Expanded outputs experiment Stamp Works
Hi,
In the Stamp Works Experiment #23 Expanded Digital Outputs with Shift Registers I am curious why there are 470 ohm resisters on the leds.
Do the two VDD inputs give the 595 10V output supply on each pin which makes sense if the leds are 2vf and 17mA rated?
But in reading about the chip the max input seems to be rated at 6V.
thanks
Post Edited (gchrt) : 9/12/2007 9:06:52 PM GMT
In the Stamp Works Experiment #23 Expanded Digital Outputs with Shift Registers I am curious why there are 470 ohm resisters on the leds.
Do the two VDD inputs give the 595 10V output supply on each pin which makes sense if the leds are 2vf and 17mA rated?
But in reading about the chip the max input seems to be rated at 6V.
thanks
Post Edited (gchrt) : 9/12/2007 9:06:52 PM GMT
Comments
Known Values: 5V Logic & 470 Ohms
Unknown Current: In this case the value computes out to be around 10 Milliamps
10 Milliamps is a safe current to operate LED's with. An unsafe current (IE; A higher or unlimited one with no resistor) makes things go POP and let's the magic smoke out.
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There's nothing like a new idea and a warm soldering iron.
But if the two VDD inputs are not accumulative and it is only 5V coming out of those pins on the 595, then I can go ahead and drop the resistance for my 3.4vf 25mA LEDs that I will be using. But I am a novice at this and want to be sure before I blow something.
Each output to·a LED is 5v. The resistor for each LED is a "safety valve" that limits the current to a safe level while still providing the voltage needed by the LED.
Looking at the diagram and making the assumption that the LED requires 1.6v then by applying the formula you mentioned
5 - 1.6 = 3.4
3.4/470 = 7mA
So it looks like the LED's in Stamp Works only require somewhere in the region of 7mA to emit light with a voltage of 1.6.
If you are making alterations to the circuit make sure the 74H595 is capable of supplying the current you need.
The resistance value for the LED's you are wanting to use works out to a prefered value of 68 ohms. Test with a higher resistance value across a 5v supply before you hook it up to your IC, the light level might be acceptable with a resistance value of 82 ohms or higher.
Jeff T.
Thank you for your detailed message. I can move forward with what you have provided.
Just curious though in the diagram they have two pins connected to VDD inputs. I will try it without the second input but I'd like to know why the second VDD connection and the corresponding second VSS connection.
hope this helps
Jeff T.
I will certainly look into it more before I attempt a connection.
Again thanks for the great info.
The reset is active low, that means whenever the reset goes low it clears the information stored in the shift register. So this pin is held high to retain the information you send.
The output enable is also active low. In order to display the information latched inside the shift register you must enable the outputs, therefore OE is connected to Vss or held low
If you don't connect these pins to anything you·can never be sure of their logical value and the circuit is going to be unreliable
Jeff T.
Once again thank you. I will now experiment.