Question to voltage regulator MAX8881 - Page 1
Forum section: electronics / beginners

author: ggeorgiev
author_rank: Contributor
timestamp: January 14, 2022, 01:28:03 am
content: Hello everyone,I am using a simple voltage regulator MAX8881(https://datasheets.maximintegrated.com/en/ds/MAX8880-MAX8881.pdf) for converting 9V to 5V for an ADC chip.Unfortunately, I am not really sure about the wiring of the internal power-OK pin.The full information about the pin is:"The open-drain POK output is useful as a simple error flag, as well as a delayed reset output. POK sinks current when the output voltage is 10% below the regulation point. Connect POK to OUT through a high-value resistor for a simple error flag indicator. Connect a capacitor in parallel with the resistor to produce a delayed POK signal (delay set by the RC time constant). POK is low during out of regulation or in shutdown and is high impedance during normal operation."Related to the description of the pin, I have three questions:1. Can I just leave the pin not connected if I do not want to use this function? Or should I connect it to GND?2. If I want to use it, what does a high-value resistor mean? 1K, 1 Mega ?3. I am not really sure what "error flag" does mean? So if I connect the POK through a "high-value resistor" to the output, then if POK is "high," the voltage is stable, and I have 5V on the output. On the other hand, if POK is low, the output is also low, as it is connected to the ground. So can I consider the POK connected to the output as an AND gate?4. I have a 9V supply from a battery, and I just want to keep the voltage power supply for an ADC on 5V stable. Should I use the POK in my case?Thank you for considering my question!

author: ledtester
author_rank: Super Contributor
timestamp: January 14, 2022, 04:35:48 am
content: 1. You can leave it unconnected.2. Pull-up resistors in these cases are usually around 10K or greater. It depends on what is going to be monitoring this signal as well as if you are going to use it in conjunction with a capacitor for the delayed signal.Btw, I think the timing capacitor should be installed between POK and ground -- not in parallel with the resistor which goes between POK and Vout.3. After installing the resistor as described by the datasheet (which is not the only way to use POK) the POK pin will be low (near 0 volts) when out of regulation or in shutdown and near Vout during normal operation. This signal could be read from a microcontroller as a digital input, for instance.I'm not sure what you mean by: "So can I consider the POK connected to the output as an AND gate?"4. Presumably a microcontroller will be interacting with the ADC. How is your microcontroller being powered -- e.g. is it being powered by the same 9V battery (like through the same or different regulator) or is it being powered from another power source?

author: jwet
author_rank: Frequent Contributor
timestamp: January 14, 2022, 08:05:08 am
content: This is a very low quiescent current regulator. In order to keep the quiescent low, the POK has a max leakage spec of only 100 nA (.1 uA). This is so you can use a very high value pullup. You could use 1 Megohm for a pullup with no trouble- it would only sit .1v below Vcc. You could probably get away with 4.7 M even. The error flag operation just warns you that the regulator is heading for dropout- 100 mV or headroom. This could be used as a low battery detector if wire to a GPIO. The other thing that you could use this for would be as a CPU reset, the would hold the micro in reset for some time after Vcc rises in order for the oscillator in your micro to start and for it to be initialized. This is what is called a supervisor function usually. If you used a 1 Meg pullup and .1 uf cap to ground, you get a reset that stayed low about 100 mS after the POK went high. 100 mS is a pretty average reset pulse width for power up. The time is basically RC if you want to play with. Use a low leakage cap, like a ceramic for the cap.

author: ggeorgiev
author_rank: Contributor
timestamp: January 14, 2022, 10:15:53 am
content: Thank you for the detailed answer ledtesterRegarding your question 4:So this will be a small circuit that I will be using with raspberry. The small PCB is connected to a 9V battery and I want the power supply for the PCB to come from the battery. The raspberry will bepowered with its own power device. So the regulator is only for the ADC.I do not really need the POK, but let say presumably I would want to read the status of the POK via GPIO on the raspberry. As the max input voltage to the GPIO is 3.3V and the voltage on the POK will be 5V, can i just use a simple voltage divider and then connect it to the GPIO?

author: ledtester
author_rank: Super Contributor
timestamp: January 14, 2022, 04:18:56 pm
content: Quote from: ggeorgiev on January 14, 2022, 10:15:53 amIn that case I would connect the resistor between POK and the Rpi's 3.3V supply. The regulator either pulls POK to ground or it leaves it high-impedance. If POK is pulled to ground the Rpi will see 0V; when left high-impedance the Rpi will see 3.3V.It sounds like the ADC is a 5V chip. How does the Rpi communicate with it, SPI? I2C? Also, what ADC chip are you using?

author: ggeorgiev
author_rank: Contributor
timestamp: January 14, 2022, 07:40:28 pm
content: Thanks for the answer ledtester and for the suggestion on how to connect the POK.Could actually my suggestion with the voltage divider even work? Or why it is not recommended to do it that way?Regarding your question, I am using the ADS8691, and it communicates with the raspy via SPI.Here is a link to the ADCs datasheet:https://www.ti.com/lit/ds/symlink/ads8691.pdf?HQS=dis-dk-null-digikeymode-dsf-pf-null-wwe&ts=1642161238052&ref_url=https%253A%252F%252Fwww.ti.com%252Fgeneral%252Fdocs%252Fsuppproductinfo.tsp%253FdistId%253D10%2526gotoUrl%253Dhttps%253A%252F%252Fwww.ti.com%252Flit%252Fgpn%252Fads8691