Making a variable power load to test a component

Thread Starter

q12x

Joined Sep 25, 2015
880
Hello and happy new year 2022 !
How to make a variable power load/consumer to test a single component? For example, I want to test a transistor heat resistance to load. I want to know at what wattage it will start worming up and then when it will heat up and then when it will boil up, near destruction. What is a cheap and efficient method ? I saw a youtube movie once with a power resistor, probably a 1 to 5W, but I dont remember the circuit or the exact context, I believe on the same 'load' testing board was a fan and a knob. I want a circuit or an idea, to make it myself with the components I have.
 

AlbertHall

Joined Jun 4, 2014
11,952
You don't need to build anything for this. You can get the temperature rise of the transistor against power dissipated from the datasheet junction to ambient thermal resistance.
 

Jon Chandler

Joined Jun 12, 2008
156
A simple method is to use an LM317 as a current regulator. In the circuit shown below, replace R1 with a pot, and the "load" with a suitably beefy resistor. I used a 10 ohm, 10w resistor as I have a bunch of them around.

I set up a nice variable load by using a number of these circuits in parallel, with fixed resistors to provide an amp or half-amp load, with each connected with a toggle switch, and a final adjustable section that can sink up to half an amp. By flipping toggle switches, I have an adjustable load from 0 - 6 amps.

SmartSelect_20211223-134632_Edge.jpg
 

Thread Starter

q12x

Joined Sep 25, 2015
880
Thanks mister @Jon Chandler but I wonder if the pot on R1 from your circuit will support 1A through it.
I remade a circuit from the datasheet of the LM317:
Original circuit:
Screenshot_4.jpg
Left img is 'like' in the original circuit but I dont trust it completely. I want it a bit more safer for the pot and not to blow it up !
Screenshot_3 copy.jpg(from) Figure 23. High-Current Adjustable Regulator Circuit
The Load (10R) should be linked to ground ? Ive added a brown line to gnd from the (10R) Load, and I believe we should cut line a from the circuit. Or add a resistor on line a? We have to experiment with this thing I guess. Output current of LM317 is 1.5A so let's use it fully.
 
Last edited:

Jon Chandler

Joined Jun 12, 2008
156
Nope. Nope. Nope. You don't want to use the regulator as a conventional voltage regulator. You need to configure it as a constant CURRENT regulator as I showed above. Note that the output terminal connects only to R1 (which would be a pot for an adjustable load), and the "load" power resistor connects to the ADJ pin of the regulator and the other end of the pot.

Note that the tab is connected to the output pin, and is not common between stages. The regulators should be mounted to a metal box or heatsink, using insulators.
 

Jon Chandler

Joined Jun 12, 2008
156
This sketch is a little hurried, but it will show you what I have used. The adjustable stage can be varied from 1mA to 500mA. The other stages are selected by toggle switches, and add a constant current load of 500mA each.

● Note the pinout of the LM317, and the arrangement of the normal output terminal.

● Note that the tabs (case) is connected to the output terminal AND IS NOT COMMON between the stages.

● Use a heatsink or metal enclosure and isolators for the LM317s.

● This arrangement works down to about 2 volts as I recall.

Screenshot_20211224-003324_Edge.jpg
 

Thread Starter

q12x

Joined Sep 25, 2015
880
Thank you mister @Jon Chandler !
R1B,R2B,R3B,R5B = 2.4R (ohm) right? not k and normal 1/4 or 1/8W resistor. My best guess.
I only have 1x 10R @ probably 3W or 5W. And some other values, like 15R ,47R also about 3W or 5W, they are big and fat and the only ones I have. But I have enough for the full thing. (and some very low values of 0.10R-0.50R at same W)
I know there are some white square bricks Resistors and they are at 5W or 10W or more, but I dont have them.
And I get all your points in the description, no problem. I had build a few circuits with LM317, but not that many, enough that I know some usual characteristics about it. Im familiar with it and I know exactly what you are saying about it. +10 from me. ;)
 

Thread Starter

q12x

Joined Sep 25, 2015
880
Edited:
I made your circuit, and I made a mistake that I repaired. Your circuit is now working OK !
But... the 1k pot is way too big. I change it to a 500R pot and still to big. But I can see it it changing the current from 0 to 270mA at 6.3V from my PSU (power supply).
And from 8.5V and over it (9,10,12,20), I get a constant 350mA. Interesting. But is not 500mA. I think is because the Load resistor I used. I bet his ass.
Our load R2 is getting very hot very quickly if I leave it on 270mA. I start to believe it is a 1W or maximum 2W even if is more fat than the usual 1/4W. The resistor I use as load, is 15mm(0.6in) long and 4mm(0.15in) in diameter.
I used this exact circuit and all it's correct values (except the R at 10W).
1640336389099.png
The conclusion: - It works !
 

Thread Starter

q12x

Joined Sep 25, 2015
880
Interesting Update !
I changed my original 500R potentiometer
(I used both models)
this1640337977929.png and this 1640338066398.png
with a 500R Multi Turn potentiometer :
1640337898426.png
Because I originally thought I dont have the FINE range with the first ones.
It was a smart decision, but it came with a twist. The amperage got even higher !!! At 12V I got an woop of 650mA and at 6.3V i got 300mA.
LM317 needs a Minimum of 4.25V for its normal operation. Thats why Im jumping over 5V all the time.
Im happy that now I got the 500mA at 9.6V (and 400mA at 8V)
I noticed that the current is 20 times than the voltage. V=20*I. I think there is another 10R inside the LM317??? and that inseries with my power 10R = the 20R that I get in the observation. It must be. (I didnt used the 2.4R) I believe you put that 2.4R like a buffer, not to get down all the way, right? Interesting if you did that.
Another success.
Now I have to find a 10W LOAD solution. Ill have to search on ebay and aliexpres for start. I am thinking on using a fan, as an alternative. Its too big though.
Again, thank you very much and it works !
 

Jon Chandler

Joined Jun 12, 2008
156
The 2.4 ohm resistor sets the MAXIMUM load current.

I = 1.2/R = 1.2 / 2.4 = 0.5 amp = 500 mA

When the resistance increases, the current decreases. If the pot has a resistance of 1000 ohms, making the total resistance 1002 ohms:

I = 1.2/R = 1.2 / 1002 =:0.0012 amps =:1.2mA.
 

ScottWang

Joined Aug 23, 2012
7,178
Another choice is that you can do as below:

VR5K → 8 bits ADC(6 Leds with resistors) → Logic level N MOSFET*6 → Load Resistors*6.

Above is a 6 bits(64 levels) Digital Load, you can expanding to 256 levels.
 

Thread Starter

q12x

Joined Sep 25, 2015
880
to mister @Jon Chandler
After some days of decisions, I ordered today a bit of diverse values:
10pcs each: 10w 5Ω ,10w 10Ω ,10w 20Ω ,10w 47Ω ,10w 100Ω ,10w 1kΩ
at around 10$ total from aliexpress. Ebay gets more expensive than aliexpress for everything, these days. And remember on ebay was like 25% cheaper than aliexpress for absolutly everything until chinese virus in spring 2020.
A bit much for a bunch of resistors, but I dont like when I dont have them when I need them.
I expect them in around 1 month, maybe 1andhalf because Christmas.
Ill update you when I will build the circuit.
I really like your circuit ! Its simple and efficient exactly how I like it.
 
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