Calculating base resistor for Darlington

Thread Starter

djsfantasi

Joined Apr 11, 2010
5,586
Hi!

It’s been a while and I’m getting a little slow.

I found that I ordered an NPN Darlington transistor. It is a Tip120 TO220. The transistor will switch 12V with a 5V input from an Arduino.

The load will draw 0.7A. The hfe is 1000. The results of my calculations don’t pass the reasonable test.

What else do you need from me? Can you guys help, please.

10PCS TIP120 60V 5A TO-220... https://www.amazon.com/dp/B079DY1RDN?ref=yo_pop_ma_swf
 

Thread Starter

djsfantasi

Joined Apr 11, 2010
5,586
Not possible to use a Logic level FET?
Max.
Maybe? I’m not sure. Now that you mention it, I seem to remember that being the correct solution. I’ve only used silicon transistors before.

I’ve gotten myself in a bind. I have a couple weeks to get this working.

I’m making 7 segment displays with LED strips. I will be switching one lead to ground to illuminate it. It draws 0.102A. The other leads on each digit’s segments will be tied together and used to multiplex the digits. An entire digit will draw 0.714A

Can you help me?
 

OBW0549

Joined Mar 2, 2015
2,982
From the data sheet, Vbe(on) is 2.5 volts max, and minimum hFE is 1000. So, subtracting Vbe(on) from the 5 volt output from the Arduino leaves 2.5 volts across your base resistor. Since the TIP120's collector current is 700 mA, dividing by the hFE gives a base current requirement of 700 μA. 2.5 volts divided by 700 μA gives a maximum base resistor value of 3.57 kΩ.

So I'd probably chose the nearest standard resistor value lower than that, or 3.3 kΩ.
 

AlbertHall

Joined Jun 4, 2014
8,333
From the data sheet, Vbe(on) is 2.5 volts max, and minimum hFE is 1000. So, subtracting Vbe(on) from the 5 volt output from the Arduino leaves 2.5 volts across your base resistor. Since the TIP120's collector current is 700 mA, dividing by the hFE gives a base current requirement of 700 μA. 2.5 volts divided by 700 μA gives a maximum base resistor value of 3.57 kΩ.

So I'd probably chose the nearest standard resistor value lower than that, or 3.3 kΩ.
The usual rule of thumb when you want to be sure the transistor is fully saturated is to use a gain of 10. As this is a darlington that would be 100 so following the calculation above but using 100 for the gain gives a resistor of 330Ω (nearest value down from 357Ω).
 

dl324

Joined Mar 30, 2015
8,879
The OnSemi datasheet uses 250 for beta:
upload_2018-9-26_8-3-7.png
upload_2018-9-26_8-5-30.png
There must be a typo on one of the graphs...
 

MaxHeadRoom

Joined Jul 18, 2013
19,017
Maybe? I’m not sure. Now that you mention it, I seem to remember that being the correct solution. I’ve only used silicon transistors before.
Can you help me?
I rarely use bi-polar transistors now, my first go to now is logic level is the 2n7000 (100ma) or IRL series, IRL510 to IRL560 for higher current.
Max.
 

crutschow

Joined Mar 14, 2008
23,316
There must be a typo on one of the graphs...
Why do you say that?
The typical gain, as shown in the graphs and what you never design to, is always much higher than what is used to insure saturation of the transistor.

Using a gain of 250 from the saturation test would mean a base current here of 0.7/250 = 2.8mA, giving a base resistor of 2.5V/2.8mA =893Ω, nearest lower std value being 887Ω.
 

crutschow

Joined Mar 14, 2008
23,316
Both graphs are for the same conditions, but show different saturation voltages.
Okay, I see that now.
I suspect that the two graphs are for different operating temperatures but are incorrectly labeled as the same temperature.
 

ebp

Joined Feb 8, 2018
2,332
One graph is for the NPN family of devices (120 series) and one is for the PNP (125 series). As usual with silicon transistors, the NPN is somewhat "better" than the PNP.

