How much current can test points handle before they get too hot?

Thread Starter

SiCEngineer

Joined May 22, 2019
337
Hi,

Basic question here, but how much current can a standard test point such as the 5005/5006 from Keystone Electronics handle before they become too hot? I have a resonant circuit that can have maximum 10A in the current loop, and I had set the layout such that test points are connected directly onto the high current traces, see below:

1631191755939.png

I am not sure whether this is the best practice - can standard test points handle 10A? I can't seem to find any answers for this online. I would expect that they would get quite hot since they aren't exactly large pieces of metal that can conduct a lot of current and heat. Is it better to place the test point somewhere just outside of the trace and connect with very thin traces such that they are not directly in the path of the high currents?
Thanks,
SiC
 

Deleted member 115935

Joined Dec 31, 1969
0
Well,
IMHO TP should , if not part of the circuit, can / should be just outside the normal track,

The limit is not the test point, more the track and PTH

Is the TP measuring the current, or the voltage ?

If the voltage, then use a high impedance tester, and then its low current.
 

Thread Starter

SiCEngineer

Joined May 22, 2019
337
Well,
IMHO TP should , if not part of the circuit, can / should be just outside the normal track,

The limit is not the test point, more the track and PTH

Is the TP measuring the current, or the voltage ?

If the voltage, then use a high impedance tester, and then its low current.
Thanks! These are all for measuring voltage. There are some current measurements, but those are made with current transformers, and therefore is measured as a voltage across the transformer anyway. Otherwise no current measurements at all. I planned to just use an oscilloscope for my measurements, which I assume is high impedance anyway?
 

Lo_volt

Joined Apr 3, 2014
235
Otherwise no current measurements at all.
If, as you say, you are only measuring voltage at the test point, you will only be drawing enough current through the test point to allow the oscilloscope to operate. Check your scope probe. Most standard use probes are 10 Meg ohm impedance.

Have you checked the ampacity of your traces? 10 amps is a lot. While I can't get an exact measurement of your trace width, it looks to me that you should have at least 2 oz copper thickness to carry 10 amps through those traces without too much heat.
 

Thread Starter

SiCEngineer

Joined May 22, 2019
337
If, as you say, you are only measuring voltage at the test point, you will only be drawing enough current through the test point to allow the oscilloscope to operate. Check your scope probe. Most standard use probes are 10 Meg ohm impedance.

Have you checked the ampacity of your traces? 10 amps is a lot. While I can't get an exact measurement of your trace width, it looks to me that you should have at least 2 oz copper thickness to carry 10 amps through those traces without too much heat.
These are 100mil traces (2.54mm) with 2oz copper, as you state. I think I used about 20 degrees rise from ambient or so, but it is a while since I checked this. I think the 2oz copper helped a lot. But even with the test point in the design the way it is, how will the 10A current not travel through it? The probe is high impedance so won't draw any current, but is the test point not effectively "in series" with the traces, so it should experience the same current as the traces?
 

Irving

Joined Jan 30, 2016
2,322
Wow, 73mW of loss per cm. I did not expect the losses to be this much in the trace. My traces are 2.54mm, so I think the temp rise and power loss will be slightly higher. Maybe I should increase trace width.
Way too small... 55degC rise, 24mV/240mW loss per linear cm (for 1oz)
Better at 2oz, only 20degC rise, but still 11.5mV and 115mW per cm loss...

The 'slice' of the test point that lies within the trace will, ignoring junction effects, experience the same current. If the test point has a higher resistivity than copper (eg it's brass typically) then it will create a discontinuity and will heat up more. Its better that test points are off-flow...
 
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Thread Starter

SiCEngineer

Joined May 22, 2019
337
Way too small... 55degC rise, 24mV/240mW loss per linear cm (for 1oz)
Better at 2oz, only 20degC rise, but still 11.5mV and 115mW per cm loss...

