# current sense resistor value

#### sdowney717

Joined Jul 18, 2012
648
I am looking at modifying the current sense resistors values.
There are 2 here.
following this one resistor might be 0.27 ohms, the other the calculator says invalid, but I think it might be 0.58 ohms
http://www.hobby-hour.com/electronics/resistorcalculator.php

Can you look at the colors and tell me what value they are? They are both under an ohm and seems to be 5 color bands. Grey and silver seem to be backwards to my eyes, and why is black band on both ends?

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#### Dodgydave

Joined Jun 22, 2012
9,870
i think they are 0.58 ohms, and 0.28 ohms, they maybe in parallel, best to remove one end, and take a reading with an ohmmeter.

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#### sdowney717

Joined Jul 18, 2012
648
Ok, I can do that to check the values. In circuit the meter reads 0.1 to 0.2, I can't get a better reading with meters I have.

What i want to do is I think lower the value of these two resistors by about 35% which will boost the inverter output by 35% from 1500 watts to 2000 watts which is still below the design capacity of the components. They made this very conservatively, it is fused for 240 amps at 12volts. 8 times 30 amp fuses is 240 amps total. And the mosfets are lots of them and all well rated beyond what this is set to shut down at.

So i was thinking to add two 0.1 ohm resistors in parallel with these both or some other values, (if they were .28 and 0.58 ohms, said values to lower both resistances by about 35%), which will lower the resistance which will lower the voltage drop across these resistors which will make the inverter circuitry think less amps are flowing which will boost the output capability to 2000 watts.

I assume a linear relationship exists between these two resistors tells the circuit how much current is flowing.
These also look like 2 watts resistors, they are thinner than the diameter of a pencil.

#### Dodgydave

Joined Jun 22, 2012
9,870
if they are in parallel,they will read 0.188 ohms approx, can you take them out and read them.

#### sdowney717

Joined Jul 18, 2012
648
I took one lead out and they both read about the same.
Top one reads 0.3 to 0.2 ohms
Bottom one reads 0.2 to 0.1 ohms
They both terminate together on the board, so they are simply paralleled.
I suppose a single resistor added parallel to these two will work.
Question is what value to determine.

This is a go power 1750 HD inverter, it is really a 2000 watt inverter they throttled low to 1500 watts for super long life.

To test my idea, I could jumper these sense resistors and then see if it can power a 2000 watt load continuously without shutting down after 5 seconds.

#### ian field

Joined Oct 27, 2012
6,539
I took one lead out and they both read about the same.
Top one reads 0.3 to 0.2 ohms
Bottom one reads 0.2 to 0.1 ohms
They both terminate together on the board, so they are simply paralleled.
I suppose a single resistor added parallel to these two will work.
Question is what value to determine.

This is a go power 1750 HD inverter, it is really a 2000 watt inverter they throttled low to 1500 watts for super long life.

To test my idea, I could jumper these sense resistors and then see if it can power a 2000 watt load continuously without shutting down after 5 seconds.
Some capacitor ESR meters will measure sub-Ohm DC resistances, and despite the high initial cost you probably wouldn't regret getting one in the long run.

Another alternative is to construct a constant current source to energise the resistor so you can measure the volt drop developed by said current.

Constructional projects for these crop up every now & then in various electronics magazines.

#### Dodgydave

Joined Jun 22, 2012
9,870
well assuming it needs 0.7V across the resistor, and the combined resistance is 0.188 ohms, the current is 3.7Amps, @ 2.6Watts, which isn't large, so you need to find out the volt drop across the resistors, then you can substitute your preferred value.

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#### sdowney717

Joined Jul 18, 2012
648
If I want 35% more watts before shutdown, it seems like this to me
The value of those two resistors together is 0.58 and 0.28 is 0.18884 ohms

So then 35% of that is 0.18884 * 0.35 = 0.066094

0.18884 − 0.066094 = 0.122746 desired final resistance

Which means about a 0.4 to 0.3 ohm resistor is needed in parallel with the other two resistors.
0.4 would give 0.12828 ohms
0.3 would give 0.11589 ohms

online calculator for parallel resistors
http://www.1728.org/resistrs.htm

1500 watts times 0.35 = 525 watts, so 1500 + 525 = 2025 watts.
Does my figuring this way make sense?

