I am using LED drivers (TLC5940) which allow one to fix the operating current \$I_c\$ flowing through LEDs. This is accomplished by connecting a resistor (of the desired value) between an output port, driven by a bandgap reference of 1.24V, and ground. The chip then amplifies the current flowing out of the reference port by a fixed factor to give \$I_c\$.

The situation is that I have several of such chips on a PCB board. To reduce component usage, I am thinking of shorting all the bandgap referenced ports and use a resistor to bias all chips in one go. The issue is that if the references on different chips happen to be slightly different, shorting the outputs may be unwise.

In that case, is shorting advisable?


I don't think this is a good idea. I agree with Russell in that I don't see this causing any damage or running the chip out of spec, so you can always try it. The reason I don't like this is because the output current is set in part from the current out the IREF pin. Each chip will have its own voltage source it tries to drive that with. The ones that happen to have a little higher internal source could drive a disproportionate share of the current out their IREF pins when you really want each IREF pin to be sourcing the same current. The chip does seem to have some impedance in series with the voltage source which will mitigate this effect. But, I just don't see the upside.

Your cure is worse than the problem. Giving each chip a separate resistor is the easy way. Resistors are cheap and small, and the traces stay local to the chip. With one global resistor you have a bigger routing problem that may end up using more board space than the local resistors.

The only way that this makes any sense is if this is a very high volume product. In that case, each resistor costs under $.01, so the savings will still be minimal. In a lower volume product it's not worth trying to play these games. You've already wasted more design time on trying to get away with this than you'll ever get back unless you're going to sell many many of these units.



  • I'd "guess" based on data sheet information that it would work OK,
    and the datasheet indicates that no damage will be done trying it -
    so I'd try it and see.

  • Worst case the datasheet can be interpreted as indicating that some ICs will drive to min or max brightness trying to compensate.

  • You are saving 1 resistor per IC - is your design really that tight?

  • Note that RIref will need to be RIref_design / N for N ICs'.

  • The datasheet indicates that doing this will not damage the IC, regardless of how successful it is or isn't (Page 2 VIref abs max ratings) so trying it should not cause damage.

  • If you took the pin equivalent circuit diagram on page 7, Fig 1 "Input Equivalent cicruit Iref" at face value it suggest that "Amp" will drive the internal MOSFET as hard as is required to mainain Pin_Iref at V= V_Iref. This implies that shorting the Iref pin to other Iref pins will lead the internal amplifiers to saturate high or low in an attempt to correct the "error". This would tend to cause each IC to be either very bright or very dim EXCEPT the IC where Iref was correcly matched.

  • HOWEVER all other information suggests that Iref setting is much less precise than the above suggests and that small differences in Iref would be compensated for by minor differences between ICs.

    eg Iolc on page 4 is specified at 54/61/69 mA for 640 ohm RIref.

  • \$\begingroup\$ +1 for the pin equivalent circuit. Never thought about using these diagrams to infer the internal circuit operation. The diagram is incomplete though, as one of the PMOS pins is left unconnected. I assume that it should be connected to VCC. Would be good if this point can be clarified. \$\endgroup\$ – Ang Zhi Ping Nov 13 '11 at 14:20
  • \$\begingroup\$ @AngZhiPing - I see the "unconnected" PMOS line as meaning that what happens beyond there is sufficiently buffered from and irrelevant to what happens at the pin that it is not shown. | In the IC the current in the PMOS transistor is THE required result of the interaction of V_Iref and the external resistor and is then used internally as the reference for the display current. Actual display currrent is scaled by a factor of 31.5. \$\endgroup\$ – Russell McMahon Nov 13 '11 at 18:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.