Does this circuit make sense?

Thread Starter

Scalpel78

Joined Aug 11, 2013
56
Hi, I'm working on a breakout board for the MC33926 motor driver.
One of the output pins, SF (Statusflag), is an active-low which will go low when there is a problem with the driver.
In the datasheet it says that the SF should be tied to VDD via a pull-up resistor.
In addition, I want to add a red LED to indicate when there is a problem (when SF goes low).
Here is a screenshot of that part of the circuit:


This is probably really basic, but do I really need the left pull-up resistor? Will the Led and right resistor also function as a pull-up resistor?
 

crutschow

Joined Mar 14, 2008
34,280
Possibly not. With only the LED and series resistor, the pullup voltage at SF will likely only be somewhere around 5V minus the LED forward voltage and that may not be sufficient to register as a logic high by the /SF input.
 

MikeML

Joined Oct 2, 2009
5,444
Just use one 470 Ohm or 680 Ohm resistor and a 20mA Indicator LED. The second, parallel resistor is not required. Use a LED with a Vf of 1.8V or so...
 

ian field

Joined Oct 27, 2012
6,536
Hi, I'm working on a breakout board for the MC33926 motor driver.
One of the output pins, SF (Statusflag), is an active-low which will go low when there is a problem with the driver.
In the datasheet it says that the SF should be tied to VDD via a pull-up resistor.
In addition, I want to add a red LED to indicate when there is a problem (when SF goes low).
Here is a screenshot of that part of the circuit:


This is probably really basic, but do I really need the left pull-up resistor? Will the Led and right resistor also function as a pull-up resistor?
Assuming that's an open collector output - its just possible that any slight leakage could produce a faint residual glow in the LED.

The circuit appears to be technically correct, but if there's no residual glow problem, you could save the cost of a resistor.

If its for production - I'd leave the extra resistor in so future tolerance spreads don't cause embarrassment.
 

MikeML

Joined Oct 2, 2009
5,444
Assuming that's an open collector output - its just possible that any slight leakage could produce a faint residual glow in the LED.
I'll bet that residual leakage spec is on Motor Driver data sheet, and I'll bet it is less than 10uA over the entire voltage and temperature range for the part. That will not cause a "faint residual glow".

Ps Here is the data:

clip.gif
 
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ian field

Joined Oct 27, 2012
6,536
I'll bet that residual leakage spec is on Motor Driver data sheet, and I'll bet it is less than 10uA over the entire voltage and temperature range for the part. That will not cause a "faint residual glow".

Ps Here is the data:

View attachment 75964
Just possible - which is a fair bit short of even likely.

The circuit the OP posted is technically correct - and advisable on a production run.
 

Thread Starter

Scalpel78

Joined Aug 11, 2013
56
Thanks for great replies.
This is just for personal hobby, so I'll try removing the second resistor, and I'll use a red 0603 20mA LED with Vf 1.95V in series with a 680 ohm resistor.
 

MikeML

Joined Oct 2, 2009
5,444
Possibly not. With only the LED and series resistor, the pullup voltage at SF will likely only be somewhere around 5V minus the LED forward voltage and that may not be sufficient to register as a logic high by the /SF input.
SF is an output...
 

Lestraveled

Joined May 19, 2014
1,946
Wait a Minute. Not only is SF an output, but the "maximum permissible load current" is 0.5 ma. 500 micro-amps!! You need a buffer to drive that 20 ma. LED.
 

ian field

Joined Oct 27, 2012
6,536
Thanks for great replies.
This is just for personal hobby, so I'll try removing the second resistor, and I'll use a red 0603 20mA LED with Vf 1.95V in series with a 680 ohm resistor.
Just a suggestion - don't use an ultra-efficient LED, some types reach full brightness at only 2mA, even though you'd only get such a faint glow as only visible in darkness, if it was my project that would irritate me.
 

MikeML

Joined Oct 2, 2009
5,444
If you are so anal that you need to worry about the 5uA leakage through the LED with SF_ high, then add the 100K resistor across the LED as shown in the first circuit. I believe the first circuit will work fine as an indicator with or without the 100K. Try it both ways, and let us know.

89a.gif


If you think you need a higher current through the LED, then the second circuit will do it.


89b.gif
 
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MikeML

Joined Oct 2, 2009
5,444
Try it on a breadboard or with some clip leads. I'll bet that it is plenty bright enough to determine the state of the logic pin at a glance...
 
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ian field

Joined Oct 27, 2012
6,536
If you are so anal that you need to worry about the 5uA leakage through the LED with SF_ high, then add the 100K resistor across the LED as shown in the first circuit. I believe the first circuit will work fine as an indicator with or without the 100K. Try it both ways, and let us know.

View attachment 76049


If you think you need a higher current through the LED, then the second circuit will do it.


View attachment 76051
Genius! - that's exactly what the OP's first post said in the first place.
 

MikeML

Joined Oct 2, 2009
5,444
Genius! - that's exactly what the OP's first post said in the first place.
Not quite, his proposed solution was a resistor in parallel with the LED and its current-limiting resistor. Either method would supply the leakage current, although we can make the case that the way I did it, the resistor can be a higher value, so wastes less power.

I went a tried a high-brightness 20mA Red LED at 5V through 100KΩ. That would be about 30uA, six times more than 5uA leakage you are worried about. Not even a detectable glow. However, with ~500uA through it (5V through 6.8K) , it is bright enough.
 

ian field

Joined Oct 27, 2012
6,536
Not quite, his proposed solution was a resistor in parallel with the LED and its current-limiting resistor. Either method would supply the leakage current, although we can make the case that the way I did it, the resistor can be a higher value, so wastes less power.

I went a tried a high-brightness 20mA Red LED at 5V through 100KΩ. That would be about 30uA, six times more than 5uA leakage you are worried about. Not even a detectable glow. However, with ~500uA through it (5V through 6.8K) , it is bright enough.
High brightness and ultra efficient are 2 different types.
 
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