Tom=E1s =D3 h=C9ilidhe wrote:

>William "Chops" Westfield wrote:
>  =

>
>>You will have trouble operating at 3V as "LED power" with
>>UL200x style (darlington) drivers.  Darlingtons have about
>>1V voltage drop across each driver.  By the time you have
>>one volt on the high side, and one volt on the low side, you
>>have one volt left for your LEDs, and it's probably not enough.
>>    =

>>
>
>There's something I misunderstand here. Now feel free to correct me, but =

>here's my current understanding:
>
>The ULN2003 lowside drivers work as follows:
>
>* You have a darlington transistor
>* The emitter goes straight to ground
>* The collector goes straight to the output
>* The base goes to a resistor that goes to the input
>
>Now as I understand it:
>* You put 5 V on the input
>* Current flows into the base resistor, into the base, across the B-E =

>junction dropping 700 mV in the process, and then straight to ground
>* Because there's a base current flowing, this causes a collector =

>current to flow (where Ic =3D Ib * Hfe)
>
>If we connect the cathode of an LED to an output of the ULN2003, then =

>current flows from the LED's cathode, into the collector, across the C-E =

>function dropping maybe 100 mV in the process, and then straight to ground.
>
>So therefore I don't see why you think I'll lose an entire volt in the =

>lowside driver. I'm thinking 100 mV tops. But please enlighten me.
>
>  =

>
Tomas,

Everything you say WOULD be true IF the ULN2003 only had one =

base-emitter drop.  But check the schematic in the datasheet. That =

double transistor is called a Darlington pair.  Ideal B-E drops are =

taught to be about 700mV.  Look in the datasheet for Vce(sat) or Voltage =

between the collector to emitter when fully on.  Also notice that this =

parameter changes with current. =


To find out exactly how much, you have to set up a little breadboard =

experiment.

Bill
-- =

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