The 'slice' of the test point that lies within the trace will, ignoring junction effects, experience the same current. If the test point has a higher resistivity than copper (eg it's brass typically) then it will create a discontinuity and will heat up more. Its better that test points are off-flow...
From the calculator you have shown, are you using the values for internal traces? It is a two layer board with just external traces. It says for external traces, 10A current, 3.7mm trace, 180mW loss per inch. Is this not very large? It seems to be extreme, since an inch is hardly a long distance, and many of my tracks are many inches long...
 
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Irving

Joined Jan 30, 2016
2,322
From the calculator you have shown, are you using the values for internal traces? It is a two layer board with just external traces. It says for external traces, 10A current, 3.7mm trace, 180mW loss per inch. Is this not very large? It seems to be extreme, since an inch is hardly a long distance, and many of my tracks are many inches long...
That's correct, for 2oz copper, 25C ambient, 10C rise, losses are 73mW/cm for external trace = 180mW/inch

1631210485491.png

Resistance = 0.73mOhm/cm = 0.73E-3Ohm/cm
Power = I^2 * R = 10 * 10 * 0.73E-3 = 0.73E-1 W = 73mW/cm = (*2.54) 185mW/inch.

A different calculator gives:
1631213995784.png
 

Thread Starter

SiCEngineer

Joined May 22, 2019
337
That's correct, for 2oz copper, 25C ambient, 10C rise, losses are 73mW/cm for external trace = 180mW/inch

View attachment 247597

Resistance = 0.73mOhm/cm = 0.73E-3Ohm/cm
Power = I^2 * R = 10 * 10 * 0.73E-3 = 0.73E-1 W = 73mW/cm = (*2.54) 185mW/inch.

A different calculator gives:
View attachment 247603
What is the reason for external traces dissipating more heat? It seems that external traces can handle more current, so my assumption would have been that for the same trace length and width the dissipation would be the same if not less for external traces. What else is at play which almost triples the dissipation in external layers?
 

Irving

Joined Jan 30, 2016
2,322
What is the reason for external traces dissipating more heat? It seems that external traces can handle more current, so my assumption would have been that for the same trace length and width the dissipation would be the same if not less for external traces. What else is at play which almost triples the dissipation in external layers?
Its not that the external traces dissipate more, its that the internal traces have to be 3 times the width to control the temperature rise so by definition have lower losses & dissipate less.

Here's the official explanation:
"In air, the external layers have better heat transfer due to convection. A good heat insulator blankets the internal layers, so they get hotter for a given width and current. Since the Trace Width Calculator tries to control the temperature rise of the traces, it makes the internal traces wider. In vacuum, or in a potted assembly, you should use the internal layer guidelines even for the external layers."

In other words, if you want to minimise losses, make the tracks wider! There's nothing stopping you making them 10mm wide, and you'll have a lower temperature rise. Setting temp rise to 1.85degC (see below) and you get 10mm tracks with 24mW/cm losses,1/3 the losses for 3.7mm tracks. Or make the copper thicker - 4oz wouldn't be unusual for a power supply. Another trick is to put a thick layer of solder over the copper. Of course, it goes without saying, that high current traces should be as short as possible!

1631217967096.png
 
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Lo_volt

Joined Apr 3, 2014
235
The 'slice' of the test point that lies within the trace will, ignoring junction effects, experience the same current.
This 'slice' also has a heatsink in the rest of the test point structure. It won't heat up at all.

If you were attaching a load and drawing 10A via an alligator clip on the test point, you would need to worry about current through the test point and the heat generated. At that point, I'd suggest that you use a 10A rated connector.

I also agree with Irving, there's no harm in making your traces wider. It doesn't cost extra and you appear to have the space.
 

Irving

Joined Jan 30, 2016
2,322
This 'slice' also has a heatsink in the rest of the test point structure. It won't heat up at all.
Not so, it will heat up, just how much depends on the quality of the rest of the pin as a heatsink. The base will be at, or very close to, the temperature of the track (it could actually be higher depending on the relative resistivity and cross-sectional area). There will be a temperature gradient along the pin, from the temperature at the pin's base to something between that and ambient at it's tip.
 

