Problem with l293d voltage drop

MikeML

Joined Oct 2, 2009
5,444
First, the output stage of an L293 looks like this (from the data sheet):

195a.gif

Here is a simulation of what happens at the L293 output pins (voltage across the motor) on two sides of the H-bridge as a function of different battery voltages (x-axis) and different effective motor resistances. Green is for R(motor)=15Ω; Red is for R(motor)=20Ω; Blue is for R(motor)=30Ω.

195b.gif

Note that the voltage drops are quite independent of the actual motor current. With a 7.4V battery, you would be getting ~5.6V across the motor. Looks ok to me...

The L293 (like this) screwed me on a project once; that is why I am down on it...
 
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MikeML

Joined Oct 2, 2009
5,444
I missed the discussion about the LM317. Are you proposing to use it in-series with the motor? If so, you would need two if you are reversing the motor? The minimum drop-out voltage across a LM317 makes this a really bad idea.
 

Thread Starter

André Ferrato

Joined Apr 5, 2015
215
My concern at the time is this one, i have a 7.4v connected to the Vcc2 pin to power the motors, the motors work in a range of 3v-6v, i'm concerned that the drop in the upper and low transistor at the l293d will be so high that the motors wont have the strength to move the car. These are the numbers i have:

MOTOR DATA:



The L293D states a typical 2.6v drop being 1.4v at the high transistor and 1.2v at the low transistor while driving a 600mA load. As i observe in the Gear Box parameters, the max load it will draw is 150mA, so i can expect a lower drop at the transistors right ? But the load of the motors will be decided by the voltage it receives right ? So how to problematize that ? I don't know the exact drop, so i don't know what voltage will they receive, so i don't know the load... I may be lacking some knowledge at this area, that's why i'm bringing this to you guys. So basically one motor per channel, 7.4v battery to feed the motors with the current data chart. Any help of what will happen ? The LM317 idea was to put it to limit the current to the 150mA and don't care about the voltage as Papabravo stated. halp :/

EDIT: I looked at the graph you sent and a motor with internal resistance of 30Ω will receive around 5.75v ? With a 7.4v battery ?
 

Thread Starter

André Ferrato

Joined Apr 5, 2015
215
Sorry, that is because i usually want to understand why and how :/. Would you mind answering some questions just for curiosity ? And about the heat, why do you say that?
 

MikeML

Joined Oct 2, 2009
5,444
P=IE = (7.2-5.7)*0.15 =0.225W
Chip power = Vcc1*Icc1 + Vcc2*Icc2 = 7.2*0.06 + 7.2*0.02 = 0.6W

Total chip power ~0.8W, so it will need some air circulation...
 

Thread Starter

André Ferrato

Joined Apr 5, 2015
215
It will be open air, i think it will suffice... I wonder why there arent resistance values for the motors in the data charts, is it variable ?
 

SgtWookie

Joined Jul 17, 2007
22,230
When you use the LM317 as a current regulator, you wind up with a minimum dropout voltage of around 3v across the regulator (~1.7v from IN to OUT, and another ~1.25v from OUT to ADJ) so you'd have to compensate for that by adding yet ANOTHER couple of 1.5v batteries in series that would dissipate their power by heating up the LM317.

You really need to look for a better H-bridge that's MOSFET-based, or your payload will wind up being a truckload of batteries.
 

Thread Starter

André Ferrato

Joined Apr 5, 2015
215
I'll stand by my LM7805, my 7.4v battery and initial diagram, i'll be making a post with a schematic and the board to ask about somethings later this day, Thanks guys, is just awesome to have a forum like this!
 

MikeML

Joined Oct 2, 2009
5,444
It will be open air, i think it will suffice... I wonder why there arent resistance values for the motors in the data charts, is it variable ?
There are voltages and currents in the motor table. I dont have a Spice model of a motor (actually too lazy to build one) so it was easier to model the motor as a resistor. Actually, as the motor load changes, as the rpm changes, the effective motor resistance (ratio of motor voltage to motor current) changes all over the map.
 
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