what's the 10ohm resistor to VCC in this bidiredtional motor driver schematic? - Page 1
Forum section: electronics / beginners

author: dentaku
author_rank: Frequent Contributor
timestamp: February 05, 2015, 04:05:08 am
content: I was experimenting with an old LB1641 Bidirectional motor driver IC. It does what it supposed to do and drives a small DC motor at 9V forwards and back just fine, controlled from my Arduino UNO.http://www.alldatasheet.com/datasheet-pdf/pdf/544103/ONSEMI/LB1641.htmlSO... I hooked uptwoof them and made them go through a sequence that drives a bi-polar stepper motor (4 wires) at 9V to 12V and it actually works because it lets me control the polarity of each winding.I know I could just use an H-Bridge but this was an experiment and learning experience.While this works, it's impractical and following the schematic from the datasheet,except for the zener which I omitted, I see that it has a 10ohm resistor between pin8 (VCC2) and the positive rail.Obviously whenab-using the LB1641 this way that 10ohm resistor gets too hot.Question: What's this resistor for? VCC1 doesn't have a resistor.

author: Skimask
author_rank: Super Contributor
timestamp: February 05, 2015, 05:27:12 am
content: Just like the datasheet shows... One feeds the motors directly, the other feeds the logic directly.What's the question again?

author: codeboy2k
author_rank: Super Contributor
timestamp: February 05, 2015, 06:29:45 am
content: that's for current limiting. The 10-ohm is only an example for their datasheet, driving a 6V motor. You need to choose an appropriate resistor for your own motor's voltage and current draw, and limit the current to something less than 1.6A (max) and Vout < 7.4V Max and transistor saturation voltage is about 2.2V + 0.6V (top and bottom BJTs), and then the device needs to also stay under it's maximum power dissipation (Pd max =1.2W)Using ohms law, you can calculate the voltage drops from VCC2 through the current limit resistor, then Vsat1 and Vsat2 and then the motor windings as the load down to ground.Figure that out, then you can also figure out the size (in watts) of the resistor you will need there to prevent the resistor from heating up too much. P = V * A , where V is the voltage dropped across the resistor, and A is the max current through the resistor.Also, since P = V * A so make sure that (Vsat1 + Vsat2) * (current) < Pd Max of the device. If you exceed Pd max you need to increase the resistor value to limit the current even more. Iterate until you get everything within the max limits.Then you might need to heat sink the device to keep operating temps below max (75C in this case)

author: dentaku
author_rank: Frequent Contributor
timestamp: February 05, 2015, 03:27:03 pm
content: Quote from: codeboy2k on February 05, 2015, 06:29:45 amOK, I guess I've got to do some calculations. I've got some big resistors that can handle much more power."Vout < 7.4V Max" doesn't help me much though because my stepper motor won't even turn unless you give it a nudge at that low of a voltage.

author: codeboy2k
author_rank: Super Contributor
timestamp: February 05, 2015, 05:37:36 pm
content: On re-reading the datasheet, the 7.4V max was in the absolute maximums section of the datasheet, so I assumed you could go no higher.But there is also a chart that shows up to 16V with a 16V zener on the zener pin of the IC.So if you go by the chart, I think you can go up to 16V output.You still have to stay within all the other maximums I mentioned, like max current and power dissipation.

author: dentaku
author_rank: Frequent Contributor
timestamp: February 05, 2015, 07:05:34 pm
content: I knew that sounded weird.The first sentence of the datasheet even says "capable of direct driving 6V, 9V, 12V motors"

author: dentaku
author_rank: Frequent Contributor
timestamp: February 06, 2015, 02:52:54 am
content: I just did some experimenting and found that the stepper I was working with is fine down to 7.5V then I dug up two big 10ohm cement resistors from one of my bins and they certainly don't get now.I'm going to stop messing with the LB1641 drivers before I blow one up and keep them as regular DC motor drivers. They work very well for their intended purpose so I might end up using them for something one of these days.I wish I could scrounge up some unipolar steppers somewhere so all I would need as a Darlington array instead of an H-Bridge. I seems like most dead printers and scanners I find only have bipolar ones.