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Basic Electronics » LED - Common Anode 8 Segment Display

September 07, 2012
by dan_engbers
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Okay, I'm not sure if this is a dumb question, but I do not know a ton about the electronics theory behind this and my google and nerdkits searches have yielded nothing of use...it's a difficult topic to search.

I purchased one of these 8 Segment Displays

They are common anode.

So I connect the power pins (1,5) to the resistor off the 9V source, and connect a single cathode wire to the MCU to test a single segment. At this point, the power to the MCU is off through the switch that came with the kit. The rest of the board is powered and grounded. I connect the cathode to any port on the MCU and it lights up.

so I program the pins to be on and off at a timed interval, and all the LED does is go from bright to brighter...it won't actually turn off at all.

I think to myself...I think I'm pushing too much power through this thing. So I disconnect the LED and connect one of these LED Matrix

Same results. I touch ground to any one of the ports on the MCU and they light up.

What am I missing? I was thinking mosfet's for the large display, but the little guy is running 2.2V...

Any help would be appreciated.

September 07, 2012
by pcbolt
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Dan -

From the looks of the 8-segment diagram, to light up any single segment you'd need to have between 7.5v and 8.9v on the 1,5 power input pins and ground any particular segment pin to light it up. This would jibe with what you're seeing when the MCU is off and with some internal resistance aside, it is at ground potential. When the MCU is on and an output pin is set to high (i.e. 5 volts) the LED segment "sees" a 4 volt drop and current still flows through the segment but dimmer than the 9 volt drop. It looks like each segment passes through 4 diodes so it surprises me that 4 volts is enough to drive the LED even dimly. As an experiment, try putting one of the Nerdkit LED's between the 8-seg cathode pin and the MCU pin (the NK LED short leg should be connected to the MCU side). That should cut the 4 volt drop down to 2.6 volts and that should be low enough to stop the current flow and turn off the 8-seg light. Not sure how the Matrix is wired so I can't help you there.

September 07, 2012
by pcbolt
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Dan -

Something else you can try. The MCU pins can be put in a "high impedance" mode (Hi-Z) which effectively makes them appear as an open circuit. To do this you need to set the pin as input and turn off the internal pull-up resistor. This may be enough to stop all current flow through the LED segment. To turn the LED back on, set the MCU pin to output then set it to low so it lets current flow to ground. Be careful how much voltage and current the MCU pin "sees" by putting the correct resistors in place.

September 10, 2012
by BobaMosfet
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Dan,

Any segment except DP is (4 LEDs in series):

R = (9VDC - (2.1VDC*4)) / 20mA
R = (9-8.4) / .020
R = .6 / .020
R = 30 Ohms.

DP only has 2 LEDS in series, so:

R = (9VDC - (2.1VDC*2)) / 20mA
R = (9-4.2) / .020
R = 4.8 / .020
R = 240 Ohms.

You'll have to choose the nearest value. If it's too far one way or the other, work the calculation backwards as I = E/R, to determine how much current will flow, and that it does not exceed 30mA max.

5VDC isn't enough to drive anything but the DP.

Lastly, whether or not common is 1 or 5, or 3 or 8, depends on the specific model and color of the 7-Seg Display. See if yours has any specific marking on it, and find the exact datasheet for it.

BM

September 10, 2012
by dan_engbers
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Thanks for the replies guys.

BobaMosfet, where did you get those numbers from? The "datasheet" was in the link in the original post, can you take a peek and revise your post...or am I missing something in your info?

I'll check our your suggestion tonight pcbolt.

September 11, 2012
by BobaMosfet
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Dan,

The 'datasheet' you have showing isn't a datasheet at all- it's just a diagram and pin layout. It has virtually no electrical spec's necessary for connecting it. For most LEDs, a good rule of thumb is to limit your output to 20mA, and not exceed 2.5V, unless you have a datasheet that says you can for that specific LED.

The part number for your 7-Seg display is: 7SR23011BS (Futurelec part number, not actual). Look on the display itself for any markings giving you actual part # information or manufacturer.

