Pull up resistor value for low power application

Thread Starter

Mayuri1234

Joined Apr 7, 2014
7
I am developing a battery powered device. In my circuit, there is a dip switch which is used as an input to micro controller pin. The supply to the switch is 3.3 v. Can I use a high value pull up resistor to minimize current in the circuit? Right now, the resistor value is 200k. Can I increase it to 1 Mega ohm.

Regards,
Mayuri
 

hp1729

Joined Nov 23, 2015
2,304
I am developing a battery powered device. In my circuit, there is a dip switch which is used as an input to micro controller pin. The supply to the switch is 3.3 v. Can I use a high value pull up resistor to minimize current in the circuit? Right now, the resistor value is 200k. Can I increase it to 1 Mega ohm.

Regards,
Mayuri
How long will it take you to try it and see if it works?
 

MrChips

Joined Oct 2, 2009
30,710
How often is the DIP switch closed vs how often it is open? Is it 50:50?
What is the function of the DIP switch?
Can you eliminate the switch?
 

AnalogKid

Joined Aug 1, 2013
10,986
You don't say explicitly, but is sounds like the switch is pulling an input line down to GND, and the line is pulled up to 3.3 V. If so, power is lost in the pullup only when the switch is closed (other than the uC logic high input current). As above, what is the on/off duty cycle of the switch, on time per hour, on time per week, etc.

One consequence of a higher pullup value is that the circuit input impedance when the switch is open is higher. This makes it more susceptible to conducted and radiated interference that can cause false inputs, and ESD events that can lower the long-term reliability.

ak
 

AlbertHall

Joined Jun 4, 2014
12,345
One consequence of a higher pullup value is that the circuit input impedance when the switch is open is higher. This makes it more susceptible to conducted and radiated interference that can cause false inputs, and ESD events that can lower the long-term reliability.
As this input is from a DIP switch, those problems could be eased by having a capacitor in parallel with the switch.
 

Picbuster

Joined Dec 2, 2013
1,047
I am developing a battery powered device. In my circuit, there is a dip switch which is used as an input to micro controller pin. The supply to the switch is 3.3 v. Can I use a high value pull up resistor to minimize current in the circuit? Right now, the resistor value is 200k. Can I increase it to 1 Mega ohm.

Regards,
Mayuri
Depending on switching time you can use one high 1 M Ohm resistor to 3v3 and put a cap 100nF at input to gnd.
the switch is has a serial resistor and is connected at input pin ( 100Ohm to dissipate power stored in 100nF.
This takes time to load the 100nF not to be used in fast switching applications.
the input current 3v3 / (10 ^ 6 + input impedance mpu) < 3.3 micro amp.
if you do the opposite and load the cap via 1M the cap might load itself with noise resulting in a high level ( activate the circuit).
Resulting in a resistor needed and you will create a current again. ( bitter or kicked current 'thou' shall use).
Picbuster
 

Sensacell

Joined Jun 19, 2012
3,432
Heavily in favor of the switched pull up idea using the extra IO pin here.

- basically zero standby current
- fast, you can poll the switch at high speed
- good noise immunity
- no extra parts
 

LesJones

Joined Jan 8, 2017
4,174
If the dip switch is close to the microcontroller then this variation on post #11 may work and avoid the use of an extra I/O pin. The internal pullups could just be enabled at the time when the I/O pin was read. The disadvantage would be that for the rest of the time the pin could be floating. I don't know if adding high value pull down resistors would draw any significant current.

Les.
 

AnalogKid

Joined Aug 1, 2013
10,986
That only works if the switch can be polled, or the application knows when to expect a switch change. If the switch is completely random, or driving an interrupt, I think the pin-driven pullup has issues. Need more info from the TS.

ak
 

MrChips

Joined Oct 2, 2009
30,710
That only works if the switch can be polled, or the application knows when to expect a switch change. If the switch is completely random, or driving an interrupt, I think the pin-driven pullup has issues. Need more info from the TS.
TS did say it is a DIP switch which I would assume is for some kind of mode selection. I like the spare output pin solution.
The other thing to mention is that in a battery operated low-power application one ought to be looking at all methods of reducing power draw:

1) Use ultra low-power components
2) Go to slower clock speed
3) Power down all unused modules
4) Sleep or Stop mode where applicable

We're talking sub-micro-amp drain currents here.
 

Tonyr1084

Joined Sep 24, 2015
7,852
It's been 35 hours since he posted his question. I wonder if he's been abducted by aliens seeking to learn more about humans and their electronics.

He said he has a DIP switch connected to 3.3 volts. A DIP is either open or closed, so if it's connected to 3.3 volts then what's the need for a pull-up resistor? If the switch is closed (on) then there should be 3.3 volts applied to whatever pin he's connected to. If it's open (off) then that pin should be floating in the ether somewhere between on and off. If anything he'd want a pull-down resistor - that is if I read his intentions correctly. In that case a 1 meg ohm pull-down would hold that pin to zero volts. 100K or even 10KΩ shouldn't present a problem. 3.3 volts on 10KΩ - that's 0.00033 amps. (330 µA) When it's on it will be drawing 11 mW.

I don't get what he's attempting to do. If his switch is truly a DIP switch then it's either open or closed. Off or on. If it's connected to 3.3 volts then it's either 3.3 volts or no volts. (respectively)

A single pole double throw switch would eliminate the need for any resistors. One throw to 3.3 volts, the other to ground. The common to the pin in question. Which he hasn't told us about yet.
 
Top