Something else to consider with shunt regulator circuits like this is the resistor. At 6 Volts with 1.05 Amps your resistor is going to be dissipating 6.3 Watts!!! Another magic smoke source unless you use a large appropriately sized wire wound resistor.
Heck that's why my Zener Diode blew up it's magic smoke when an inline fuse of the load blew because I slipped with a small screwdriver 🥴... Thanks again for a very clear and thorough explanation!
Thank you so much for this explanation. I was loosing tracking following the linear power supply series and this extra more detailed explanation made much more sense. Thank you again
Very informative videos and probably the most accessible explanations I ever had. This is not an easy topic but the videos make it more understandable. Thank you very much! Many thanks!
Perfect explaining, like an electrical course. Thing is - I want to hold the voltage at power transistor no higher then 14.7v to battery from solar. Will that do it?
Great explanation! I have a 10W (36 cell) solar panel that will charge a small 12V battery. To prevent it from overcharge I connected a 15V zener diod as shunt regulator. The diod is 5W and the panel doesn't manage more than 300mA so it should be fine. But I will try the series setup just for fun.
Thank you for the explanation sir. I think they might be a problem sir. When the load is taken off, you are considering that the current demand will still be the same whereas that is not true. When the load is off the current will obviously reduce given that the resistance of the circuit has been increased. The increase in resistance comes from the fact that parallel load creat smaller resistance. Sir please if I'm wrong do me well by illuminating my dark spots
Interesting thought. I think you have a point there. The current would probably be a little lower as you say. The resistor is still 5.71 ohm and the total voltage is still 18V . The voltage over the zener diode will increase say to a maximum of 14V ( I don't know if it can reach those voltages, but just to explain my point ) and as an effect of that makes the voltage over the resistor to 4V. 4V divided by 5.71ohm equals 700mA. The power dissipated from the zener dioder is 14V times 0.7A equals 9.8W, so a lot lower but still too much :)
The Zener holds the base of first transistor at 12.7 V. Each transistor needs 0.7 volts across base emitter junction to turn on since they are two that will be 1.4 V. So 12.7-1.4=11.3V at the output. The current gain comes from the HFE of both transistors
Something else to consider with shunt regulator circuits like this is the resistor. At 6 Volts with 1.05 Amps your resistor is going to be dissipating 6.3 Watts!!! Another magic smoke source unless you use a large appropriately sized wire wound resistor.
The transition from Zener to serie Regulator is excellent ! Very good explained
Heck that's why my Zener Diode blew up it's magic smoke when an inline fuse of the load blew because I slipped with a small screwdriver 🥴...
Thanks again for a very clear and thorough explanation!
Perfect explanation, watching for the second time; informative and somehow joyful
Thank you sir, great job in explanation supported by calculation.
Your elucidation was greatly appreciated, sir. Thank you.
Thank you so much for this explanation. I was loosing tracking following the linear power supply series and this extra more detailed explanation made much more sense. Thank you again
Very informative videos and probably the most accessible explanations I ever had. This is not an easy topic but the videos make it more understandable. Thank you very much! Many thanks!
Thank you for the time spent here for us!❤
Another excellent explanation, Bob!
Good explanation. Thanks for this video.
Releasing the magic smoke..😀🤣. Great video👍
Always take a moment to think about heat and smoking devices at a cost.
Perfect explaining, like an electrical course. Thing is - I want to hold the voltage at power transistor no higher then 14.7v to battery from solar. Will that do it?
Great explanation!
I have a 10W (36 cell) solar panel that will charge a small 12V battery.
To prevent it from overcharge I connected a 15V zener diod as shunt regulator.
The diod is 5W and the panel doesn't manage more than 300mA so it should be fine.
But I will try the series setup just for fun.
What is controlling the current?
@@commonsenseisdeadandgone The weak panel
@@rwd1973 .3A can still overcharge a battery in time, no?
About terminology: the "series" in series regulator is the diode and resistor? The network is different but isn't the load still in parallel?
Was it was really clear, I appreciate your explanation very much.
Thank you.
p.s
Actually I am in master studying in engineer of electronics
Great video
Thank you for your efforts
Very nice video
Thank you for the explanation sir.
I think they might be a problem sir.
When the load is taken off, you are considering that the current demand will still be the same whereas that is not true.
When the load is off the current will obviously reduce given that the resistance of the circuit has been increased.
The increase in resistance comes from the fact that parallel load creat smaller resistance.
Sir please if I'm wrong do me well by illuminating my dark spots
Interesting thought. I think you have a point there. The current would probably be a little lower as you say.
The resistor is still 5.71 ohm and the total voltage is still 18V . The voltage over the zener diode will increase say to a maximum of 14V ( I don't know if it can reach those voltages, but just to explain my point ) and as an effect of that makes the voltage over the resistor to 4V. 4V divided by 5.71ohm equals 700mA. The power dissipated from the zener dioder is 14V times 0.7A equals 9.8W, so a lot lower but still too much :)
WHAT IS THE DIFF BET BRUSH MOTOR AND A NON BRUSH MOTOR
Thank you. It is easy to understan.
Still bit unclear how that power transistor on top right corner in your sketch brings voltage down from 18v to 12v?
The Zener holds the base of first transistor at 12.7 V. Each transistor needs 0.7 volts across base emitter junction to turn on since they are two that will be 1.4 V. So 12.7-1.4=11.3V at the output. The current gain comes from the HFE of both transistors
The resistor brings the voltage down....
I think?
great explain, thank you, may allah bless you.
nice!
❤
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ahhh