Timeline for Voltage protection of PNP Open-Collector Pull-Down Resistor Circuit
Current License: CC BY-SA 3.0
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May 10, 2017 at 19:29 | comment | added | rioraxe | When the sensor output is high, while ignoring the NI9401, the current through the resistor divider would be 12V/(2.2K+1.5K) = 3.2mA. Now back track and compare that to the NI9401 ±250uA input current, which is less than 10%. The possibility of NI9401 sourcing +250uA can be ignored because it cannot source current above 5V (its supply voltage). If the NI9401 sinks -250uA, it would reduce the divider output voltage by less than 10%, which would still be a solidly high level with a big margin. | |
May 10, 2017 at 19:18 | comment | added | rioraxe | page 5: Input current ±250 uA. The NI 9401 can be sourcing or sinking per the spec. The sensor cannot sink current per your "I/O Circuit Diagram" at the top. So when the sensor output is low, the sensor is not sourcing or sinking any current, the pull down has the task for ensuring a low level. The sensor can source up to 50mA. So when the sensor is high, it would source at 12V (your chosen supply voltage) of up to 50mA. The resistor divided can be chosen to give the desired voltage. | |
May 10, 2017 at 11:45 | comment | added | Discbrake | (2) Can your reasoning be summarized as follows? (a) Assume that the sensor is sourcing 250uA, which go through R2 and then split into R1 (most of it) and into the NI 9401 (just a fraction due to the high input impedance). (b) The voltage drop across the pull-down and thus across the NI 9401 should still be able to get a solid LOW signal, thus about 0.4V. (c) With V = R*I follows R = V/I = 0.4V/250uA = 1600 Ohm -> 1.5k (d) If we were to choose a higher R, we might get a HIGH signal just from the voltage drop over R1 alone, without even having a HIGH voltage from the sensor. | |
May 10, 2017 at 10:58 | comment | added | Discbrake | Sorry, I'm still not quite there yet. (1) "So the worst case calculation for the pull down would be when the input is sourcing 250uA, ..." -> But the NI 9401 is always sinking, isn't it? Or did you mean the sensor when you wrote "input" there? | |
Apr 11, 2017 at 0:15 | comment | added | rioraxe | Looking up the NI 9401 Datasheet, on page 5: Input current \$\pm\$250 uA. What that means is that within the input voltage range of 0 to 4.5V, the NI 9401 input can source or sink up to 250uA. So the worst case calculation for the pull down would be when the input is sourcing 250uA, it must be of low enough resistance to pull the voltage to below certain threshold which was chosen to be 0.4V. The NI 9401 does not care how much current the resistor sinks when the output from the sensor is at 5V. | |
Apr 10, 2017 at 20:02 | comment | added | Discbrake | But why do you use VIL minus the margin for the design of R1? If I understand it correctly, the NI9401 cannot handle more than 250uA. So with the formula V = R*I -> R = V/I, why do you not use VIH = 5V plus some safety margin to calculate R1, which would then be 5.25V/250uA = 21kOhm? That would give a current of 238uA at 5V and 19uA at 0.4V, both values below the 250uA. But with R1=1.5k, the current at 0.4V is 267uA and at 5V a 3.33mA - both values (much) higher than the specified 250uA! What do I miss? | |
Apr 10, 2017 at 19:17 | vote | accept | Discbrake | ||
Feb 2, 2017 at 18:55 | comment | added | rioraxe | The input low threshold VIL for TTL is usually 0.8V, so set the resistor to pull below that. 0.4V is a common number that gives 0.4V of margin. My version looks better :), why all the extra stuffs. | |
Feb 2, 2017 at 14:15 | comment | added | Discbrake | Also, why 0.4V/250uA and not 4.5V/250uA or 4.7V/250uA? | |
Feb 2, 2017 at 8:26 | comment | added | Discbrake | Thanks. Now I tried to adapt my schematic based on your answer in the last section "Update" of my question. Does that look better now? | |
Feb 2, 2017 at 7:44 | history | answered | rioraxe | CC BY-SA 3.0 |