I got the L298N motor driver. I now (after ordering it) found that it has voltage drops of about 1.9 volts. I have a 12v power supply for this 12v motor so this driver is not going to work. Also, the stall torque of the motor is 3.6 amps and this motor driver can't handle that either (definitely did not do enough [any] research on the driver before ordering it). So I found this new (overkill) driver.
This new driver can obviously have a lot of current drawn through it. Also, it can handle 12 volts. My question, being a beginner in power electronics is, how much voltage drop does it have (that is, of course, a big thing to know considering my power source and motor have the same voltage ratings), and how do I find this out using the data sheet? In addition, how many volts below the nominal voltage of the motors is acceptable? My tolerances in terms of torque and speed are pretty tight so how much of a voltage drop is okay to have small performance drops in the motor specs? Should I have planned ahead and bought a battery with than 12v to account
Finally, I am simply wondering if anyone happens to know of a driver that meets these threshold of specs but also provides a 5v regulated output like the L298N does? This seems like a very convenient feature I would be willing to pay more for. This new driver I found does not have that feature.
This TLE5206-2G IC seems good. I just wanted to check and make sure I understand its operation before I destroy something. On page 15 of the datasheet it says:
Sign/Magnitude Control
For this mode two ports with PWM capability are necessary. Motor turns clockwise
(current flows from OUT1 to OUT2; means: OUT1 is switched HIGH continuously and
OUT2 is PWM controlled.
Then it shows a diagram. Am I correct in saying the following knowing that my power supply is 12v?
The difference of IN1 and IN2 is proportional to that of OUT1 and OUT2. For instance, if IN1 is set to 5v and IN2 is set to a PWM signal of 1v, this is a difference of 4v out of the 5v. Proportionally to 12v, OUT1 and OUT2 now have a difference of 4/5 * 12 or 9.6v or 80% power.
PureStress:
This TLE5206-2G IC seems good. I just wanted to check and make sure I understand its operation before I destroy something. On page 15 of the datasheet it says:
Then it shows a diagram. Am I correct in saying the following knowing that my power supply is 12v?
The difference of IN1 and IN2 is proportional to that of OUT1 and OUT2. For instance, if IN1 is set to 5v and IN2 is set to a PWM signal of 1v, this is a difference of 4v out of the 5v. Proportionally to 12v, OUT1 and OUT2 now have a difference of 4/5 * 12 or 9.6v or 80% power.
Is this correct?
I'm not sure what you are trying to say but I think you have it all wrong.
For motion in one direction you hold IN1 LOW and apply PWM to IN2 (and vice versa for the other direction). The motor speed depends on the PWM duty cycle in the normal way. Look at the Input Logic Truth Table in Table 1.6
You can also hold IN1 HIGH and apply PWM to IN2 in which case analogWrite(0) will represent maximum speed.
You can't have a PWM signal of 1v. A PWM signal is a sequence of HIGH and LOW signals (5v and 0v).
Thanks, I’m pretty sure we have the same idea. By 1v PWM I meant a PWM signal with an average voltage of 1. So if IN1 is set to 0v, OUT1 will have 0v. If IN2 is 5v, OUT2 will output the full 12v I supplied the IC. If I supply IN1 with 0v and IN2 with a 50% duty cycle 5v PWM (average of 2.5v), OUT1 will output 0v and OUT2 will output an average of 6v (half of 12v).
Averaging with PWM and motors only happens if synchronous rectification(*) mode is used which is
basically linear in response. This is the mode to use for position feedback if a stable positio PID loop
is desired.
slow or fast decay modes are non-linear and many factors affect the mapping from PWM value to
"effective voltage".
For routine speed control decay modes are often used because they are simpler in hardware and
somewhat less lossy than sync rectification.
(*) in this mode the windings are always voltage driven, never floating. You get active braking
for free this way, note.
how much voltage drop does it have (that is, of course, a big thing to know considering my power source and motor have the same voltage ratings), and how do I find this out using the data sheet?
The BTS7960 module is a MOSFET H-bridge, so you calculate voltage drop from the on-resistance which
is 16 milliohms. 3.6A x 0.016ohm = 60mV or so of loss at stall - utterly negligible.
In addition, how many volts below the nominal voltage of the motors is acceptable?
Totally depends, the speed is a linear function of voltage in a DC motor - how slow is acceptable?