Darlington current gain calculations

Thread Starter

Mark1015

Joined May 15, 2020
22
I am having difficulty calculating the current gain for the attached image. This is a Toshiba darlington array IC. I want to attach 100ma leds to the chip via a 9volt power supply on the common and input will be 5V from an arduino pin. I have read everywhere i can find on the calculations and so far it has not helped. Any help is appreciated
1651691349834.png
 

Thread Starter

Mark1015

Joined May 15, 2020
22
Maybe I am misunderstanding the Darlintons exact purpose. Is it not to allow more milliamps of load to be connected to the "switching" device-in this case the arduino pin which allows 40ma of current? I am trying to figure out how many 100ma leds I can connect to the darlington chip, then reverse engineer my power supply requirements.
 

crutschow

Joined Mar 14, 2008
29,810
Maybe I am misunderstanding the Darlintons exact purpose. Is it not to allow more milliamps of load to be connected to the "switching" device-in this case the arduino pin which allows 40ma of current? I am trying to figure out how many 100ma leds I can connect to the darlington chip, then reverse engineer my power supply requirements.
Yes, it will do that, but you don't need the transistor current gain for that, which is used for analog circuits, not switching.
You use the base current that they show in the data sheet (as shown in post #3) for the amount of current you want to switch.

I would recommend you don't go over 300mA per output.
That will dissipate about 0.4W in the chip per output that is on, so you must be aware of how hot the chip will get (as determined by its thermal resistance), depending upon how many outputs are on and the current conducted by each output.
 
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dl324

Joined Mar 30, 2015
14,475
Maybe I am misunderstanding the Darlintons exact purpose. Is it not to allow more milliamps of load to be connected to the "switching" device-in this case the arduino pin which allows 40ma of current?
A single BJT used in saturation mode (i.e. as a switch) will use a "beta" of 10 for general purpose transistors (it's 20 for some BC5xx transistors). The beta you should use should be given in the datasheet; typically specified with the saturation voltage specifications.

A Darlington uses an additional transistor to multiply the overall beta. But, again, for saturation mode, you use the data specified in the datasheet.

In your case, Texas Instruments specified the beta to use for 3 different currents; one of which happened to be the current you desired. Had that not been the case, you could use the more conservative number for the datapoints that bracketed your desired current, interpolate a value from the data provided, or used the graph for typical devices:
1651696966631.png
The safest approach would be the first option mentioned.

Also note that ULN2803 are designed to operate directly from 5V logic.
 

Papabravo

Joined Feb 24, 2006
18,431
When you use a transistor, or a Darlington, as a switch you force the device(s) into saturation hard, by supplying much more base current than it needs. This forces the current gain to be much lower than it would be otherwise, since the current is limited by external components. This is called a "forced beta" and it typically has a value in the range of 10 to 20. This number is probably 2 orders of magnitude smaller than the gain you would use in an amplifier. That is why you could care less what the actual current gain of the device is.

Being careful on the power dissipation of your components is a valuable suggestion which you should not ignore.
 

BobTPH

Joined Jun 5, 2013
4,923
The current limiting base resistor is built in, just connect 5V to turn it on.

And the max current output is specified as 500mA, but as already mentioned, be conservative and use 300mA max.

And, you do need a current limiting resistor for each LED.

Bob
 

LvW

Joined Jun 13, 2013
1,438
When you use a transistor, or a Darlington, as a switch you force the device(s) into saturation hard, by supplying much more base current than it needs. This forces the current gain to be much lower than it would be otherwise, since the current is limited by external components. This is called a "forced beta" and it typically has a value in the range of 10 to 20. This number is probably 2 orders of magnitude smaller than the gain you would use in an amplifier. That is why you could care less what the actual current gain of the device is.
OK - from the "practical point of view" I agree, , of course.
However, in case somebody wants to understand what really happens, one should not overlook the definition of "saturation".
And - as a consequence - we must realize that a large base current is the RESULT of saturation and not its cause (or precondition).

Saturation is defined as a state where the B-E junction as well as the B-C junction is biased in forward direction (due to a large voltage drop across the collector resistor which makes Vc<Vb).
And this effect adds an additional current through the base node - with the consequence that the definition Ib=Ic/B does not apply anymore.
Hence, any resistor between the base node and the controlling voltage must allow (not supply) this enlarged base current - and create, at the same time, a voltage Vbe which is sufficient to allow the necessary collector curent Ic.
 

MrAl

Joined Jun 17, 2014
8,996
Hello,

Knowing the current gain of the transistor will tell you the load it can drive effectively. That's really the minimum current gain though.
If the transistor output can output say 100ma, then it should be able to drive a 50ma load effectively but of course a test is in order.
If you know the load impedance and the min current gain then you can calculate the output voltage but you also have to know the input current.
 
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