When you change the load resistor, for example to a lower value, the current increases, this means, the base-emitter current increases, which causes the Ir current to increase as well, which in its turn increases voltage drop over the resistor, which can be shift the zener diode into non stabilizing range. So, you should count for this min/max values as well. Can you describe how should you fully calculate this circuit?
The voltage drop over the resistor and its current should first stay constant, with more current flowing to the base then to the Zener. But that then can, as you are correctly worried about, effect the Zener.
also for the power dissipation of the zener diode, u said before the zener diode safe operation is 10mA and 20mA is current that flowing the resistor, so should we multiple the Vd with 10 mA instead of 20mA?
12v voltage regulator with good efficiency which will not give up and i would be able to apply heatsink on it with good 2 Amp at 12v output with input varing from 12v to 15v?
I agree. need this circuit for lead acid batteries & solar. Sick of blowing things you cant fix or dont do what you want voltage wise. Im talking 40 amps & deep cycle on solar.
@@kennmossman8701 what about C out and yr talking amps on load I assume. Lots of caps for 40 amps to regulate 14.4 volts for my project. Otherwise its arduino.
Why is the power dissipation of the transistor = (7.5-6) x 0.5? Can you explain where did the 0.5A comes from and why is the voltage across the transistor consider as Vout -Vin? In addition, why did you use 20mA instead of 10mA in the calculation for the zener and resistor power dissipation?
The 0.5A current comes from the chosen load current, and since the emitter current is the sum of all currents passing through the transistor, it represents the total current that generates power through the transistor. The voltages across the collector and emitter are 7.5V and 6V respectively, so the voltage drop would be 7.5V-6V (according to Kirchhoff's Voltage Law). Plugging those values into the power equation, P=IV, we come to P= (7.5V-6V) * 0.5A. For the zener diode and resistor power dissipation calculations, the current used in calculations was 20mA for the resistor because we choose the resistor's value to force that current, so that is known by the design. For the zener diode, we use 20mA as a worse case where all the current is passed through the diode (for reliability purposes). In reality, 10mA of that current should be used to control the base of the transistor, but it is almost always better to use higher power-rated components in the event that the worse case scenario occurs.
I will recommend you to use lower capacity output capacitor like 200uf or something and use a load resistor in output with a value like 1 kOhm or 500 Ohms, or reducing voltage will be a trouble, don't be like you have to follow 100% Video, if you have knowledge, apply it or you'll never learn
When you change the load resistor, for example to a lower value, the current increases, this means, the base-emitter current increases, which causes the Ir current to increase as well, which in its turn increases voltage drop over the resistor, which can be shift the zener diode into non stabilizing range. So, you should count for this min/max values as well. Can you describe how should you fully calculate this circuit?
very clear observation! I need to think on that.
The voltage drop over the resistor and its current should first stay constant, with more current flowing to the base then to the Zener. But that then can, as you are correctly worried about, effect the Zener.
basically that is true in any circuit
Great job!
Thank you so much, Glad you liked it!!
excellent video sir! but i just want to know, don't we need to put a resistor in Ib to drop it from 6.8 to 0.7v? i am sorry if i am wrong
also for the power dissipation of the zener diode, u said before the zener diode safe operation is 10mA and 20mA is current that flowing the resistor, so should we multiple the Vd with 10 mA instead of 20mA?
why is the emitter is connected to the +ve in the output end as we not it should be always reverse biased.
Fan regulator what is out put voltage in lowest knob
12v voltage regulator with good efficiency which will not give up and i would be able to apply heatsink on it with good 2 Amp at 12v output with input varing from 12v to 15v?
what? use a boost/buck converter
I agree. need this circuit for lead acid batteries & solar. Sick of blowing things you cant fix or dont do what you want voltage wise. Im talking 40 amps & deep cycle on solar.
Simulation and explanation thanks
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Excellent Explanation, Good Animation , Simplicity
Glad it was helpful!
how do you calculate for the size of the capacitors?
rule of thumb for Cin is 1000uF per Amp though some say 2000uF per Amp depends on the load
@@kennmossman8701 what about C out and yr talking amps on load I assume. Lots of caps for 40 amps to regulate 14.4 volts for my project. Otherwise its arduino.
@@davidjackson2115 49A ???? thats a very different beast
Why is the power dissipation of the transistor = (7.5-6) x 0.5? Can you explain where did the 0.5A comes from and why is the voltage across the transistor consider as Vout -Vin? In addition, why did you use 20mA instead of 10mA in the calculation for the zener and resistor power dissipation?
The 0.5A current comes from the chosen load current, and since the emitter current is the sum of all currents passing through the transistor, it represents the total current that generates power through the transistor. The voltages across the collector and emitter are 7.5V and 6V respectively, so the voltage drop would be 7.5V-6V (according to Kirchhoff's Voltage Law).
Plugging those values into the power equation, P=IV, we come to P= (7.5V-6V) * 0.5A.
For the zener diode and resistor power dissipation calculations, the current used in calculations was 20mA for the resistor because we choose the resistor's value to force that current, so that is known by the design. For the zener diode, we use 20mA as a worse case where all the current is passed through the diode (for reliability purposes). In reality, 10mA of that current should be used to control the base of the transistor, but it is almost always better to use higher power-rated components in the event that the worse case scenario occurs.
Good question
Good answer
Can you also explain the same, if we replace zener with TL431 for precise tegulation?
Yes, we can do that!
trouble is the low current capability
Thank You My Indian Brother! 👋
Thank you so much, Glad you liked it!!
Well, seems the engineer's not so foolish after all...
Damn, that is some good explaining
How does this man this compare to the Linear Regulator in terms of noise and PSRR?
This regulator is very helpful for such application where input voltage ripple is very low & voltage range is also low.
How can you change ac voltage from a motor to DC to charge a phone
yes
At least transistors generally dissipate heat better then the diodes, more surface area, like to TO-3 type.
excellent explanation about series regulation. thank you.
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Can i use n channel mosfet instead of bjt
Yes you can do that! normally, this has more advantage than the bjt. because bjt has more temperature dependency and values will vary.
I will recommend you to use lower capacity output capacitor like 200uf or something and use a load resistor in output with a value like 1 kOhm or 500 Ohms, or reducing voltage will be a trouble, don't be like you have to follow 100% Video, if you have knowledge, apply it or you'll never learn
well there is an argument for a lower Cout but not for that reason
Super
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Thanks for video.
"Small ball of guts" what does it mean.
thanks ! I got important role of Z-diode .
The problem is the voltage drop in load at some amount of current
same is true in 'amy' regulator
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Great !
the background music is quite distracting esp when listening through headphones
Sorry for the inconvenience
can you please controlle 100 volt to 30volt
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😍🙏 नमस्कार
First u study English