michael word
Newly Enlightened
- Joined
- Mar 3, 2006
- Messages
- 58
What is the proper way to measure the resistance of the mag switch with a DMM?
Measuring Switch Resistance
- Thread starter michael word
- Start date
Help Support Candle Power Flashlight Forum
the absolute best way is to use a constant current source (like an LM317 configured as a current source) to run a set amount of current through the switch, then use your DMM to measure the voltage drop across the switch. Then, the resistance of your switch is R = Vdrop / I where Vdrop is the voltage measured, and I is the test current the LM317 is configured to provide.
andrewwynn
Flashlight Enthusiast
similarly.. i use my bench supply.. dead-short the supply across the switch.. and measure the milivolt drop across the switch.. set the bench supply to 1.0A (you can add'l run through an ammeter DMM to ensure accuracy if needed)..
It is very important that you clip your test leads onto the switch and NOT your power supply clip leads or you'll be measuring the contact resistance.
You can get close with measuring under load with any current.. for example.. if you measure tailcap current at 3.3A with 1185 lamp.. you can take your bat. pack out and your switch.. and measure the vDrop from the VBat+ to Vlamp+.. and add in the vDrop from Vlamp- to the ground contact on your mag switch.
A stock mag switch will have about 160-180mohm (0.160-0.180) ohm.. if you solder a solder wick (braided copper wire) through the spring it'll drop in half to about 70mohm.. do the same with the ground sliding contact and you get down to about 50mohm.. and if you solder the spring to the top and bottom contacts and pro-gold even the internal contacts of the magswitch.. i've measured below 40mohm.
http://prfix.rouse.com shows how.
upgrading to a KIU kit will drop resistance to about 35-40 mohm.
upgrading to a hotdriver will drop resistance below 20 mohm.
-awr
It is very important that you clip your test leads onto the switch and NOT your power supply clip leads or you'll be measuring the contact resistance.
You can get close with measuring under load with any current.. for example.. if you measure tailcap current at 3.3A with 1185 lamp.. you can take your bat. pack out and your switch.. and measure the vDrop from the VBat+ to Vlamp+.. and add in the vDrop from Vlamp- to the ground contact on your mag switch.
A stock mag switch will have about 160-180mohm (0.160-0.180) ohm.. if you solder a solder wick (braided copper wire) through the spring it'll drop in half to about 70mohm.. do the same with the ground sliding contact and you get down to about 50mohm.. and if you solder the spring to the top and bottom contacts and pro-gold even the internal contacts of the magswitch.. i've measured below 40mohm.
http://prfix.rouse.com shows how.
upgrading to a KIU kit will drop resistance to about 35-40 mohm.
upgrading to a hotdriver will drop resistance below 20 mohm.
-awr
mudman cj
Flashlight Enthusiast
A lot of people have access to a DC voltage source (such as a battery) and a DMM, but not a constant current supply. They could also perform this measurement using a shunt resistor.
First, choose a resistor that will be able to dissipate the heat generated according to Power=I^2*R. The current that flows through the resistor can be estimated given the battery drive voltage (under load - so if measuring Vbatt open circuit remember it will drop under load). You will of course have to dissipate much less power with less current so choose a low voltage source such as a NimH. Don't forget to check out the current capacity for your chosen battery (CPF has info on this!)
Assuming your switch has a resistance less than one Ohm (I will assume 0.1 Ohms as a "worst case" in terms of allowing higher current flow) and you have access to a resistor with a value of 1 Ohm that can dissipate about 5W, then a circuit with the switch in series with the 1 Ohm resistor driven by a 1.2V battery will allow about 1.2=I*(1+.1) or I=1.09 Amps to flow. (Remember to make sure the battery can supply this!) From P=I^2*R the shunt resistor will dissipate about 1.09W in this case.
Once you have determined it will be safe (you will not exceed the resistor specifications), connect the circuit and measure the voltage drop across the resistor (the difference in voltage from one side of the resistor to the other while current is flowing) to calculate a more accurate value for the current, again from V=IR. Let's say you measure 0.92V and you therefore calculate you have a current of 0.92 Amps. (For the curious I assumed the switch resistance in this example and a battery voltage of 1.1V under load)
Now, with the circuit still connected, measure the voltage drop across the switch of interest. Plug that voltage into V=IR with the current you just calculated to get your switch resistance. Let's say you measure 0.184V across the switch, then at 0.92 Amps your switch has a resistance of 0.2 Ohms. Remember that a switch is an Ohmic device, and as such it will exhibit a slightly higher resistance at higher current levels.
It is possible to determine the amount of error in your measurement given the error in the resistor value (stated on the resistor or in the specifications) and the error in your DMM readings of the voltage drops. That is beyond the scope of this discussion, but the point is that you can get a pretty good idea of your switch resistance this way even if you lack a fancy 4-point DMM that can measure very accurately at low resistances (in which case you would probably have access to a constant current power supply anyway!)
I hope I don't just generate a lot of :huh2: with this posting...
First, choose a resistor that will be able to dissipate the heat generated according to Power=I^2*R. The current that flows through the resistor can be estimated given the battery drive voltage (under load - so if measuring Vbatt open circuit remember it will drop under load). You will of course have to dissipate much less power with less current so choose a low voltage source such as a NimH. Don't forget to check out the current capacity for your chosen battery (CPF has info on this!)