Juhahoo

Joined Jun 3, 2019
253
Why would your track get hot in the first place?, your copper is too thin, too narrow and you have exceeded its ability to carry current, you will eventually damage the PCB with excessive heat. PCBs can handle 130-150celsius before they deform.

Your PCB needs a rework, not your test points. Apply thicker copper and widen the traces to the maximum.
 

Thread Starter

SiCEngineer

Joined May 22, 2019
337
Why would your track get hot in the first place?, your copper is too thin, too narrow and you have exceeded its ability to carry current, you will eventually damage the PCB with excessive heat. PCBs can handle 130-150celsius before they deform.

Your PCB needs a rework, not your test points. Apply thicker copper and widen the traces to the maximum.
I have done. I made the traces a mixture of 4mm and 3.7mm wide as recommended here by others in this thread. According to the calculations this should be sufficient. The question originally wasn't about the width of the traces but the actual ability of the test pins themselves to get hot. For example, even if the traces were 10mm wide and the test pin would only be able to conduct say 5A of current, the traces would be fine but the test pin would be very hot.
 

Lo_volt

Joined Apr 3, 2014
235
@Irving, sorry to belabor the point, but the testpoint cross-section is huge compared to the trace cross-section. Assuming similar material conductivity and given the large cross-section, the testpoint resistance from one side to the other is a small fraction of the resistance of the trace. You will get far less i^2r loss through the test point than through the trace and therefore less heat.
 

Irving

Joined Jan 30, 2016
2,322
For example, even if the traces were 10mm wide and the test pin would only be able to conduct say 5A of current, the traces would be fine but the test pin would be very hot.
@Irving, sorry to belabor the point, but the testpoint cross-section is huge compared to the trace cross-section. Assuming similar material conductivity and given the large cross-section, the testpoint resistance from one side to the other is a small fraction of the resistance of the trace. You will get far less i^2r loss through the test point than through the trace and therefore less heat.
NB Edited to correct transcription error in equivalent circuit and text

I think you're looking at it from the wrong perspective. Consider the diagram (not to scale) below showing a 1.2 mm sq tinned brass pin centred in 1cm of a 3.6mm wide track. The pin has a higher resistivity than the copper track (about 5.3x). Assuming perfect connectivity, current flows along the track and through the slice of the pin coincident with the track. Ignoring minor edge effects, no current flows in the pin above or below the track. The approx equivalent circuit for this is shown below. If you do the math, ~4.5A flows in each copper track around the pin, while the slice of pin carries ~1A. Over the 1cm of track there is a 4% increase in resistivity and therefore a similar increase in power loss from 70mW to 73mW. However, in the 1mm of track surrounding and including the pin, there is a 37% increase in resistivity and power loss (to 9.6mW from 7mW) resulting in an increased rise in temperature to around 17degC mitigated by the increased thermal mass and surface area of the pin.

The overall impact is, I agree, negligable! But bad practice as poor solder joints can have a negative effect.

1631812510796.png1631875655965.png
 
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Thread Starter

SiCEngineer

Joined May 22, 2019
337
I think you're looking at it from the wrong perspective. Consider the diagram (not to scale) below showing a 1.2 mm sq tinned brass pin centred in 1cm of a 3.6mm wide track. The pin has a higher resistivity than the copper track (about 5.3x). Assuming perfect connectivty, current flows along the track and through the slice of the pin coincident with the track. Ignoring minor edge effects, no current flows in the pin above or below the track. The approx equivalent circuit for this is shown below. If you do the math over the 1cm of track there is a 4% increase in resistivity or a 0.4degC rise in temperature which I accept will to some extent will be offset by the increased thermal mass of the pin.

The overall impact is, I agree, negligable! But bad practice as poor solder joints can have a negative effect.

View attachment 248131View attachment 248137
This is incredibly informative Irving, made better by the equivalent circuit in LTSpice. This question resulted in a very good discussion between all parties for which I am very appreciative. Thank you!
 
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