Voltage divider and the Propeller.

lyassa

I know the propeller is 3.3V. Suppose I have 20 Volts going through a voltage divider as shown below. Is it safe to connect the 10v midpoint directly to the Propeller because the 100k resistors are high enough to provide enough protection? Or will this damage the Propeller?


20v
+
|
R 100k
|
| 10v
+

---

[P0]
|
R 100k
|
_|_
///


Will it be safe if the resistors are 17k and 3k, thus producing 3v at the midpoint?

Mike Green

It's all Ohm's Law and the limit of 500uA maximum current through the protective diode to Vdd.

In the first case, the 100K resistor to ground is not a significant factor in the circuit.

You've got a 100K resistor between 20V and 4V (3.3V + one diode drop). I = E / R = 16V / 100K = 160uA which is well below the limit for the diode.

You may still want to use an optocoupler here. What if the 100K resistor fails? It's very unlikely for it to fail with a short circuit, but the Propeller is expensive by comparison. It depends on what kind of 20V source you have on the other end of the 100K resistor. Could the 20V occasionally become 30V or 40V?

In the case of the 17K and 3K resistors, your major risk is that the 3K resistor fails by opening. The 17K resistor would allow over 1mA to flow through the protective diode which would destroy it and possibly damage the Propeller.

It really all depends on Ohm's Law and the risk you want to take if something goes wrong.

lyassa

Thanks Mike for the clear answers.

Rayman

I think you'd be fine with 100k between 20V and the Prop.
It's often been said that 10k is enough to connect to 5V. That would be a Prop current of (5-3.3)/10k = 17/100k.
Here you're looking at (20-3.3)/100k = 16.7/100k