• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Subscribers 63 subscribers
  • Views 1440 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

ULN2003B: ULN2003B input pull down?

Dong Ba
Intellectual 1055 points
Part Number: ULN2003B

The input to my ULN2003B is connected to a switch.

When the switch is OPEN, the input to the ULN2003B is therefore OPEN and float also.
Do I need to add a pulldown resistor on the ULN2003 input to assure it is OFF, or can I just leave it open and float?

  • 0
    TI__Genius 10652 points

    Hi Ba, 

    ULN2003B is a darlington pair and is a current controlled device. When you open the switch to a channel, you stop the base current to that darlington pair, which completely turns off the **output current**. Thus, you do not need a pulldown resistor on the inputs. Though you may desire a series resistor depending on your output current need. 

    please let me know if any additional questions, but if this answers your question please let me know by pressing the green button. 

    best

    dimitri

  • 0 in reply to dimitri james
    TI__Intellectual 1055 points

    Could you please give an example how to select the series resistor and Vb?
    For example, Iout=100mA @25Celcious is needed.
    How to understand Figure 5. hFE vs IOUT？ Iout =hFE*Ib, according to Figure 5, Iout is Monotone increasing with hFE.
    IF Iou=100mA, hFE=1000, so the Ib=100uA, right?

    then how to apply Vb and the external series resistor R to realize this.

  • 0 in reply to Dong Ba
    TI__Genius 10652 points

    Hi Ba,

    Ba Dong said:
    How to understand Figure 5. hFE vs IOUT？ Iout =hFE*Ib, according to Figure 5, Iout is Monotone increasing with hFE.

    You are correct, and it is measured at input pin to output pin. and is the LARGE signal current gain. For darlington pairs, you consider the hFE1*hFE2=hFE of the darlington pair.

    Ba Dong said:
    IF Iou=100mA, hFE=1000, so the Ib=100uA, right?

    Right. 

    Ba Dong said:
    Could you please give an example how to select the series resistor and Vb?

    You don't really need an extra series resistor if you can select VB as the part has built-in base resistor. Selecting base resistor is helpful if you can't tune VB. In that case you can either estimate the current by considering the VBE drop (base of NPN1 is ~1.4V higher than emitter by rule of thumb). 

    A better option is you can use Figure 7 in the d/s. By knowing hFE at the output current need -> you get input current. From input current you need-> you get input voltage at the pin. From input current requirement + input voltage as a result of that current you can fine tune VB/RB_ext. If VB is fixed, then you can change RB os that VB-Vin(at pin) drop across your RB resistor gives you that input current you need. 

    If you can tune VB , even easier, you can directly get the voltage you should apply to the input based on the figure you shared from d/s + this figure. 

    Hopefully this helps, if you have any other questions on the design process please don't hesistate to reach out. And if this answers all your questions, please let me know by pressing the green button  .

    best

    dimitri