According to my meter, I do not read the '0.58' ohm resistor as 0.58 ohms, but it makes some sense that it is 0.58 ohms from the color bands. I suppose I could buy five 5 watt 0.1 ohm resistors and place them in series to see what it does, that will give me 0.1, 0.2, 0.3, 0,4 etc... Also makes a big ugly mess on the board. I see some of these are cheap on ebay. Of course soldering in series also will add some resistance in the joints.

I wonder, could a wire of a long length substitute for a resistor with this very low value?
Insulation would have to hold up to the temp, what temp range does heat shrink tubing work over?

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#### Dodgydave

Joined Jun 22, 2012
9,870
What is the maximum current output, your best way is to load the inverter and draw current at 1amp, 5amps, 10amps, and measure the voltage across the resistors.

#### sdowney717

Joined Jul 18, 2012
648
Ok, I can do that tomorrow and report back some findings.
You mean the amps of the AC output, I can use my clamp on meter for that and measure the voltage across the resistors with the other meter.

Using a wire as a shunt just wont work for this, wire would need to be too long and too small I think for the current flow.

#### sdowney717

Joined Jul 18, 2012
648
Success, I can run the big 1600 watt microwave from this inverter. And a 1500 watt toaster at the same time.
Although not very well with both together, the car battery runs out of power as it was only drawing 165 dc amps and a round trip DC wire of 2 gauge 12 feet limits the available current.
Those DC supply wires did not get more than warm.
The inverter gets warm, not hot, but the fan never turns on.
I have never seen the fan turn on with this inverter.

I created a steel wire shunt 0.027 inch diameter from a Christmas tree ornament hanger. I soldered that across the resistors, put some wheel bearing grease on the wire as it might rust, and put some white heat shrink tubing on it. The wire does not get hot. It still drops DC millivolts across this wire.
The other mod I had done before was to add 8 DC 12 gauge wires attached to the DC positive and negative bus. I had noticed earlier in testing under heavy loads the DC buss bars got hot, they don't get hot anymore.
That may have been one reason Go Power downgraded the capacity of this inverter.
I tested the millivolt drops across the resistors using a small fan, big fan, weller solder gun, toaster, microwave.

small fan 0.40 amps = 48 watts
weller solder gun 100watt and 140 watt trigger = 0.83 amps, 1.16 amps
big fan 1.45 amps = 170 watt
toaster 1500 watts 4 slice, so running only one half is 750 watts 12.5 amps, 6.25 amps
microwave 1.6kw, 1600 watts 13.3 amps

Now some measure readings
inverter idle
0.2 mv measured across shunt regardless of modification

small fan
2.4 mv low speed, oem shunt
1.6 mv low speed, added wire shunt
3.1 mv high speed, oem shunt
2.2 mv high speed, added wire shunt

weller solder gun
11.5 mv low trigger, oem shunt
9.0 mv low trigger, added wire shunt
14 mv high trigger, oem shunt
11.2 mv high trigger, added wire shunt

big fan drew 11 amps DC
13.6 mv oem shunt
8.7 mv added wire shunt

Toaster 750 watts drew 62 amp DC
80 mv oem shunt
54 mv added wire shunt

Toaster 1500 watts drew 125 amps DC
154 mv oem shunt
107 mv added wire shunt

Microwave 1600 watts drew 145 amps DC
162 mv oem shunt
115 mv added wire shunt

Anyway, you can see the pattern developed. With the microwave high amp device, the battery power does drop out some so I think it affected the mv readings. It is just a single old car battery I used for this test.
Any thoughts on my added steel wire of 3.25 inch length and 0.027 inch diameter as a shunt?

Here is a view of the entire inverter, you can see my added in DC bus wires, which it really needs. It needed those before I did any other changes. I think the small buss bars are barely adequate to flow the DC current. and they are long away from the source.
But otherwise, the inverter is quite large and seems quite capable.
16 input fets FQP 50N06
16 input caps
8 transformers
8 output fets IRF 640
8 high voltage side diodes
1 fet in the middle (between diodes and irf640's) = IRF 680

Adding, this online calculator seems to work with these color bands, except no 5 colors only four, but it does not error
black green silver gold is 0.05 ohms
black red silver silver 0.02 ohms
I also wonder about the silver and grey, they are harder to tell apart and one of these looks like it is grey on the resistor but makes no sense as silver makes more sense to what they read with meter.
What ever they truly are, they are very low ohms.

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