L293D unused pins

[Dermotx]

Hi,
I'm using an L293D to drive a solenoid. I'm using pins 2 and 7 to trigger the pulse and 3 and 6 are connected to the solenoid.4,5,12,13 are of course grounded but I am not using 9,10,11,14,15 and I don't have these pins connected to anything. Pins 1,8,16 are powered at 5V. Should I have the unused pins grounded or what should I connect them to?

Thank you,
Regards
Dermot

 

[manuka]

Welcome Dermotx! Guess you're familiar with L293D resources such as <A href='http://www.me.umn.edu/courses/me2011/robot/technotes/L293/L293.html ' Target=_Blank>External Web Link</a>? You're only using half of the L293D of course, so AFAIK it's quite OK to leave the other terminals unconnected. In fact just driving a solenoid is rather a costly no brainer for such a versatile IC- have you considered the dirt cheap <A href='http://www.picaxe.orcon.net.nz/npnamp.gif ' Target=_Blank>External Web Link</a> or even a simple DIY Darlington pair? What rating is your solenoid ? The L293D is only good for &lt;1A &amp; some solenoids are real power hogs. Hence does it get warm/hot to touch at present when activated?


Edited by - stan. swan on 27/05/2006 18:22:55

 

[Dermotx]

Hi Stan,
thanks for the reply. The links you gave me were very useful. It said that I should leave pins 10 through 15 disconnected and showed how to parallel the 4 channels. I did this i.e. connected the inputs and outputs of channel 1 to channel 4 and channel 2 to channel 3 and am now getting consistant outputs to my solenoid.
I am still experiencing a problem however.
The voltage I am using to power pins 1,8,9 and 16 is 5V.
When I input 5V into input 1 (pin 2) I get voltage of 5V at output 1 (pin 3) and 0.1V at output 2 (pin 6). So far so good.

When I connect the solenoid across pin 3 and pin 6 I get only 4.1V at pin 3 which I would expect as this drop is probably due to the flyback diode in the L293. The problem is a get a voltage of 0.8V at pin 6 which is attached to the other solenoid lead! Although the solenoid still clicks I now only have 3.3v across it and it is a 6V solenoid I'm worried that when the battery voltage falls a bit there won't be enough voltage to drive the solenoid.

The solenoid has a resistance of 15 ohms. It is a single coil latching bistable and I energise it with a 15 millisecond pulse to turn it on or off by reversing the direction of the current with the L293D.
The reason I use the L293D is because it takes up a lot less space than four transistors, four resistors and four diodes which I would need to make a H-bridge out of discrete components.

Thanks for your time and thanks for your help.

Regards
Dermot

 

[manuka]

Aha- a latching solenoid. You'd not mentioned this initially &amp; I'd though just a simple type. As 5V/15 Ohms = .33Amp, the current is L293D tolerable. Does the solenoid get warm? What battery are you using - an SLA? Checked supply voltage ? What is pulled in solenoid doing ? Tell us as much about your circuit as you can. Digital pix upload somewhere ?

Thought -perhaps rustle up a more detailed ASCII art schematic using the well thought of <b>AACircuit </b> <A href='http://www.tech-chat.de/download.html ' Target=_Blank>External Web Link</a>? Things look in this style <code><pre><font size=2 face='Courier'>









__ /
+---------+o| |o- o o _
-o| |o- |(
-o| |o- |(
-o|__|o--------| |(_
|
|

GND </font></pre></code>


Edited by - stan. swan on 28/05/2006 00:47:55

 

[Michael 2727]

If you are getting a voltage drop like that
it sounds as if your battery may be a little
small for the application. what sise or mAh
is it ?
Some solenoids will also need twice the holding
current to initialy pull in. If you run a few of
these solenoids you may need to upgrade the
supply bypass/filter capacitors around the
picaxe chip to avoid dropouts etc.
What duty/time are these solenoids required
for and how big is the battery would be the most important
question.
Is the battery backed up from mains or solar
etc.

