Connecting Voltmeter/Ammeter

Following on with a scale model build.
http://www.boatdesign.net/forums/projects-proposals/fast-build-scale-model-40905-4.html

I purchased a cheap Ammeter/Voltmeter to use during testing. I would like it to display the Amps and Volts during operation with a 7V battery. Hopefully the Amps being used will give an indication of the power being required at different speeds.

Being cheap and Chinese, it arrives with no instructions.

I would very much appreciate any details on the best way to wire it.

I am particularly curious about the use and purpose of the accompanying 'shunt resistor'. Is it needed in all installations, or just with higher voltages I wonder ?



On the Ebay Page, it has some 'hints'

http://www.ebay.com.au/itm/DC-4-5-3...28022346&tpos=unknow&ttype=price&talgo=origal




I have also attached a PDF of the details


PS - I googled a bit, and came up with a chinese wiring diagram. Does this look right ?

 

No need to study Chinese! Black to ground, red to + battery gives you the voltage reading. The user (load) is connected to the shunt resistor instead of ground, the other end of the shunt goes to ground.
The yellow lead goes to the load side of the shunt; the circuit sees the voltage drop across the shunt. For a 0.075 volt drop the display will show 100 Amps.

If your circuit handles smaller currents you will need a more modest shunt, this one has a resistance of nearly 0 Ohms (0.75 milli-Ohms). Use Ohm's law to calculate the proper value: i x r = 0.075, so 75 milli-Ohms for 1A, 7.5 milli-Ohms for 10A. These are not resistors you can buy, they are made from a short piece of copper wire.

 

Talk about me, the uneducated novice moving into the mysteries of electronics


I started to educate myself, and looked up shunt resistors on the internet, I got the most simple explanation at
http://www.ebay.com/gds/A-Guide-To-Current-Shunts-/10000000001675505/g.html

which also had some shunt resistors for sale at various ratings, including wire only versions.

But, I am still unclear on how I can calculate if I even need a shunt, let alone what resistance to choose.



Maybe I should get an electrical friend to run a multimeter across the electric motor under load to find out what amp range I should expect, so I can choose the right resistance. There is absolutely no documentation for the electric motor I bought.

I looked up some specs for other brushless motors ( example pdf attached )
http://www.hobbyking.com/hobbyking/store/__438__59__Electric_Motors-Boat_Motors.html

and got motor spec results like

Max voltage: 22.2V (6S)
Max Current: 160A
Max Watts: 3100W
Resistance: 0.014ohm
No Load Current: 1.4A

So, using advice from the internet
"How do I choose the right shunt?

The shunt you chose should always be rated higher than the maximum amount of current you’ll be measuring. "


I am using 7 volt batteries, and guessing perhaps 100 amps draw ?

To my uneducated mind, that seems to indicate that the supplied resistor might be OK ?



If anyone can show the workings of the calculations, I would be very appreciative.

 

I assume that the shunt delivered with the meter is for 100 Amp if the meter is rated max 100 Amp. The only way is to measure the voltage over the shunt , by placing a known current of a 60 watt automotive globe/bulb and get a reading in milliVolts. Having done that , the resistance is then millivolt divided by the current gives you the resistance in milli Ohms. If your shunt is for 100 Ampere, you need a micro voltmeter not milliVolts reading.
Bert

 

Ray, here are the specs for the motor.

Specs:

RPM/v: 1050kv
Winding: 1.5Y
Max voltage: 22.2V (6S)
Max Current: 160A
Max Watts: 3100W
Resistance: 0.014ohm
No Load Current: 1.4A
Can Diameter: 39mm (actual motor diameter)
Can Diameter inc. Water Jacket: 49mm
Can Length: 84mm
Shaft Size: 5mm
Weight: 490g
Motor Connector: 4mm bullet plug

Thus at 7 Volt you will rotate the shaft at about 7000 revolutions per minute (1050kv x 7 Volt = 7350 )
The current will be at 7 Volt approximately 50 Ampere. thus approx 350 watt. Is this what you had in mind?
Bert
p.s. nice little water cooled motors.

 

Ray,
What controller are you using?. If you use 3 x 7 Volt batteries in serial, you need a shunt for approx 150 Ampere. Your motor will be close to the 3 kw. This in view that your 7 Volt batteries when fully charged will be close to the max 22,5 Volt the motor can handle. If you use 7 Volt batteries parallel, you motor power is only 350 watt and 50 ampere.
Bert

 

Thanks for the help Bert.

Those specs I posted are for a motor that seems to be similar to mine ( a real worst case example ), but I have no reliable way of telling the actual stats for my motor - yet.

Also, my motor is only 7V, so the higher current draw wont be such a headache.

The first thing I am going to do, is to see if I can find any markings on the motor that will provide some performance info, when I get back to the workshop.

Added info, just found -
http://www.rcnz.com/Shop/Service/De...ili_Model___Quick_540_Brushless_Outboard.html
Quick 540 outboard
2858B brushless motor
Voltage: 7.2-12V
Output: 2800KV


It appears that until I do this, trying to mount the ammeter might cause problems.

In the meantime, i am still trying to get my non-electrical head around the formulas, principals that I will need to master at that point.


For my own reference, I came across this info on a model site.

"Deciphering the numbers.
Brush-less motors usually have several figures describing their characteristics, as they tend to be much more standardised over all the brands. It makes it simple to select a motor to suit an application.

