Why does the Darlington transistor get so hot?

Thread Starter

Don J.P

Joined Mar 26, 2020
16
I have a simple question. I am trying to drive a darlington transistor (TIP120) from an IR sensor (HC-SR501). See diagram. The load is a 300 LED strip using about 5 amps from a 12 v power supply. The circuit works, but the TIP120 gets pretty hot (I have it on a heat sink). I am just wondering why so much energy is being wasted in the heating. I've adjusted R1 selecting both 100 ohm and 150 ohm since I think that is the ballpark I want to be in for delivering 20mA to the transistor base and getting the transitor to act like a switch basically. The PIR puts out a 3.3v signal when detection occurs.
IMG_8803.JPG
 

Papabravo

Joined Feb 24, 2006
12,975
Twinkle, twinkle
little star,
Power's equal
I squared R

This bit of doggerel encapsulates the answer to your query. The effective resistance of a Darlington, if it were saturated (forced β of 10), will be the voltage drop divided by the current. Suppose that Vce is 0.35V and the collector current is 5 Amperes. That means the effective resistance is 0.35/5 = 70 mΩ = 70e-3. The power dissipation is (5)²(.070) = 1.75 Watts. A typical thermal resistance for a TO-220 package is 62.5°C/Watt from junction to ambient. So the temperature rise from ambient to junction will be 1.75 * 62.5 ≈ 109°C. If the ambient temperature is 20°C then the junction will be a 129°C which is probably too hot to survive for very long.
 
Last edited:

AlbertHall

Joined Jun 4, 2014
9,331
You could use two separate transistors connected like a darlington except the first transistor's collector is connected to the supply by a resistor. That way the second one can fully saturate.
Or, probably better, a logic level MOSFET but you must be sure it will be fully turned on by 3V.
 

Thread Starter

Don J.P

Joined Mar 26, 2020
16
You could use two separate transistors connected like a darlington except the first transistor's collector is connected to the supply by a resistor. That way the second one can fully saturate.
Or, probably better, a logic level MOSFET but you must be sure it will be fully turned on by 3V.
I was also leaning towards a logic leve MOSFET, but finding one that saturates with 3.3v is not easy. I did see a IRLB8721PbF that seems like it might work?
 
LTspice says...
Base current = 1.2 mA
Vbe = 3.18
Vce = 3.81
Wattage = 15.6
Load current 4A

Darlington_Wattage.JPG

So I'm guessing you need more base current. Or the model is wrong.

Measure the voltage across the transistors collector and emitter.
 
I could not find a detailed datasheet for the HC-SR501 PIR. They say "3.3V TTL" but TTL high is nowhere near a current as high as 20mA.

The datasheet for a TIP120 shows a maximum output saturation voltage of 4V when its input current is 20mA and its output current is 5A so then its heating is 20W! The max base-emitter ON voltage is 2.5V when it conducts only 3A and is higher with 5A so the resistor from the 3.3V PIR must be about 25 ohms but we no not know if the PIR can do it.
So replace the TIP120 with the Mosfet shown.

EDIT: Simulators use "typical" spec's. Where can you buy a TIP120 that has typical spec's? You get whatever they have which might have minimum spec's.
 

Thread Starter

Don J.P

Joined Mar 26, 2020
16
I could not find a detailed datasheet for the HC-SR501 PIR. They say "3.3V TTL" but TTL high is nowhere near a current as high as 20mA.

The datasheet for a TIP120 shows a maximum output saturation voltage of 4V when its input current is 20mA and its output current is 5A so then its heating is 20W! The max base-emitter ON voltage is 2.5V when it conducts only 3A and is higher with 5A so the resistor from the 3.3V PIR must be about 25 ohms but we no not know if the PIR can do it.
So replace the TIP120 with the Mosfet shown.

EDIT: Simulators use "typical" spec's. Where can you buy a TIP120 that has typical spec's? You get whatever they have which might have minimum spec's.
Thanks. I can see most of your logic: I do see the 4 volt Vce(sat) for a 5 amp current and a base current of 20mA., so it makes sense regarding the 20W dissipation. I can also see the Vbe(on) being 2.5 V I am having a bit of trouble backing out the 25 ohms for the resistor. I would expect the resistor to be set such that I have 20mA dropping 0.8 V (3.3-2.5), which makes it 40 ohms? Regardless, I will switch over to the MOSFET, but just trying to educate myself based on y our logic.
 
When changed the base resistor to 25 ohms in the sim...it still didn't work.
Does your Sim program know what is a TIP120 and what is the output high voltage of the PIR when driving the resistor and TIP120?
The Sim uses parts with "typical" spec's that you cannot buy. Many parts have spec's that are guaranteed only to be above minimum spec's.
 
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