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Topic: 4" common anode 7-segments: driver reference to buy in 2014 (Read 27634 times) previous topic - next topic

Grumpy_Mike

I would not use 8V as this is too close to the forward voltage drop to allow a resistor to be effective, try 9V instead.

Okay,  thanks for your advices!
I finally order TPIC6c595 and 680R  :)

Now I have to work on the wiring and code
yours
Mat

PaulRB


>If not, you will need to add a regulator.
I have found these step-down regulators that I would fed with 12V to output 8V regulated
http://www.dx.com/p/dc-dc-27v-to-24v-12v-3v-buck-car-power-constant-current-regulator-circuit-board-285001#.VAqrf2Pisxk
or this:
http://www.dx.com/p/mini-dc-dc-adjustable-voltage-regulator-module-blue-151211#.VAqrlWPisxk

Do you have an advice?
Thks


Those look expensive for what you require. DC-DC converters are great idea if you need to maximise battery life, but when using a plug-in power supply, just use a normal regulator like 7809 to regulate the 12V down to 9V.

But you only need to do this if the PSU is unregulated. Did you do the test I suggested yet?

Thanks PaulRB to stay tuned!

I'm hacking a PC PSU to get a regulated 12V in order to do the tests you suggested  :)

PaulRB

#34
Sep 09, 2014, 02:41 pm Last Edit: Sep 09, 2014, 02:48 pm by PaulRB Reason: 1

I'm hacking a PC PSU to get a regulated 12V in order to do the tests you suggested  :)


I don't understand why you are doing that. You just need a multimeter to test the psu you posted the picture of.

Set the psu to the 12V position, then measure the actual voltage with the meter and no circuit connected. If it reads 12V, it is regulated. If it is not regulated, it will read much higher, probably 15~20V.

Ok, I have finally understood!
The power supply was unregulated: I had to use 10V to get 12V output...

With the regulated PSU, I have 11,45V output, and with 680 R, I have 7,35V across the segment.
-> 6 mA current and still a bright light! looks good to me

I have read the specs trying to figure out the schematic:
When CLR is low, the input shift register is cleared.
When output enable (G) is held high, all data in the output buffers is held low and all drain outputs are off.
Here is what I have deduced:

I'm really unsure... Can you give me your opinion?

I haven't found the Output Enable (OE) for PWM...  :~
Thanks!
Mat

Grumpy_Mike

Looks good to me.

Quote
I haven't found the Output Enable (OE) for PWM.

Pin 9

CrossRoads

G/ is the output enable that you drive with PWM.
Quote
When output enable (G) is held high, all data in the output buffers is held low and all drain outputs are off.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

#38
Sep 15, 2014, 09:07 pm Last Edit: Sep 16, 2014, 09:51 am by Mat13 Reason: 1
Thanks guys! I understand that I have to drive G ( pin 8 ) in PWM
I'll post a new schematic.

@CrossRoads : The sentence comes from the specs, I don't understand why you crossed some words...



and with PWM dimming:


now I'll look for coding Arduino
from CrossRoads message http://forum.arduino.cc/index.php?topic=264333.msg1866410#msg1866410
and Arduino standard SPI library http://arduino.cc/en/Reference/SPI

PaulRB

Those look fine, except that your display is common anode, isn't it? So there will be only one connection between 12V and the display, and the 7/8 series resistors will be between the display and the '595.

Something to understand: because the "G" or "Output Enable" pin is shown as "/G" this means that it is "logically inverted" so a LOW output from the Arduino enables the display and a HIGH disables it. Because of this, when you use analogWrite() to control the brightness with PWM, a value of 0 will be maximum brightness and a value of 255 will be dark. (So this is the opposite of what you would see if you connected a simple LED/resistor between the PWM output and 0V, but exactly what you would see if you connected an LED/resistor between the PWM output and 5V).

Do you plan to set the brightness with a pot or an ldr?

yes, I have to correct this!  :P

I think I'll use the pot because I will have no unwanted variation. But LDR is smart as well.

Reading the SPI arduino library, I understand that:
    MISO (Master In Slave Out) will not be used
    MOSI (Master Out Slave In) is Serial In ( pin 3 ) of 6c595
    SS (Slave Select) is Output Enable ( pin 8 )
    but SCK (Serial Clock) is Shift Register Clock or Register Clock?...

I have to think about this...

Grumpy_Mike

There is no need for any great speed on this, you can just use the normal shift out code:-
http://arduino.cc/en/Reference/shiftOut

PaulRB

#43
Sep 16, 2014, 08:47 pm Last Edit: Sep 16, 2014, 08:48 pm by PaulRB Reason: 1
Mike is correct, there is no need for the extra speed that the SPI provides. But, since you have done most of the research...


   but SCK (Serial Clock) is Shift Register Clock or Register Clock?...


It is the Shift Register clock. You will need to use another output as the "register clock", usually called the "latch".

Grumpy_Mike

Quote
I understand that dimming with a resistor (versus PWM) is not efficient (higher power consumption?)


No for any given brightness the power consumption is the same.

You still need a resistor with PWM to limit the peak current.

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