Kv. RPM per Volt, the most important figure, gives the no load RPM motor as a constant, so a 1000 Kv motor will be 6000 RPM at 6 volts, 12000 RPM at 12 volts etc (Kv x V = RPM)
Io. No load current, at 10 volts usually.
Rm, Internal resistance of the windings.
These two figures can be used to calculate amp draw of the motor
Size, most out-runners state the stator size not case size, but the overall dimensions are usually stated in the specifications as well. The motor is usually labeled something like this, 22-20 or 28-36 with some letters added to the front and back. The 22-20 has a stator size of 22mm diameter and 20 mm long the 36-40, 36 mm diameter and 40 mm long, you get the picture.
A further note the larger the diameter or the longer the stator size increases torque but usually lower Kv, Motor turns (number of windings) also affects Kv.?

 

There is some misunderstanding here.
Only if you want your "Amps" display to work, you need a shunt. The one you bought on Ebay is matched to the display, when the voltage drop is 0.075 volts, the display shows "100 Amps".

If your motor pulls more than 100 amps you can still use the same shunt, but you must connect the yellow lead to a voltage divider: two equal (small) resistors, for example 100 ohms each. Now the display will read 100 (%) when 200 Amps flow through the shunt. The voltage drop is then 0.15 volts.

 

Hi Ray, Good thinking. I am a little concerned about the 2800 KV, it means that by 7 Volt, the rpm will be +/- 20.000. versus the 7000 you have selected. I don't know in what state your old motor is, but maybe you can measure the resistance by placing the faulty motor in serial with a headlight globe/bulb of 60 watt 12 Volt. You measure the voltage over the globe and you measure the very low voltage over the old faulty motor. You do this with a multi meter which can measure up to 12 Volt and can measure milli or even better micro volts. 12 Volt/60 watt (headlight globe) = 5 Ampere, 12 Volt /5 ampere = 2,4 Ohm . It may be at a tolerance of 10%, but it gives you an idea. Now you measure the millivolts or micro Volts over the old motor. If it is 30 milliVolt. , your internal resistance will be 0.030 / 5 Ampere = 0.006 Ohm. This you compare with the specs of your new motor. Off course the motor must not have a internal short circuit. Yolu cannot damage your old motor, in view that the glob/bulb is protecting it.
Bert
If you measure

 

Thanks for the input CDK.

You have raised an interesting point.

I thought that the amp display would be actual AMPS - not a percentage of the maximum.

For example, the picture below is showing 12.3 Volts, but .3 amps

Are you saying that .3 represents .3% of some amount ?

This is totally not what I was expecting.

 

Ah no, I have confused the issue. The high power 12 volt motor I was mentioning was me trying to find some kind of comparison equipment.

The 7 Volt 'old' motor is the one I am still using - hopefully it is in good condition.

The 7000 rpm figure isn't a worry to me.

Your info on potential 20,000 rpm is of real interest though. This gives me some indication of the maximum prop speed that will be possible. In another thread, I worked was given a formulae for the power generated, and water moved.

This may be very handy for a variety of purposes, measuring the performance of the motor.

I even had to look up the definition of milli and micro-volts - I am learning from a low base point

 

With the shunt and display as supplied, you get 99.9 Amps as the maximum reading. Using a voltage divider over the shunt in case the load exceeds 100 Amps, you get a percentage. Without it, the display probably says "OVL" or just blinks.

 

Ray, the best is to compare a water tap with electricity.

The pressure on your closed water tap is the Voltage. How higher the pressure, how higher the Voltage and thus more dangerous. As soon you turn your tap slightly open and the water is trickling out, that is compared with current (Ampere, milli Ampere, Micro Ampere) thus the resistance which the tap creates by only slightly being opened is the resistance it forms to let the water go out. Thus Voltage divided by ampere = resistance.
In your case your motor is 7 Volt. But the resistance is very low, thus the amperage will be high. Volt x resistance = current.
On your display you see 0.3 Ampere and Voltage is 12.3 Volt. Thus your resistance of that measured device = 12.3 divided by 0.3 = 41 Ohm Should the resistance had been 0.2 Ohm , you would have seen with that exact same shunt 12.3 / 0.2 = 61.5 Ampere.
Should the resistance had been 0.1 Ohm the meter would have shown most likely HHH as the reading is too high and it should have been 123 Ampere. By taking CDK suggestion and you place 2 resistors over the very same shunt, and connect the yellow wire to the middle point of the two resistors, you would now see 123 Ampere. Because your digital input is no longer overloaded !!.
Hope this helps.
Bert

 

After you have absorbed the previous reply, we go one step further.
How to determine the value of those 2 resistors?

Simple, you need to know the input resistance of your digital ampere reading side. AND THAT IS MOST LIKELY unknown, but we do know the input resistance is high in relation to the shunt resistance.

Thus if we take 2x 10 Ohm resistance, the 20 Ohm (10 + 10) over this probably very low 0.01 Ohm shunt resistance, has only a minute influence on the accuracy of the reading. We could also take 2 x 1000 Ohm resistors. But now the influence on the shunt resistor is absolute to ignore, but it start to have an influence on the input resistance of the input of the ampere side.

Moral of the story, it does not matter what 2 resistors you take, as long the two are the same and 1 percent or 2 percent tolerance and substantial higher than the shunt resistance and substantial lower than the input resistance of the ampere meter input.
Bert

 

A funny and interesting site explaining electricity:

http://www.hydrogenappliances.com/ohms_law/ohmslawcalculator.html