ULN2003 sinking 5 and 12 volt signals. Where to tie common pin.

Dave E

Hello all.
I am using a ULN2003 IC to boost the power of my stamp outputs. The issue is that I am sinking both 5 and 12 volt signals. I assume that I tie the kick back diode common pin (pin 9 on ULN2003)·to +12 volts since that is the higher of the two voltages used.

See page 4 of the attached N&V article for a functional drawing of the IC.

The reasoning is that if the diode common pin is at +12 volts then under normal circumstances , that is with the output off so no pins actively sinking current,·depending on weather the output is tied to a 5 or 12 volt inductive load (relay coil), the·diode will have either 5 or 12 volts at the anode and 12 volts on the cathode so the diode will not be forward biased.

However, if the common pin is tied to +5 volts, then the diode associated with the 12 volt relay coil will have 12 volts on the diode anode and 5 volts on the diode cathode. This will forward bias (with gusto) the diode.

So, can someone confirm that as a general rule of thumb, when using the ULN2003 ICs, the common cathode pin should be tied to the higher supply voltage when using more than one supply voltage on the load side of the outputs.

Thanks for the input.
Dave E

Post Edited (Dave E) : 4/23/2009 1:07:52 AM GMT

kwinn

You are correct. The common pin for the diodes must go to 12V or you could have 12V going back to your logic voltage ad most likely cook things. The only connection common to the logic and load voltage would be the ground.

PJAllen

I disagree with the premise.· A free-wheeling diode is to be connected across·a coil, that's what's important, that's how it effects protection and·not by some connection to some or no·particular·voltage source.

Dave E

Thanks for the replys.
I thought I had things right but this was the first time that I was using two different voltage levels and I do not remember seeing anything about this in the data sheet.

Thanks again,
Dave E

kwinn

The ULN2003 is an open collector output with a diode from the collector to ground plus the diode we are discussing, which goes from the collector to the common pin. To drive a relay coil the common pin and one side of the coil go to the supply and the other side of the coil goes to the collector and lo and behold the diode is across the coil.

PJAllen

Neither lo nor behold.
If the COM pin is at 12V and·some coil's source voltage·is a separate 5V supply then the diode for that gate is not across that coil.

kwinn

That's a bit of a stretch. If the coils were 5V then the common pin would be connected to 5V. If you were to drive coils with 5V and 12V ratings from the same ULN then you would need to put external diodes across the 5V or 12V coils and tie the common pin to the other supply voltage.

PJAllen

Cheese & Crackers, guy, there's no stretch where I'm standing.·

The OP proposes exactly to,·owing to·some false notion of economy,·run 5V and 12V inductive loads from the same 2003, mixing it up:

Dave E said...
The issue is that I am sinking both 5 and 12 volt signals.

It's not me, it's him.· OK?

kwinn

You are correct. I did not realize he intended to use both 5V and 12V coils on the same ULN. I should have read the post more carefully.

Dave E

Either way, I think that there is not a realistic problem with using both 5 and 12 volt inductive loads with the same ULN.

Correct me if I am wrong, please, but the kick back diodes are for bypassing spikes caused by collapsing magnetic fields. These spikes can be in the form of many tens or hundreds of volts. After all, spark plug sparks are generated with collapsing magnetic fields. When the 5 volt coil collapses it generates a neg. spike, that voltage is placed on the anode of the bypass diode. If the cathode is at +5 then the output of the ULN will see a max of +5.7 ish volts. If the cathode is at +12 then the output will see a max of 12.7 volts. This is the same voltage that would be seen at the output if that output were used to sink a 12 volt signal. Per the specs, the output is good to 50 volts. The only difference I can see is that it might take a few milliseconds longer to collapse the field. Other than that, I see no issue as long as the load is purely inductive.

Dave E


Post Edited (Dave E) : 4/25/2009 1:21:31 AM GMT

PJAllen

Sorry, no, Dave E, that's just incorrect.· It's a very realistic problem being that the coils (inductive loads) powered from the supply which isn't connected to COM will therefore·not have a free-wheeling (clamping) diode across them.

Go ahead and do what you want to do, though, my work here is done.

kwinn

Add diodes across the coils connected to 5V.

Chris Savage

The easy solution to this is to simply connect individual diodes across the coils of each device. This solves the problem where the common point is only in reference to one supply. Leave the COM pin on the ULN disconnected and use your own diodes.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Chris Savage
Parallax Engineering

Dave E

Thanks for the replies.

Will add individual diodes to the +5 coils.



Dave E

Chris Savage

Dave, this is only effective if you do it for both sides. If the common pin for the diodes is connected to either voltage source, that source will be connected to both sides. You need to leave that pin connected and put diodes across the coils of all inductive devices being switched by the ULN. It is not necessary to use diodes for non-inductive devices such as lamps and LEDs. Take care.

▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Chris Savage
Parallax Engineering