I'm making a circuit to control two motors and now i am using a 6v motor, and as the datasheet of the l293d says, we'll get a typical 2.6 voltage drop across the output pins right ? So if i put a 7.4v battery, it will receive 4,8, which is not a very good value. There are 8.6v batteries and 8.4v in the stores, but for me they're really expensive, i was looking for an alternative way to get around 5.5-5.8v in the motor with a 9v battery that is cheap, can you guys help me ? The diagram i have for now is this one:

http://i.imgur.com/suqL0vg.png

 

There are better H-Bridge chips available.

 

I am limited to this

 

Can you put two of your 7.4 volt batteries in series for the motor supply?

 

I could, but i dont see the point, it would just require a heatsink in LM7805 and the drop would still destroy my motors...

EDIT: The motors i'm using use around 150mA, is there a way to know the exact drop voltage on the L293D?

 

Well I guess you're well and truly up against it then. Without a diagram it is hard to know why you think using a higher voltage battery won't give you the same problems that using two of the ones you have will. But never mind what I don't understand. Why can't you run the 7805 off of the lower one and the motors off the two in series. BTW motors hardly care about voltage, what they care about is current.

 

The specs says the motor can only deal with voltages between 3 and 6. I intend to use a 7.4v battery and expect a drop voltage at the l293d output pins of around 2 volts, that will give me 5.4v for the motors, which is a good value... But i wanted to know how to calculate the drop voltage at this IC, is it possible ?

 

The specs says the motor can only deal with voltages between 3 and 6. I intend to use a 7.4v battery and expect a drop voltage at the l293d output pins of around 2 volts, that will give me 5.4v for the motors, which is a good value... But i wanted to know how to calculate the drop voltage at this IC, is it possible ?

Have you studied the data sheet?

 

Yes, but i am just a begineer at this, if you could advise me. The datasheet says that with a 0.6A load there is a typical drop at the ouput pins for the motors of high 1.4v and low 1.2v at the transistors, so it sums up to 2.6v, but my motors will use at 6v 150mA, so two motors 300 mA, that's half the load at the datasheet, BUT i don't know if it is a linear relation between the load and drop, it think its not, it is ?

 

No, there is a small resistive component, but the bulk of the drop is caused by the Darlington pairs (both P and N)

 

So i should expect a minimum drop... ?

 

The specs says the motor can only deal with voltages between 3 and 6. I intend to use a 7.4v battery and expect a drop voltage at the l293d output pins of around 2 volts, that will give me 5.4v for the motors, which is a good value... But i wanted to know how to calculate the drop voltage at this IC, is it possible ?

The reason they say that is because people hook motors directly to power supplies. The motor will survive this if the voltage is limited. You have some flexibility to use higher voltages if you limit the current so that the power is similar. For example if the motor draws 150 mA @ 6 volts, that is 900 mWatts. If you want to use 9 Volts and limit the current to 100 mAmps the motor should be fine with that. I'm trying to give you options here because you can't use a better driver.

 

So i could limit the current with a resistor ? Counting the drop voltage at the l293d output of course. Should i use for calculations the typical value of 2,6v drop?

 

So i could limit the current with a resistor ?

That would be one very inefficient way of doing it. Are you trying to run a motor or make a heater? You're already losing voltage in the L293 and I'm trying to get around that. You could use an LM317 as a current regulator. You goal would be to have enough voltage from two 7.4 batteries in series to overcome the L293 limitations and supply up 150 mA to the motors at whatever voltage that requires, so long as the motors don't have a power dissipation problem. In that case you back off on the current and the voltage across the motor for the lower current will go down.

Also don't forget that the resistance in the motor windings will increase with temperature further limiting the current.

Will these motors be running continuously?

 

They are the wheels of a RC car. I dont know the lm317 component

 

The LM317 is a three terminal device that can be used as a voltage regulator, similar to the 7805 except adjustable. It can also be configured as a current regulator. Google "LM317 datasheet", or better yet:
http://lmgtfy.com/?q=LM317 datasheet
http://users.telenet.be/davshomepage/current-source.htm

 

I'll take a look at this and return. Whats you suggestin is thst i still use the LM 7805 to adjust the main battey for 5v and then use the lm317 to regulate the current for the motors??

 

Essentially the 7805 is only connected to one of the two batteries. Two batteries in series provide the supply for the motors and the LM317's, one for each motor will regulate the current.

 

I would humbly ask of you to make a diagram for that... i tried to attach the lm317 to the output pins of the l293d but i dont think this is the way to go. Can you show me Papabravo ?

 

I would humbly ask of you to make a diagram for that... i tried to attach the lm317 to the output pins of the l293d but i dont think this is the way to go. Can you show me Papabravo ?

The second link in post#16 has a drawing which shows the LM317 with its input connected to a power source and the output with a feedback resistor to the adjust pin.