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Quick Question About a LM2901N

Supercap2F

Mar 22, 2014
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Hey Everyone!

First of all it appears that the LM2901N is not able to put out a high level voltage. Is that true? If so then I will need a pull up resistor to drive a CMOS input. What value should the pull up resistor be? Is there a formula to calculate it?

Thanks Everyone. :)

Dan
 

Harald Kapp

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The LM2901 is a so called open-collector component. This means that it can pull to ground (low), but positive potential (high) has to be provided externally, typically by a resistor.

Advantages of this type of circuit are:
  • can be used to level-shift by using a different voltage for the output circuit
  • can be used to vreate a so called wired-or (Google) to combine several outputs into one logic signal
The value of the pull-up resistor is entirely up to you. Probably the two most important issues to consider are:
  • power requirements
  • speed
  • output current in high state
Power requirements:
In the low state of the output the resistor takes the full voltage from Vcc to Gnd and will consequently dissipate power. You want to have a large resistance to dissipate very little power.

Speed:
The resistor forms a low pass filter together with all the capacitances that are connected to the output (wire traces, input circuits etc.). You want the output to switch fast, therefore you make the resistance small to minimize the RC time constant.

These requirements are contradictory: a large resistance will dissipate little power, but slow down the circuit. And vice versa. Therefore you have to find a suitable compromise, taking into account factors as e.g. the expected mean on-time of the output transistor (aka duty ciycle of the output signal). If the output is on for a comparatively short time and off for a long time, you may be able to tolerate higher dissipation during that short time much easier than if the output is low for a long time.

Output current in high state:
The load current (from other inputs connected to the output) will drop some voltage across the load resistor. This voltage drop has to be small enough to still ensure a logic high level. With CMOS gates the input current is (almost) negligible and therefore not an issue. If you drive a load that needs considerable current (e.g. a bipolar inpu circuit) you have to take into account the input current of this circuit as it will typically not be negligible.
 

Supercap2F

Mar 22, 2014
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It does not need to be very fast... So how does a 5K6 resistor sound?

Thanks

Dan
 

Arouse1973

Adam
Dec 18, 2013
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Hi Dan
The device can sink 20mA and has a internal base current of 80uA so that's a gain of 250. The value of resistor will depend on the desired response time of the output. But there are min and max values, you will have to choose one in between that fit the requirements. Say 4K7 to start with.

To work out the min and max values as follows.

VinH = ( High Level of gate your driving)
VinL = (Low Level of gate your driving)

I pull up min = Leakage through comparator when off + Leakage current through logic gate your driving

Rmax = V+ - VinH / I pull Up min
Rmin = V+ / 20mA(max current of comparator) - Leakage through logic gate.

Thanks
Adam
 

Arouse1973

Adam
Dec 18, 2013
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You beat me again Harald :) I must have been typing this as you posted.
 

Supercap2F

Mar 22, 2014
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Hey Adam!

Can you explain where to find the "leakage through comparator when off" and "leakage current through logic gate" on a data sheet? :D

All the best

Dan
 

Harald Kapp

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The "leakage through comparator output when off" would be the high-level output current, 0.1nA for this comparator.
The "leakage current through logic gate" would be the input current of the logic gate, which in the case of a CMOS gate is a leakage current because the input is voltage drivven and input current is ideally 0mA. Other logic families, e.g. the venerable 74xx series, may have much higher input currents in the mA range.
 

Supercap2F

Mar 22, 2014
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OK heres what I came up with.

Gate #1 = 1uA leakage, 3.5V high, 1.5V low
Gate #2 = 1uA leakage, 4V high, 1V low
Supply = 5VDC

Gate #1 max resistance = 999,900Ω
Gate #2 max resistance = 1,499,850Ω
Gate #1 min resistance = 250Ω
Gate #2 min resistance = 250Ω

So a 2KΩ resistor will work good. I picked that value because I think the bands on it are nice looking. LOL Remember that Harald? :p

Thanks so much for the help! :D

Dan
 

Harald Kapp

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I'd prefer black - red - yellow, unfortunately this is an invalid color code for resistors ;)
 

Arouse1973

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The "leakage through comparator output when off" would be the high-level output current, 0.1nA for this comparator.
The "leakage current through logic gate" would be the input current of the logic gate, which in the case of a CMOS gate is a leakage current because the input is voltage drivven and input current is ideally 0mA. Other logic families, e.g. the venerable 74xx series, may have much higher input currents in the mA range.

Thanks Harald
Adam
 
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