Protection circuit ULN2803

[jorginho1877]

Hello Everyone

In this post I want to ask if you know of any circuit that can protect the integrated ULN2803, because when I make the connection to a stepper motor it overheats and stops working correctly, or I advise to use another reference to control the motor Step by Step?

ULN2803

I am using a voltage of 9v on the COM pin with a current of 0.6 A

 

[gophert]

Hello Everyone

In this post I want to ask if you know of any circuit that can protect the integrated ULN2803, because when I make the connection to a stepper motor it overheats and stops working correctly, or I advise to use another reference to control the motor Step by Step?

ULN2803

I am using a voltage of 9v on the COM pin with a current of 0.6 A

The negative wire (ground) should be on the common pin. The various outputs should connect to the motor (-) and the motor (+) should be connected to 9v supply.

This is known as "low-side switching" (The Transistors from the uln2803 is between motor and ground).

 

[gophert]

See images...

 

[audioguru]

The ULN2803A has a maximum allowed current from one output of 500mA so you are overloading and cooking yours with 600mA.
Is more than one output turned on at the same time?
What feeds the inputs and at what voltage?

 

[jorginho1877]

The ULN2803A has a maximum allowed current from one output of 500mA so you are overloading and cooking yours with 600mA.
Is more than one output turned on at the same time?
What feeds the inputs and at what voltage?

Hello

the eight outputs are turned on, because I'm controller 2 stepper motors, the ULN input is 5v

 

[Nigel Goodwin]

Hello

the eight outputs are turned on, because I'm controller 2 stepper motors, the ULN input is 5v

So you're massively overloading and cooking the chips.

Do you have a decent heatsink on the chip?, at least that would help a bit - as would using two different chips, one for each stepper (and perhaps paralleling pairs of drivers in each chip as well, although I'm sure how well the drivers would current share - but perhaps worth a try?).

 

[KeepItSimpleStupid]

The negative wire (ground) should be on the common pin.

Nope! Common is common of all of the free-wheeling diodes. You have in and ground used so you get inputs to the base of the transistors. Out goes to one end of the solenoid and that has the cathode of the diode connected.

Now where would the anode of that built-in diode go?

The other end of the relay/solenoid which happens to be the +supply to the relay/solenoid and they can go to a COMMON point, the +supply of the solenoids/relays.

 

[gophert]

Sorry

Nope! Common is common of all of the free-wheeling diodes. You have in and ground used so you get inputs to the base of the transistors. Out goes to one end of the solenoid and that has the cathode of the diode connected.

Now where would the anode of that built-in diode go?

The other end of the relay/solenoid which happens to be the +supply to the relay/solenoid and they can go to a COMMON point, the +supply of the solenoids/relays.

, I mis-named the pin (in the text, change common to ground) but the accompanying schematic is correct.

 

[KeepItSimpleStupid]

Sometimes I hate words.