4050 cmos question

[paulsantangelo]

Hey All:

Can someone explain why a 4050 would be used? I'm building a circuit using a LM3914 to drive LEDS.
What I would like to do is switch a circuit on when a particular LED is turned on. Currently, I'm using a couple of PNP transistors which is working well. But now after reading some posts, I'm thinking about removing one transistor and using a cmos AND chip, after the first transistor, would I need the 4050? I read the specs, and it looks like I would, for TTL and cmos compaibility. In kind of simple terms, can someone tell me why?

-Thanks
paul

[L. Daniel Rosa]

The 4050 gives the current gain that you're using the transistor for. It also has some drive capability near both rails, which the transistor probably lacks.

As for TTL compatibility, use two buffers. One to drive LEDs and one to drive logic, both with the same input and nothing else (except more buffers) attached to the input (to avoid loading it) and you'll probably be okay.

[philba]

Note the 3914 sinks current so you'd need to take that into account. I think (i.e. a superficial look says so) that you can leave the LEDs on the 3914. Take the input from the 3914's LED pins into the 4050 - they should be at Vled (off) or Gnd (on). Since the LEDS are driven by comparators, I am presuming that they are pulled high internally to the chip. At the worst case, you can add pull ups externally. The reason for the 4050 (I believe you mean 74HC4050) is that the Vled can be fairly high - up to 17V so you may need to do level shifting.

However, if you use Vled at 4.5 to 5V, I don't see the need for the 4050. As long as the high and low logic levels are within HCMOS spec, you will be fine (for 4.5V supply, it would be < 1.35V for low and > 3.15V for high). If your transistors do that, they ain't broke...

[ecerfoglio]

Philba posted:
Note the 3914 sinks current so you'd need to take that into account. I think (i.e. a superficial look says so) that you can leave the LEDs on the 3914.

The 3914's outputs are have current regulation to drive leds without external resistors. One of its pins "program" the output current from 2 mA to 30 mA.

The leds are connected between +V and the 3914 outputs, so the output will be at +V (LED off) or at (+V - VLed) (Led On).

If you want to drive both the LED and a logical input, you will have to add a resistor in series with the led so that [+V - (Vled + Iled x R)] is considered a "low" by the logical buffer

Or, as Daniel Rosa suggested, use the 3914 to drive two buffers, one driving the leds and the other driving your logic. But then you loose the constant current drivers and have to provide a current limiting resistor for each led (in bar mode, if you use dot mode you can have only one resistor).

You can get the 3914's data sheet Here (National Semiconductor)

[paulsantangelo]

Ok, Thanks for these, I currently have a resistor to the base of the transistor. I used a preset to actually figure out the nominal voltage to turn on the transistor and still keep the LED lit.

Just to ask a stupid question (because I should know this), Vled and Iled those can only be figured out by using a meter in parallel for vled and in series for iled, correct?

-again
thanks
paul

[Michael J]

The outputs of a 3914 can be tricky to use, e.g
reverse leakage etc.
You can use an Optocoulper 4N25/28 etc.

[ecerfoglio]

That´s a good idea. If the optocoupler's LED and the "visible" led are in series, the 3914's constant current "source" (sink) will lit both of them.

[paulsantangelo]

Does the 4N25-28 need to be in series? Can it also work in parallel with the current LED?

[rshayes]

You definitely don't want to parallel the optocoupler with an LED.

The optocouplers use infrared LEDs, which have a different forward voltage than a visible light LED. The infrared LED will conduct all or nearly all of the current before the voltage across it gets high enough to start conduction through the visible LED.

Basically, the optocoupler will work, but you will get little or no visible indication.

[paulsantangelo]

Maybe I should use an SSR instead. Reason being is that I currently divide the current between an LED and a resistor to the gate on an PNP transistor. The 4N25 seems to operate an NPN transistor. I'm using the PNP transistor to switch a circuit on. When using the 4N25, I'm not sure what to connect to the gate.