The NPN graph for typical performance at 25°C shows there is next to nothing to be gained by using a base current of greater than 1 mA for collector current of 2 A. The temperature graphs predict gain is reduced by a factor of about 2 at -55 °C. The tables show a minimum hFE of 1000 at 25 °C and 3 A, so another gain reduction factor of about 2 from the typical. If I were operating that transistor at 2 A and down to -55 °C, I'd use a base current of about 4 to 5 mA, so a gain of about four to five hundred. At 700 mA collector current at no less than about 0 °C, I'd run a base current of 2 mA, and feel entirely comfortable that it would perform well. I actually have designed this transistor into harsh environment production equipment.

The notion that that the characterization conditions for saturation must always be used must die, except in cases where the last few millivolts is super critical. And of course if you really want that, then you must recognize that saturation is characterized at 25 °C, and in consequence does not reflect performance at -55 °C.

Returning to the original question, one thing that I don't think has been mentioned is the fact that the output from the processor will not be equal to the supply voltage when sourcing some current. And supply voltage at the lower margin must be considered.
 

Thread Starter

djsfantasi

Joined Apr 11, 2010
5,586
Thanks for all the help. I've attached my resultant schematic. Can I get a couple of extra eyes to review it?
scoreboardSchematic.PNG
 

neonstrobe

Joined May 15, 2009
73
Thanks for all the help. I've attached my resultant schematic. Can I get a couple of extra eyes to review it?
View attachment 160623
The PNP device does not need as low a resistor pull-down if it is to 5V. You will need to buffer it if that is from a micro at 5V, using an additional transistor or if the Arduino has an open collector output that would be OK (but I'm not sure it does).
If 5mA is assumed for the base then the PNP resistor should be about 2k, say 1.8k for preferred lower value.
How are you connecting this to 5V drive - it should not be directly to a 5V output pin, but something that is on (near 0V) or off (e.g. an additional transistor).
 

Thread Starter

djsfantasi

Joined Apr 11, 2010
5,586
Why do I need buffer transistors? The total current draw of all pins will be less than 50% of the Arduino’s capacity. In all of my previous circuits, I have driven the transistor switch directly from a digital I/O pin on the Arduino. What makes this circuit unique?

@OBW0549 @MaxHeadRoom
 

OBW0549

Joined Mar 2, 2015
2,982
Thanks for all the help. I've attached my resultant schematic. Can I get a couple of extra eyes to review it?
Just noticed: what are the voltage levels applied to the ssel input? With +12V on the emitter of the TIP42C, you can't drive ssel with a 0-5 volt logic signal because regardless of the input state, the transistor will always be on.
 

danadak

Joined Mar 10, 2018
3,577
or if the Arduino has an open collector output that would be OK (but I'm not sure it does).
The Arduino can be configed to be open drain, but there still is a parasitic diode
at the pin such that the pin V can never exceed Vdd + Vdiode, or for 5V Vdd ~ 5.7V
So you will never get the PNP shutoff this way. As O-Bee-Wan-549 points out.

This is basically the way -

upload_2018-9-28_19-29-31.png


Regards, Dana.
 

ebp

Joined Feb 8, 2018
2,332
If your Arduino is running at nominally 5 volts but actually at 4.75 and the logic 1 output while sourcing 5 mA is 3.75 V (estimate, very likely low, since I don't know what processor is being used), the TIP has a Vbe of 1.5 V and the load current is 700 mA, a suitable base resistor would be
(3.75 - 1.5)/(0.7/1000) = 3200 ohms
For great gobs of margin, even at -55 °C, a quarter of that would be suitable, so 750 or 820 ohms for standard values. If very low temperature operation is not required, I'd use 2k or 2.2k.

The PNP gain is lower but the base drive voltage will be at least 10 V, so 4.7k would yield lots of margin.

The base-emitter resistors are not required because suitable resistors are built into the transistors. This is usually but not always the case with power Darlingtons. Signal types rarely have built in base-emitter resistors. An external resistor can slightly speed up turn-off, but the base of the power device is not accessible - which is why Darlingtons generally are rather slow.
 
Top