I'm new to electronics and I have a problem with driving several (66) 4" high 7 Segment display which requires a 12V supply but are controlled through a Arduino +expander I/O. I have purchased a transistor array but found that it has a common positive V. I need individual grounded emitters. I cannot find a common-ground transistor array on the market so I will probably have to use individual transistors (i.e. 66 x 7 = 462 transistors) . I am struggling to specify the correct transistor to use. Can anybody point me in the right direction.

 

ULN2003 array?

 

hi Phil,
Welcome to AAC.
What type of transistor array have you bought, ie; type number.?
E

 

AND, are the 7-segments common cathode (supplying +12VDC to a segment turns it on; the common of the display is connected to ground)? ...
...Or are they common anode (connecting a segment to ground turns it on; the common of the display is +12VDC)?

From your description it appears to be the latter. Do you know how much current one segment draws?


PS, I just completed a project with 18” tall 7-segment displays and an Arduino

 

hi Phil,
Welcome to AAC.
What type of transistor array have you bought, ie; type number.?
E

Thanks for responding
I bought UNL-2004A but I have continued my research and found
UDN2981A which is expensive from most suppliers but is only 99p from China

 

Hello,

The UDN driver is used for high side driving:



The ULN2004 is a low side driver:



Bertus

 

ULN2003 array?

Bingo, this is the corrrect chip to use. However, it should be understand that it is open-collector, so pullups need to be on every collector to hold it high by default. I suggest 4609X-101-183LF SIP Resistor.

 

AND, are the 7-segments common cathode (supplying +12VDC to a segment turns it on; the common of the display is connected to ground)? ...
...Or are they common anode (connecting a segment to ground turns it on; the common of the display is +12VDC)?

From your description it appears to be the latter. Do you know how much current one segment draws?


PS, I just completed a project with 18” tall 7-segment displays and an Arduino

You are right they are common anode. Each segment draws 0.02mV and I amusing a 55 Ohm resistor on each segment with constant voltage supply adjusted to 11.5V This will change when I establish the voltage drop on the chosen transistor.

 

Bingo, this is the corrrect chip to use. However, it should be understand that it is open-collector, so pullups need to be on every collector to hold it high by default. I suggest 4609X-101-183LF SIP Resistor.

Each segment draws 0.02mV and I amusing a 55 Ohm resistor on each segment with constant voltage supply adjusted to 11.5V This will change when I establish the voltage drop on the chosen transistor. You suggest 4609X-101-183LF. I am using a 4608X-101-561LF resistor array which is only 560 Ohms compared to your 18000 Ohms. Is this a problem?

 

Hello,

The UDN driver is used for high side driving:

View attachment 177008

The ULN2004 is a low side driver:

View attachment 177009

Hello,

Thank you I am probably going to use a UDN2981A

The UDN driver is used for high side driving:

View attachment 177008

The ULN2004 is a low side driver:

View attachment 177009

Bertus

Bertus

 

Sorry for the typo I should have said 20mA not 0.02 mA

 

I don't see the need for pull up resistors.

 

I don't see the need for pull up resistors.

Hope your right as I need to stick with 20mA

 

Each segment draws 0.02mV and I amusing a 55 Ohm resistor on each segment with constant voltage supply adjusted to 11.5V This will change when I establish the voltage drop on the chosen transistor. You suggest 4609X-101-183LF. I am using a 4608X-101-561LF resistor array which is only 560 Ohms compared to your 18000 Ohms. Is this a problem?

Sorry- just meant my SIP resistor as an example. I use what I use because I'm pulling 18V signals up at a desired output of approx 1mA, so 18K-Ohm was right for me.

 

I don't see the need for pull up resistors.

You always use some means to control the default signal state on an open collector output, otherwise it can float to erroneous values.*

*See my thinking further down in this thread. If output isn't actually disconnected, the pullup is probably not necessary. Thanks ElectricSpidey.

 

A pullup resistor on the outputs of the ULN2803 are just additional loads in parallel with the LEDs.

 

Hope your right as I need to stick with 20mA

11.5/560 = 20.5mA. You don't need it that high. You only need to pull up at about 1mA. What the pullup does has nothing to do with the LED segment current-wise.

Now, 20mA is quite a bit of current all things considered. It would help if we had a part number on your 7-seg display because you're probably using a lot more curren than necessary. Most LEDs are rated for a max 20mA, but only need 8mA or so to do their job properly. Never use more current than is absolutely necessary. Where there is current, there is junction temperature heat.

 

A pullup resistor on the outputs of the ULN2803 are just additional loads in parallel with the LEDs.

R = E/I
R = 11.5/0.001
R = 11.5K-Ohms. a 12K-Ohm resistor would work fine.

Yes, it's in parallel, but it's purpose is to keep voltage high, while still being overcome by the transistor. In considering what you've said, because the transistor output isn't just floating, but is tied to a guaranteed load, I suspect one could get away with no pull-up.

I'm leaning towards your thinking on this. I'm just used to dealing with outputs that aren't always guaranteed to be tied to a load so use a resistor as a default safety measure.

 

But why, the LED display wont light when the Darlington is in cutoff.

 

11.5/560 = 20.5mA. You don't need it that high. You only need to pull up at about 1mA. What the pullup does has nothing to do with the LED segment current-wise.

Now, 20mA is quite a bit of current all things considered. It would help if we had a part number on your 7-seg display because you're probably using a lot more curren than necessary. Most LEDs are rated for a max 20mA, but only need 8mA or so to do their job properly. Never use more current than is absolutely necessary. Where there is current, there is junction temperature heat.

I have done some research and it seems that the correct calculation for resistor value is the difference between the supply voltage and the LED forward voltage / i. My segments are 5 leds in series each requiring 1.7V = 8.5V so the formula should be (11.5-8.5)/.02 = 150 Ohm as a minimum value for 20mA
I have also discovered that although the brightness is nearly proportional to the current it is not very noticeable to the human eye above 15mA so I think 15mA will be my new target as these need to be clear in bright daylight at 40 metres
So I will recalculate based on this but taking into account the resistance offered by the transistor(as yet unknown)