Using transistors to increase the voltage with grounded emitter

Thread Starter

Philatphil

Joined May 10, 2019
15
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.
 

djsfantasi

Joined Apr 11, 2010
5,955
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
 

Thread Starter

Philatphil

Joined May 10, 2019
15
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.
 

Thread Starter

Philatphil

Joined May 10, 2019
15
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?
 

BobaMosfet

Joined Jul 1, 2009
939
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.
 

BobaMosfet

Joined Jul 1, 2009
939
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.
 
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BobaMosfet

Joined Jul 1, 2009
939
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.
 
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BobaMosfet

Joined Jul 1, 2009
939
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.
 

Thread Starter

Philatphil

Joined May 10, 2019
15
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)
 
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