Yes, the minimum input voltage at the lowest point of ripple from the main filter capacitor is 4.75V for good regulation when its load current is 800mA. Its dropout voltage is 1.2V with an input voltage of 4.5V but then it has already lost regulation.
Yes, the minimum input voltage at the lowest point of ripple from the main filter capacitor is 4.75V for good regulation when its load current is 800mA. Its dropout voltage is 1.2V with an input voltage of 4.5V but then it has already lost regulation.
And what do you mean please when you say that it has lost its regulation?
I dont understand it from datasheet that when Vin-Vout = 1.2V, then it has lost its regulation. (meaning it doesnt guaranty 3.3V outpunt for 4.5V input?)
It means that if the input drops by 0.1v, the output will possibly drop by this amount too.
You are sitting on the very edge of the capability of the regulator.
It means that if the input drops by 0.1v, the output will possibly drop by this amount too.
You are sitting on the very edge of the capability of the regulator.
I'm very sorry that i dont get it right.
What is dropout voltage? and why do they mention next to it "measured at Vout - 100mV"?
If the dropout voltage is 1.2V than it means that it is actually 1.3V?
I'm very sorry that i dont get it right.
What is dropout voltage? and why do they mention next to it "measured at Vout - 100mV"?
If the dropout voltage is 1.2V than it means that it is actually 1.3V?
Dropout voltage is the difference between the output voltage and the input voltage at a point where the output voltage drops to a specified point at a specified current, in the present case defined as an output voltage drop of 100 mV, at three different output current levels.
When used a regulator, the device is operated above the maximum dropout voltage for all possible values of input voltage so that a maximum regulation percentage is maintained.
In some applications, the device may be used as a voltage limiter (like a zener diode) in which case the point where output voltage will drop, by how much and what current, is important.
Thank you! I think i understood it.
So the "measured in Vin-100mV" means that the maximum output ripple was 100mV or that the the whole level got down to 100mV?
It's Vout-100mV. It means that the output level is down 100mV from what it would be if the device was not dropping out. The output ripple spec is a different matter that is provided seperately and is given with a Vout-Vin well above the dropout voltage. In addition to providing a regulated output voltage, when operated well above the dropout, the device does a better job of reducing input ripple voltage than a reasonably sized filter capacitor could.
It's Vout-100mV. It means that the output level is down 100mV from what it would be if the device was not dropping out. The output ripple spec is a different matter that is provided seperately and is given with a Vout-Vin well above the dropout voltage. In addition to providing a regulated output voltage, when operated well above the dropout, the device does a better job of reducing input ripple voltage than a reasonably sized filter capacitor could.
That's approximately correct. It all depends on the current you are drawing.
What we are saying is this:
If the minimum input voltage is say 4.7v for the voltage you want to have on the output of a regulator, always add about 0.5v to the input voltage so that you never get to a stage where the regulator is “dropping out.”
By dropping out I mean the output voltage “jumps up and down” a small amount. All sorts of funny things happen when the regulator “hunts” and tries to produce the required output voltage. The hunting is due to the electrolytic on the front-end.
This is all getting far too complicated for a beginner. Simply provide adequate input voltage and you will not experience any problems.
A regulator will reduce the ripple on the input by about 1,000 times.
This means a voltage such as 4.7v that rises to 5.7v has a 1v ripple and this will pass though the regulator and the regulator will deliver a voltage of 3.3v with 1mV ripple.
You can find out exactly how much ripple your regulator will deliver when 1v ripple is on the input.
Dropout voltage is just how much "overhead" input voltage above the output voltage that the regulator needs to work properly. If this overhead is met, 3.3V voltage regulator will output 3.3V.
If this overhead voltage margin is not met, it might be designed to shut off. Otherwise it will just output a voltage as close to the input voltage as it can (there will be a small constant voltage drop due to the regulator's internal transistors).
Therefore it will operate properly and output your expected 3.3V.
If you were drawing 500mA from the regulator the input voltage would need to be greater than or equal to 3.3V + (1.01V to 1.15V) to output your expected voltage.
I think they are considering the minimum acceptable output voltage for proper operation is 100mV below the specified output voltage (like 3.3V), so any of the current conditions specified apply for an output voltage of 3.2V.