R and R ?
Hi Nabil. That is a great question. I tried to simplify the explanation by changing the first resistor from 2R to R. You are correct that when the error amp COMP will go to around 4.2V when the PWM is at full load. However, we usually set the CS line to about 80% of 1V or 0.8V. That way you have 20% overcurrent margin. That also means that the COMP will drop down to 80% of 4.2V to around 3.2V. This is where the 3.2V comes from. Great question! RB
I had been waiting for this video since long. I had requested twice for the video on this topic. It is very informative & very useful to me as many doubts are cleared especially explanation on how the opto emitter current & voltage remain constant even in aging. Teaching method is also very effective & explanation is very nice which I have not seen in any book or on web. Thanks for taking efforts for making the topic simple & teaching it to our level. Now I am eagerly waiting for next video on compensation part. Thanks once again.I have a question that my out-put voltage will be 320V DC then can I connect it to the opto LED or I have to connect from other low voltage (12v) on secondary side? Is there any effect on feedback?Thanks.
Thank you Satyavijay for the feedback. It's took a long time. Last year was a bad year for me and did not have much time for making videos. I lost my father on September 30, and my wife got diagnosed with cancer. So most of my spare time was taking care of my wife. My wife is doing much better now and has finished her chemo. Hopefully, I will have more time to make videos. Thank you for your support. Robert
Sorry to hear. My sincere sympathies to you and family. I hope you and your family will come out the bad time very soon.
Please guide me how to connect LED of opto to 320V DC output ?
Thanks.
The same way. Rup=(320-1-3)/(1mA + 30mA)=10.2 kOhms.
Thanks Catalin. Required resistor will be of more wattage of about 10watts. Instead, if I connect it from 12v from secondary side (but voltage sensing from resistor divider from 320 VDC) will it do and is there any effect on feedback control?
How much of current flow to control pin of ic through optocoupler and what is calculation.
How you got opto VE=1.9V, can you explain that.
Hi Giridhar, Sorry I missed you question. The Ve = 1.9V comes from the fact that the internal PWM has a comparator that switches at 1 Volts and has two diodes in series and I make an assumption for the purpose to try to make the explanation simple. So the output of the op amp has to output 3.1V at steady state once everything is in regulation. If you look at the op amp, the noninverting is connected to the a reference of 2.5V. Therefore, then inverting input should also equal to the 2.5V. That means that a current flows from the output to the inverting of (3.1V-2.5V)/100kohm =6uA. Current of 6uA cannot flow into the input of the inverting node so it has to flow thru the input resistor. Therefore you will have a drop of 6uA*100kohm = 0.6V which is subtracted from the inverting input of 2.5V - 0.6V = 1.9V. I hope this make sense. Regard, Robert Bolanos. Commenst can be sent to rbola35618@aol.com
The bandwidth of Optocoupler is measured. What is its significance and afterwards where it is used ?
Hi Satyavijay, I have travelling for my daughter volleyball tournaments. I am hoping to make the final video which explain everthing including the bandwidth of the opto coupler. I hope to do it this weekend on Saturday May 5. Robert
Ok thanks
Hi, Thanks for the tutorial... Which book/note by Basso are you referring to ?
Hi Raed. I have the first edition. This the second editionwww.amazon.com/Switch-Mode-Power-Supplies-Second-Simulations/dp/0071823468/ref=la_B001IOH604_1_1?s=books&ie=UTF8&qid=1520820825&sr=1-1
Robert Bolanos , thanks...
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