27 replies to this topic

[Stefan]

Sure Stefan don't believe me if not check the tech spec page. analog input is from 0 to 3.3 V at a max of 2mA per pin.

I believe you, thats why I'm not going to solder anything yet
If I didn't believe you I already tried it all out and perhaps would be virtually crying now because of broken analog inputs

Already thought about modifying the board by disconnecting the +5V pin from the PCB and connect it to the +3V. Only downside would be that the led resistors will be a bit too high, although it will be hardly noticed I suppose. But it would be amazing if it worked without modifications, that's why I am asking.

[Mario Vernari]

Terry is right: the overvoltage is allowed only for the logic inputs, but not for the analogs. However you may scale the voltage using two resistor per analog channel. The ratio is: 3.3/5=0.66. Then fix the overall resistance as load for the Danger output, let's say 50K. At this point you will have: 0.66 = R1 / (R1+R2) = R1 / 50K So, R1 = 33K and R2=50K-33K=17K. Where R2 is a resistor connected from the Danger output to the N input, and R1 is connected from the same N input to ground. Ciao

[CW2]

Sure Stefan don't believe me if not check the tech spec page. analog input is from 0 to 3.3 V at a max of 2mA per pin.

From the datasheet: Absolute Maximum Ratings
Voltage on Input Pins with Respect to Ground...........................-0.3V to + 5.5V

[Terry Massey]

From the datasheet: Absolute Maximum Ratings
Voltage on Input Pins with Respect to Ground...........................-0.3V to + 5.5V

CW2: If you look at page 37 of the summary datasheet is says 8 independent 3.3 v Analog inputs.

[CW2]

CW2: If you look at page 37 of the summary datasheet is says 8 independent 3.3 v Analog inputs.

You are right, but could not this mean that the input range is 0 - 3.3V (0 - ADVREF, to be precise) and still 5V tolerant? I think I'll let Stefan to figure this out - he is the one with the fire extinguisher

[Stefan]

You are right,

Glad to see that the wise elders all agree

but could not this mean that the input range is 0 - 3.3V (0 - ADVREF, to be precise) and still 5V tolerant? I think I'll let Stefan to figure this out - he is the one with the fire extinguisher

Best things of fire extinguishers is to test them after they're expired. I prefer not to use them before that time


Now a proper solution. I'm going to breadboard it, to work on a fully 3.3V, so no 5V on the board. I think that would be the most easy solution.
The only downside I can think about would be that the leds are less bright (although it won't be much of a different I suppose) and the buzzer is less annoying. Oh that's no downside

[Mario Vernari]

CW2, the specs is clear about: "max" means max... Anyway, I think I have had a chance to feed an analog input over VRef...the ADC is still reading correctly (as far as "correctly" could be an acceptable term). Ciao

[Stefan]

Without taking any risks, I'm going to disconnect the +5V pin from the board and connect +3.3V to it instead. When I look at the schematic, it only contains a few resistors, leds, two buzzers, two sensors, a 74hc595, a few buttons and three potmeters. For the leds there are 330Ω resistors onboard, for the buttons 10kΩ resistors, and for the knock sensor/buzzer a 1MΩ resistor. I could replace the 330Ω for 220Ω with 3.3V but I don't see why I should. The others, well, their value doesn't matter that much I suppose. All those components can operate at 3.3V as far as my knowledge goes. Leds will be a bit less bright and the buzzer volume will be a bit lower. I think that will be all.