# Triangle Resistor Network

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#### electrocub

##### New Member
I am building a charging circuit using the Max712. The thread is 12 Volt Gel Cell Charger. It requires a .62 ohm resistor which I don't have at the moment. I am going to order the correct resistor, but I also wanted to breadboard the circuit while I am waiting for the resistors. I got on the internet and found a parallel resistance calculator. Using the calculator, I discovered that I could use a one ohm and a two ohm resistor in parallel and come up with resistance of .66. Unfortunately, I didn't have a two ohm resistor, so I came up with the idea of putting two-1 ohm resistors in series then wrapping another one ohm resistor in parallel around the resistors in series. I figured this would be the equivalent of the resistance needed. I soldered the resistors together and it came out more like a triangle and when I put a ohm meter on it, the resistance came out as .9 ohms. If I put two one ohm resistors together, it comes out at .5 ohms which I think will be close enough for what I want to do.However, my question is , using Ohm's law, how do we calculate the circuit correctly at .9 instead of .66? the tolerance is 5%k but I don't think that is the issue. Thanks in advance.

#### MikeMl

##### Well-Known Member
None of my Ohmmeters, including a Fluke 75 and a Fluke 87 are capable of accurately reading resistances below two Ohms.

If I need to measure such a low resistance, I set my lab supply to deliver 1.00A constant-current, and the use the Fluke in Volts mode to measure the drop across the low value resistor. R=E/I=E/1.00=E

Your resistor network should be 1//(1+1) = 1//2 = (1*2)/(1+2) = 2/3 = 0.667.

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#### smanches

##### New Member
Also, are you taking into consideration the tolerances of the resistors? Did you measure them, or just use the stamped value?

#### birdman0_o

##### Active Member
If you want to test the resistance of your leads, measure the total resistance of a 1 ohm, 2 ohm , 3 ohm, 4ohm , however many it takes to see the pattern if you get 1.24, 2.24,3.24,4.24 , X.24 then your first experiment was right....

#### electrocub

##### New Member
Thanks for your replies. Since my Radio Shack meter isn't that accurate, I guess I can't measure the resistance. However, none of you said that my theory was incorrect. 3 - one ohm resistors will produce a resistance of .66 +/- when two are in series and the leads of one are soldered onto either side of the two. So I guess I can continue with the breadboarding. What I will do is build it and measure the output in amps. If it is around 400ma, then it is correc and is close enough. I might even incorporate it in my design since it is easier to get 1 ohm resistors. Thanks again.

#### MrAl

##### Well-Known Member
Hi,

The resistance should measure 0.666666....
If it doesnt, your meter is wrong or the resistor values are off by too much.
If they were all off (high) by 5 percent the resistance would read 0.7
ohms, and that is a max.

#### ericgibbs

##### Well-Known Member
hi electro,

In order to do a quick project board test, a 1R resistor would make the circuit work, the Icharge would be reduced to about 60% of the required value
[ when the correct 0.62R is used]

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#### Willbe

##### New Member
Use 6.2' of #30 AWG, trim length to suit.

#### MrAl

##### Well-Known Member
Hi,

One thing i dont see mentioned here yet is the required current flow.
We need to know this in order to calculate the total power rating
of the final resistor, be it one resistor, two resistors, three, etc.,
or even a length of wire. If any of these overheats its resistance
will go up and may even burn up and start a fire.

The power is simply:

P=I*I*R

but we need to know I (the current) to calculate this way.

#### ericgibbs

##### Well-Known Member
Hi,

One thing i dont see mentioned here yet is the required current flow.
We need to know this in order to calculate the total power rating
of the final resistor, be it one resistor, two resistors, three, etc.,
or even a length of wire. If any of these overheats its resistance
will go up and may even burn up and start a fire.

The power is simply:

P=I*I*R

but we need to know I (the current) to calculate this way.
The circuit diagram text says 400mA max falling to 50mA.

I make that 0.4^2 * 0.62 ~ 0.1W.

#### MrAl

##### Well-Known Member
Hi again,

connected in series with this resistor too. Know what current the

What else is strange is that for a circuit like this i would like to see
a 0.1 ohm sense resistor. Someone might be able to look up the
chip and see if that chip can be configured to use only 0.1 ohms,
which would lower voltage drop.

#### ericgibbs

##### Well-Known Member
Hi again,

connected in series with this resistor too. Know what current the

What else is strange is that for a circuit like this i would like to see
a 0.1 ohm sense resistor. Someone might be able to look up the
chip and see if that chip can be configured to use only 0.1 ohms,
which would lower voltage drop.
hi Al,
I have seen this 0.62R value used on other current limiting circuits, it 'fits' well into the Vbe ON voltage for most sensing transistors for a 1amp limit.

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#### MikeMl

##### Well-Known Member
Hi again,
...
What else is strange is that for a circuit like this i would like to see
a 0.1 ohm sense resistor. Someone might be able to look up the
chip and see if that chip can be configured to use only 0.1 ohms,
which would lower voltage drop.
Or you could use a standard 1mΩ shunt (1mV/1A) along with a ZXCT1009 High Side Current Monitor to create an amplified ground-referenced voltage proportional to the current. I have used these to connect a current monitor to the 0-5V A/D input on a PIC.

#### MrAl

##### Well-Known Member
hi Al,
I have seen this 0.62R value used on other current limiting circuits, it 'fits' well into the Vbe ON voltage for most sensing transistors for a 1amp limit.
Hi eric,

Yes, that is a good value for that purpose, but it is kinda high for a
modern circuit.

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