Old Analog Ammeter - Minimum Current - Page 1
Forum section: electronics / beginners

author: luiz_aug
author_rank: Newbie
timestamp: December 13, 2018, 02:18:01 am
content: Hi There.I have here a bunch of old analog ammeter and its maximum scale value (here in Brazil we call it "fundo de escala", witch term is used for this in english?) is 10 A.I want to use these analog ammeters as a weather display, connected to some uP over the wifi.But to do so, as a energy efficiency concern, I need to be able to deflect the pointer over all the scale by using the minimum current possible.I have already opened one and removed the "shunt" resistance, and by applying a 3.3V signal through a potentiometer, I discovered that the new value of the full deflection current is about 5.18 mA.Now comes my questions:(1) - Is there a way toreduce even more this current? I thought about it for a while and Im guessing that the only way is by reducing the spring resistance that opposes to the pointer movement (aka: changing to another analog ammeter).(2) - As a magnetic circuit, is there any way to"recycle" some energy?I know these are very specific (and maybe newbie) questions, so I appreciate a lot your comments here.Thanks!

author: Nusa
author_rank: Super Contributor
timestamp: December 13, 2018, 03:31:55 am
content: Energy (power over time), power (rate of energy use, P=IE) and current (I) aren't the same things. 3.3 Volts * 0.00518 Amps = 0.017 Watts for the circuit. The bulk of that is being converted to heat at the resistor (potentiometer in your case) in the circuit, not the meter itself. If you measure the voltage drop across the meter alone you will discover the voltage at FSD, which will then let you calculate the internal resistance of the meter. Measuring that directly with a DMM is not advised, as the test voltages used could destroy a sensitive meter.Here's an excellent series of articles on analog meters:https://www.embedded.com/electronics-blogs/embedded-down-under/4439175/Analog-meters---The-joy-of-movement

author: Brumby
author_rank: Unknown
timestamp: December 13, 2018, 05:11:58 am
content: @Nusa - while what you say is not wrong, the OP appears to have put together a test setup that would seem to have been effective. Voltages aren't all that important - but the current IS. So long as the combination of voltage and resistance allows control of an appropriate current, then all is well.@luiz_aug - 5mA is fairly high for FSD of a meter movement. A 50uA movement shouldn't be hard to find.But to answer you questions(1) - Possibly - but the spring is a very carefully manufactured part. I would not suggest you go any further than a thought experiment.(2) - No. The meter movement consists of a balance of magnetic fields between the permanent magnet and the energised coil against the spring. You could improve sensitivity by using a stronger permanent magnet - however, the geometry of the magnet and its field is important - and critical for accuracy across the scale. Again, I would not suggest you step beyond the thought experiment.I seriously suggest looking for a meter with greater sensitivity. An affordable 50uA should not be too difficult to find (Hmmm... not sure about Brazil) but I think the movement out of an old multimeter would be an option.Edit: Typo

author: Nusa
author_rank: Super Contributor
timestamp: December 13, 2018, 07:23:23 am
content: @BrumbyYes, his test circuit works, although there's a risk to the meter being tested if he ever sets the pot to a low or zero value.He asked two questions. Other than terminology, my explanation was to point out that the power cost of the circuit is negligible at the meter itself, compared to the rest of the circuit. For instance, if one changes the source voltage from 3.3 V to 1 V, while changing the resistance to get the same result, you've cut your power expenditure by over a third. That's the math.Current, Voltage Drop, and Meter resistance are all related in the usual way. You can't change one without affecting another. The articles I linked really are a good explanation of it all. The last part has some really practical stuff.

author: Brumby
author_rank: Unknown
timestamp: December 13, 2018, 10:52:11 pm
content: I'm not saying what you said was wrong - just irrelevant.It is perfectly clear to me that the OP was talking about theirtesting setupto determine the meter's sensitivity. There has been NO description of the actual circuit into which they want to put such a meter.

author: george.b
author_rank: Frequent Contributor
timestamp: December 13, 2018, 11:12:05 pm
content: Quote from: luiz_aug on December 13, 2018, 02:18:01 amIt's called "full scale".

author: luiz_aug
author_rank: Newbie
timestamp: December 28, 2018, 12:27:41 am
content: Thank you guys, I'll read the articles posted by Nusa and take Brumby`s advice and forget about changing the spring or the permanent magnet of the movement mechanism.Again, my first post here, I'm very happy with the detail of the answers given !

author: spec
author_rank: Frequent Contributor
timestamp: December 29, 2018, 12:26:12 pm
content: Hi luiz_augQuote from: luiz_aug on December 13, 2018, 02:18:01 amI presume that you want to limit current consumption to extend battery life in the weather display unit.Yes, there is a way to effectively reduce the current for the full scale deflection (FSD) of the meter. The power used by the meter is 5.18mA * (whatever voltage was across the meter terminals at FSD). This is a constant power (I*V) and cannot be changed practically. But what you can do is to change the I/V ratio to suit the voltage of your power line (battery).Take an example:Meter FSD current: 5mA (round figures)Meter FSD voltage: 100mV (assumed)Thus meter power requirement = 5mA * 100mV = 500uWPower line = 3.6V (single LiIon battery cell assumed)If you drove the meter directly from the 3.6V power line the current would be 5mA and the power dissipation would be 5mA * 3.6V= 18 mW. So 17.5 mW would be wasted in heat.But, on the other hand, if you used a micro-power switch mode power supply (SMPS) to generate 100mV from the 3.7V supply line, the wasted power would be zero (assuming for the time being that micro-power SMPS were 100% efficient). So now the current taken from the 3.7v power line for meter FSD would only be 500uW/3.6V = 139uA. In practice, I would imagine that you would be able to reduce the current drawn from the 3.6V supply line, for meter FSD, from 5mA to 200uA.And the micro-power SMPS could be implemented by a microcontroller PWM output pin and a voltage monitoring input pin. A small inductor, small Schottky diode, and a small capacitor would complete the micro-power SMPS (roughly speaking). But there are many ways to realise the micro-power SMPS.By the way, depending on the total power consumption of your weather station unit, you could think about adding a small solar panel to effectively make battery life infinite. Small solar panels are dirt cheap these days.

author: luiz_aug
author_rank: Newbie
timestamp: February 27, 2019, 07:32:20 pm
content: Hi Spec!I take a while to read your comment, but thank you so much for the valuable information!I'll study some SMPS topologies and try to apply your advices.Again, thanks!