I understand that CE is some kind of a enable pin. The datasheet actually provides the left circuit as a "typical application circuit".
But regarding the right circuit, I suppose VCC-3V3 refers to a 3.3V so what happens if I put another voltage to BL_CTRL?? what happens if I put GND there? and what happens if I apply a PWM signal there??
I understand voltage dividers when they go from VCC to ground so I guess R8 and R9 are somehow dividing the voltage, but ok, if this is so, then what happens when this divided voltage enters CE???
My understanding of the circuits is stated above. Plus the datasheet puts the internals as :
I will greatly appreciate if someone can explain the role of the circuit to me.
This is simply a 3.3V LDO regulator with logic level input for Chip Enable, LV type CMOS . Have a series R ratio and values given allows logic from say 5V to drive the CE line with current limit to ESD diodes typ. 5mA.
The part number designates the voltage of the chip.
The low (1V) dropout minimum depends on current and ultralow Vout as shown in table specs.
One of the unique features for an LDO is Foldback current limiter
The XC6209/6212 series includes a combination of a fixed current limiter circuit & a foldback circuit, which aid the operations of the current limiter and circuit protection.
This is simply a 3.3V LDO regulator with logic level input for Chip Enable, LV type CMOS . Have a series R ratio and values given allows logic from say 5V to drive the CE line with current limit to ESD diodes typ. 5mA.
The part number designates the voltage of the chip.
The low (1V) dropout minimum depends on current and ultralow Vout as shown in table specs.
One of the unique features for an LDO is Foldback current limiter
The XC6209/6212 series includes a combination of a fixed current limiter circuit & a foldback circuit, which aid the operations of the current limiter and circuit protection.
By reading the datasheet, I kind of understand that for an input of say 5V I get and output of 3.3V when the chip is enabled (CE=5V) .
What I don't understand is the circuit on the right, where CE is not 5V or GND but connected to what I think is a voltage divider. What implications has this on the circuit??
CE is indeed a Chip Enable, and it's threshold voltages are listed on page 6 of the datasheet as CE "High" Voltage, and CE "Low" Voltage. Pulling that pin below 0.25V will turn off the output. It's quite possible that PWMing that pin will have a some dimming ability.
But the ratio of 10K to 1K will put the minimum voltage at that pin at 0.3V when BL_CTRL is zero. Which, if you go by the stated numbers from the datasheet, should never turn off the LDO since CE Low is listed at 0.25 volts minimum. The designers of the circuit on the right have probably tested it and found that the values they have chosen work, but that is very fuzzy engineering, since a different part, or batch pf parts may not work at those levels.
CE is indeed a Chip Enable, and it's threshold voltages are listed on page 6 of the datasheet as CE "High" Voltage, and CE "Low" Voltage. Pulling that pin below 0.25V will turn off the output. It's quite possible that PWMing that pin will have a some dimming ability.
But the ratio of 10K to 1K will put the minimum voltage at that pin at 0.3V when BL_CTRL is zero. Which, if you go by the stated numbers from the datasheet, should never turn off the LDO since CE Low is listed at 0.25 volts minimum. The designers of the circuit on the right have probably tested it and found that the values they have chosen work, but that is very fuzzy engineering, since a different part, or batch pf parts may not work at those levels.
Thank you very much for your reply.
Reading your reply, and considering that even 0V would not turn off the LDO, does that mean that any voltage in CE (including 0) will turn on the LDO??
Why would PWM made it dimming if 0 does not turn it off??
I think it is indeed a PWM signal and will allow the brightness to be controlled from "just on" to "fully on".. There is no need really to turn the backlight off on an LCD as you can't see jack... So it would be the same as turning off the whole thing!!!!
I think it is indeed a PWM signal and will allow the brightness to be controlled from "just on" to "fully on".. There is no need really to turn the backlight off on an LCD as you can't see jack... So it would be the same as turning off the whole thing!!!!
You are right!
Tomorrow I will take the solder, connect some pins to the interface board I am building, connect it to a breadboard and try
Thanks again!
By reading the datasheet, I kind of understand that for an input of say 5V I get and output of 3.3V when the chip is enabled (CE=5V) .
What I don't understand is the circuit on the right, where CE is not 5V or GND but connected to what I think is a voltage divider. What implications has this on the circuit??
i already explained]
a series R ratio allows logic from 3 ~5 V to drive the CE line with current limit to ESD diodes which are usually rated typ. 5mA. as a clamp for DC
I have to report that I made a mistake in my previous post. I didnt connect the pin properly
so the results are the following
1) with the pin unconnected , the LCD is always on
2) with a potentiometer, the LCD turns off abruptly
3) with PWM, it dims slowly to off and back to on...
Thank for your help
Now my only thing would be theoretical. According to the analysis and the circuit, even with 0 V it shouldnt have turn off but it did... ummm well. I dont know why.
Time to continue the development of the main project.
Well, that was already explained, and that was because the data sheet spec is probably not accurate. Whoever heard of a logical low signal having to be as low as 0.25v ?
It's a CMOS input too.
So this chip is working because either the data sheet is wrong or the chip just happens to be better or worse than most chips are.
However, you can bypass all this by creating your own network that DOES work with ANY chip. Simply lower the 1k to a lower value that allows a voltage of less than 0.25v. Alternately, just remove the 10k and drive it with a HIGH and a LOW (both) and you'll do pretty well. Just dont ever run it with the pin completely open as that is not good.
Yep, in both cases it will be on. Someone goofed. With tolerences, it may or may not work. You can't reliably pull CE below 0.25V
Note that the chip can be had with different logic and internal pull up/pull down resistors.
Depending on the suffix, use a high value pull-up/pull down. I THINK the datasheet said something like 2M would work.
Yep, in both cases it will be on. Someone goofed. With tolerences, it may or may not work. You can't reliably pull CE below 0.25V
Note that the chip can be had with different logic and internal pull up/pull down resistors.
Depending on the suffix, use a high value pull-up/pull down. I THINK the datasheet said something like 2M would work.
I made a mistake the first time. For reasons I dont know it actually turn off when the signal is Ground.
Anyway, are you telling me to put a pull up in the BL_CTRL pin??
Thanks
What I'm saying is to replace R9 with a 10K to 2M resistor and jumper out R8. I think anything within that range would work. You have to pull CE below 0.25 V to turn the BL off. You also have to be able to sink the current from R9+the few uA the CE requires. My brain isn't in the mood to study the datasheet to get an exact value.
As it is, it "might" work or might work sometimes. By work, I mean, pulling CE below 0.25V turns off the BL. I don't see any reason why PWM could not be used to control brightness.
I'd also need more info about the characteristics of whatever port your using.