Assuming your switch has a resistance less than one Ohm (I will assume 0.1 Ohms as a "worst case" in terms of allowing higher current flow) and you have access to a resistor with a value of 1 Ohm that can dissipate about 5W, then a circuit with the switch in series with the 1 Ohm resistor driven by a 1.2V battery will allow about 1.2=I*(1+.1) or I=1.09 Amps to flow. (Remember to make sure the battery can supply this!) From P=I^2*R the shunt resistor will dissipate about 1.09W in this case.
Once you have determined it will be safe (you will not exceed the resistor specifications), connect the circuit and measure the voltage drop across the resistor (the difference in voltage from one side of the resistor to the other while current is flowing) to calculate a more accurate value for the current, again from V=IR. Let's say you measure 0.92V and you therefore calculate you have a current of 0.92 Amps. (For the curious I assumed the switch resistance in this example and a battery voltage of 1.1V under load)
Now, with the circuit still connected, measure the voltage drop across the switch of interest. Plug that voltage into V=IR with the current you just calculated to get your switch resistance. Let's say you measure 0.184V across the switch, then at 0.92 Amps your switch has a resistance of 0.2 Ohms. Remember that a switch is an Ohmic device, and as such it will exhibit a slightly higher resistance at higher current levels.
It is possible to determine the amount of error in your measurement given the error in the resistor value (stated on the resistor or in the specifications) and the error in your DMM readings of the voltage drops. That is beyond the scope of this discussion, but the point is that you can get a pretty good idea of your switch resistance this way even if you lack a fancy 4-point DMM that can measure very accurately at low resistances (in which case you would probably have access to a constant current power supply anyway!)
I hope I don't just generate a lot of :huh2: with this posting...
andrewwynn
Flashlight Enthusiast
this is an excellent way to do what we want to do.. i hope people can understand what you are saying.. i'm going to try to make an extremely abbreviated description here:
hook up a battery in series with your 'test switch' and a low resistance power resistor... measure the voltage drop on the power resistor and then the voltage drop on the switch in relatively quick succession..
From the known value of the power resistor you can calculate the current through the circuit (all in series so the same everywhere).. and knowing the current and the vDrop on the switch, calculate the Rswitch.
do some of the quick math shown above to figure out that you'll be in a ball park of 1 to 3A through the circuit and you won't blow up the resistor as shown above..
Very good stuff.
Oh.. when measuring these vDrop.. make sure you measure as close as possible to the device.. and absolutely NOT across ANY connections.
It's not possible to direct-read the low resistance of a high power switch.. modamag did a test at .3A and the readings were off dramatically! Apparently and this is speculation on my part.. when there is a lot of current going over the switch.. my theory is that enough electrons are flowing across the switch they sort of create a dynamic bridge.. and the resistance drops.. this is what i've measured in the lab.. for example.. using the method described above.. you might measure 0.3ohm at 0.3A.. but the resistance might drop to 0.1ohm at 3.0A.. it's not by any means linear.. usually there is very little difference between 1A and 3A.. but in the fractional amps.. and especially the nearly zero amps of a direct reading.. huge differences!
-awr
hook up a battery in series with your 'test switch' and a low resistance power resistor... measure the voltage drop on the power resistor and then the voltage drop on the switch in relatively quick succession..
From the known value of the power resistor you can calculate the current through the circuit (all in series so the same everywhere).. and knowing the current and the vDrop on the switch, calculate the Rswitch.
do some of the quick math shown above to figure out that you'll be in a ball park of 1 to 3A through the circuit and you won't blow up the resistor as shown above..
Very good stuff.
Oh.. when measuring these vDrop.. make sure you measure as close as possible to the device.. and absolutely NOT across ANY connections.
It's not possible to direct-read the low resistance of a high power switch.. modamag did a test at .3A and the readings were off dramatically! Apparently and this is speculation on my part.. when there is a lot of current going over the switch.. my theory is that enough electrons are flowing across the switch they sort of create a dynamic bridge.. and the resistance drops.. this is what i've measured in the lab.. for example.. using the method described above.. you might measure 0.3ohm at 0.3A.. but the resistance might drop to 0.1ohm at 3.0A.. it's not by any means linear.. usually there is very little difference between 1A and 3A.. but in the fractional amps.. and especially the nearly zero amps of a direct reading.. huge differences!
-awr
mudman cj
Flashlight Enthusiast
The differences in switch resistance at low current levels reported by andrewwyn are very interesting. I speculate that this effect may be due to contact resistances that are easily overcome at higher voltages (which would be needed to generate a higher current) thanks to some level of arcing. Using a DMM to measure resistance will only drive a very small current using a very small voltage (probably micro or even nano level). Driving 3 Amps through a 0.1 Ohm switch requires 0.3 Volts; that doesn't sound like much, but some contact resistances due to surface corrosion of metal contacts are caused by only a few atomic layers of troublesome non-metals. The arcing occurs through these surface layers to the underlying metal. This is why gold is commonly used to plate contact surfaces in audio connectors and the like - it is a very stable metal.
andrewwynn
Flashlight Enthusiast
yup i think you are spot-on.. at some atomic level.. this is what's happening. Modamag in his first experiment did use a 'driving current' to do the test but used 300mA and his results were way off the mark.. he re-did them afterwards to confirm there is a big difference. I've done the tests at 1A and 3A and there was not a huge difference.. which is nice because using unity in a number in a forumula makes it go away.. hence.. your reading in mV = your resistance in miliohms!
-awr
-awr