I suspect it's actually a knock-off of a Kingbright SA23-11EWA High Efficiency RED (GaAsP/GaP), Common Anode; Right Hand Decimal 7-Segment Display. Because it's a RED colored display, common anode is pin 1. If it were green, it'd be pin 5. That's why they list pin 1 & 5 on the common anode connection.

Voltage required changes based on type and color of LED. Range is from 1.85V to 2.2, not to exceed 2.5V, and PRV should not exceed 5V. High-RED is 2.1V @ 30mA. So if 20mA doesn't work, rework the number for 30mA-- just rework my calculations above for 30mA, instead of 20mA.

You must always consider both voltage and current. They are not mutually exclusive. Voltage is just attraction-- how hard current is pulled on. Current is what does the work. A 9V battery can instantaneously dump about half-an-amp at 9V across something. If that something can't resist the voltage, it will fry.

Hope that helps.

BM

September 11, 2012
by BobaMosfet
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Dan,

I did a little looking around, based on the Futurlec drawing to see if I can find a real datasheet at a manufacturer where all specs seemed to match.

Closest I could find was at digikey, from one of their vendors: Avago.

Part #: HDSP-C2A1

http://www.avagotech.com/docs/AV02-2496EN

The link opens a PDF datasheet.

Hopefully, this gives you all kinds of information to help you make & understand what you're working with better.

BM

September 13, 2012
by dan_engbers
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I found the datasheet. You were right, there was a part number on the side of the unit (not sure why I didn't search that already).

NFD-23011HS-11

Datasheet is here

It still looks like the same information that was on the futurlec sheet. 7.2 forward voltage per segment. 20mA forward current per seg. This comes out to 90ohm resistor right (or whatever is closest)?

Posts 1 and 5 need to be connected for all the segments to work. I'll hook it up again tonight to confirm this but when I was trying to use it I did have to hook up post 1 and 5 in order to light up all segments.

I haven't tried anything new since my first post, I just haven't found the time.

One question, when the datasheet states "Forward Voltage VF/Seg.(DP)", what is DP?

September 13, 2012
by BobaMosfet
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Dan,

R = 9-7.2 / .020
R = 1.8 / .020
R = 90 Ohms

So, if you use a 100-Ohm Resistor (since I don't believe a 90-ohm resistor exists), you get this:

I = E/R
I = 1.8 / 100
I = .018

So 18mA at 7.2V, should power a segment.

I would connect EITHER 1 or 5, not both. Normally, the type of display (or some characteristic of it) determines whether you use pin 1 or pin 5. The way they describe it on the datasheet is just a 'shorthand' method.

DP = Decimal Point. Don't feel embarassed, we all do it :P

It's telling you that a whole (4 LED) segment takes 7.2 volts, and the DP (using only 2 LEDs) uses 3.6V. Remember to recalculate R for the different voltage for the DP.

I note on the datasheet, that none specifically match your "HS" designation- you might send ningbo an email asking them if they can tell you exactly what that means- it may require different voltage/current levels than the datasheet you found.

BM

September 13, 2012
by dan_engbers
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Update.

The closest resistor I had was 120ohm. It powered up just fine with that.

I can connect either pin 1, or 5, and all the segments will work. I do not need both like you said.

I followed pcbolt's suggestion of throwing another LED in between the MCU and the display and it dimmed considerably when I switched the pin. I'll program a blink again this weekend and see if it lights up as bright as expected.

September 13, 2012
by pcbolt
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Dan -

Wow, it still didn't turn off completely? If I understand correctly, when the MCU pin is at 5 volts the segment "dims considerably" but does not turn off? That's only a 4 volt drop from the 9v supply going through 4 internal diodes and 1 external LED. Surprising to say the least.

I was actually more curious if the other solution worked (trying to "tri-state" or disable the port pin giving no path to ground and turning off the segment). Having the LED in the final circuit would be too hack-y.

September 14, 2012
by BobaMosfet
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pcbolt, dan--

Basics. An LED is a diode. It will begin to conduct at .7V (approx.). Current will flow, so 4V is adequate. Just can't draw enough current to make it bright.

BM

September 20, 2012
by pcbolt
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Boba -

Gotcha. When Dan added the external LED I thought maybe that would push it over the limit. Thanks for the info.

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