Expanded outputs experiment Stamp Works

gchrt

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

pwillard

Might be related to Ohms Law, which resolves the relationship between Volts, Resistance and Current

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.

gchrt

I understood to determine the resistor's value you subtract the forward voltage of the LED from the supply voltage then divide by the LED's current. At 10mA the forward voltage of the LED would be around .5V. Seems a bit low on both numbers for an LED. Whereas 2vf and 17mA made more sense, but would work out to have a 10V supply.
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.

Unsoundcode

gchrt, The 74H595 has only a 5v supply, positive connected to pin 16 and 0 connected to pin 8. Any other connections you see in the diagram are acting as inputs or outputs to the IC.

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.

gchrt

Jeff,

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.

Unsoundcode

The reason there are two additional connections from the power supply to the 74H595 IC is that one is to hold the reset pin high (RST @ 5v) and the other is to hold the output enable low (OE @ 0v)

hope this helps

Jeff T.

gchrt

If you'd rather not go further that is fine but I don't understand why that is necessary, specifically necessary for my setup.
I will certainly look into it more before I attempt a connection.

Again thanks for the great info.

Unsoundcode

The reset and output enable are going to have to be connected whichever setup you use.

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.

gchrt

Jeff,

Once again thank you. I will now experiment.