 

[Dermotx]

Hi Stan and Michael,

Thanks for reply.

The situation has moved on somewhat. In order to get rid of as many variables as possible I have taken the L293D out of the circuit and now have it on a CADET breadboard. This is a mains operated educational breadboard setup with a stable and well regulated 5V DC supply.
I have the 16 pins of the L293D connected as follows:

Pin 1..........5V

Pin 2..........I connect the 5V supply to this pin to trigger pin 3 high. In the untriggered state it is at 0.1V.

Pin 3...........Output to solenoid 4.1V when pin 2 is triggered. In untriggered state it is 0.1V.

Pin 4.............Grounded.

Pin 5.............Grounded.

Pin 6.............output to solenoid. When pin 2 is triggered it goes up to 0.8V. Otherwise it is ar 0.1V.

Pin 7..............Normally at 0.1V but when I want to trigger current to go in opposite direction to reverse solenoid I connect it to the 5V supply. When pin 7 is triggered I get the opposite voltages at pin 3 and pin 6 than when pin 2 was triggered; i.e. pin 3 goes to 0.8V and pin 6 goes to 4.1V.

Pin 8............connected to 5V supply.

Pin 9............connected to 5V supply.

Pin 10...........paralled with pin 7.

Pin 11...........paralled with pin 6.

Pin 12...........grounded.

Pin 13...........grounded.

Pin 14............paralled with pin 3.

Pin 15.............paralled with pin 2.

Pin 16.............connected to 5V supply.

Stan, I tried the ASCII software and it's great but unfortunately with all the lines crossing it obscured the L293D pins.

In addition I replaced the solenoid with a 15 ohm resistor just in case that was the problem. Problem still persists.

As I have paralled the inputs and outputs floating pins are not the problem as all pins are connected to something.

So all I have in the circuit now is the L293D and connections between the pins and connections to ground and a stable 5V supply.

The solenoid doesn't get warm but I don't leave it on for long. That won't be a problem in my circuit as it will only have to be energised for 15 milliseconds anyway every few minutes at most.

The batteries I was using in my circuit were four AA NiMh 1700mAh which gave me a voltage of 5.2V. This is pretty much irrevalent now because I am still experiencing the same problem with a stable 5V power supply and nothing in the circuit except a 15 ohm resistor and the L293D.

Just for info the original circuit is used to turn on and off the supply of water to a central heating radiator on a timed basis and on a temperature basis.

Thanks again for the help. It really is quite perplexing.

Regards
Dermot

 

[Dermotx]

One other thing I forgot to mention. I replaced the original L293D with a new one just in case I had damaged the original. Same problem exists.
Dermot

 

[Dermotx]

Hi All,

Just to close off this topic, the problem with the fall off in voltage that I was experiencing when using the L293D is in fact normal. Please see excerpt from article below.

Also I have discovered from an internet search that if you only want to use one circuit in the L293 that you leave pins 9 through 15 disconnected. I have tried this and it works fine.

Regards

Dermot


&quot;When the L293 chips drive a motor, there is a voltage drop across the transistors that
form the H-bridge. The transistor connected to motor ground (0 volt potential) might
drive the motor at some voltage between .2 and .8 volts; the transistor connected to
the positive terminal of the battery (say it's at 6 volts) might drive the motor between
5.2 and 5.8 volts.
The amount of this voltage drop is proportional to the amount of current being
supplied by the motor-driving transistor. When more current is being supplied, the
transistor drops more voltage.
This undesirable property of the L293 transistors is exploited to give a crude mea-
surement of the amount of current being driven through the motor. A fundamental
property of motors is that as the amount of work they are performing increases, the
amount of current they drawn also increases. So the current measurement yields data
on how hard the motor is working|if it is turning freely, if it is stalled, or if it is
working somewhere in between.&quot;