Hello Valentina,

what happens if you switch the outputs of your microcontroller (slowly, only a few Hz) between logical 0 and 1? Do the relays follow?  If so, the ULN2003 is working fine, and you just need to switch the outputs of the dsPIC to the 0-state, initially.

But I wonder about the relays: In the ON-state, the voltage at the outputs of the ULN2003 will be 1.3 V or more. So, if one side of a relay coil is connected to +5 V and the other side to an output, that leaves only 3.7 V or less for the coil. Do the relays really work at such a low voltage? (My suspicion is that you made a mistake when connecting the relay coils; sorry if I'm wrong.) Did you measure the voltage at a coil?

Hi Valentina,

"The power supply of the uln2003 is 15v . . . . . These relays have their coil connected to a 5V power supply. "

Unless you have freewheeling diodes (or at least RC combinations) connected to your relay coils, you may have "shot" the ULN. Usually one would connect the ULN supply to the same supply line as the high-side of the loads - the 5 V rail in your case.

When it comes to bipolar transistors, "shot" usually means shorted. Which would explain the immediate switching when powered. And you should be aware that the internal freewheeling diodes if the ULN series are not too robust; an external freewheeling diode does aid in "survival" of these chips when driving inductive loads.

I do not completely agree with Joerg as your relays will hardly consume beyond 40 mA, so the saturation voltage of the outputs will be lower than 1 V. And these relays switch at voltages well below the nominal 5 V as they are intended to be operated by some semiconductor output stage. (3/4 of the nominal voltage would not come unexpected.)

Regards

Hello Valentina and Ulrich,

provided each output has just to switch 40 mA, I agree with Ulrich's less than 1 V. But there are 5V-relays with a nominal current of roughly 100 mA, too, and Valentina wrote: "... i have connected 3 relays in each output.", so that would add up to 300 mA ...

I understood that the freewheeling diodes are connected to +15 V in order to ramp down the current in short time; this would also relieve the diodes, power dissipation-wise. The only problem that might occur in this case is the need for some kind of power sink at the 15 V line.

BTW, the rule of thumb: "Any electronic device that gets warm isn't completely defect." leads to the question: When powered for some minutes with all these relays switched on, does the ULN2003 warm up?

That's great: "Any electronic device that gets warm isn't completely defect." :)

In the past I've encountered bipolar transistors as well as some MOSFETs that had turned LET (Light Emitting Transistor). They were no more than bondwires with some piece of silicon between - still getting very, very "warm".

Yes, those LETs are the reason it's a RoT and not a law of nature. ;-) It depends on the surrounding circuit, of course.

But in the actual case (3 (?) parallel relay coils in series with each Darlington transistor) I bet that a ULN2003 with shot transistors is considerably cooler (only in the literal sense) than a faultless one in the ON state.

Now let's see what Valentina is answering.

BTW: I was assuming the 3 relays occupy 3 outputs - not a single one.

Regards

Hello everyone,

thank you so much for your answers! Here is a schematic for my circuit. What do you think? Should I supply the uln2003 and the relay with the same voltage 5V to work right? Moreover, the uln2003 is driven by the dspic30f4011 which outputs 5V.

Hello Valentina,

your circuit looks fine to me, provided that the energy transmitted to the 15 V supply while switching a relay off is somehow led off (e. g. by a load resistor) otherwise the voltage would rise beyond 15 V and / or your 15 V power supply might be damaged.

Thank you all very much! I will try to implement your suggestions. You help me a lot!

Have a nice day :)

I mostly agree with Victor - - - with some deviations:

In this state an connected relay is expected to be "on".

Good luck with that!

Hello, could anyone  explain to me why this happens: When  I have not put the ic circuit uln2003 on my pcb, the 15V supply reaches my pcb and with the 7805 I convert  it from 15V to 5V. However, when I connect the uln2003 on my pcb, the supply voltage drops to 5V from 15V.

Measure the vootages (5 V, 15 V) with an oscilloscope!

Could be that you've got an oscillation due to overloading the power supply (15 V) - especially if current limited. It would also help if you supplied your complete schematics.