How to protect analog DC ammeter? - Page 1
Forum section: electronics / projects

author: 001
author_rank: Super Contributor
timestamp: September 13, 2018, 07:57:14 pm
content: I have analog panel ammeterIt is basicaly 50mA + shuntBurden voltage is about 200mVHow I can protect it from overload? Parallel Shottky Diode? Mosfet?

author: Zero999
author_rank: Super Contributor
timestamp: September 13, 2018, 09:56:06 pm
content: How much current can it stand before being damaged?Nothing will provide total protection. A Schottky diode will provide some level of protection, but it might start to conduct a little too much a 200mV and affect the result. A small diode, such as the BAT54 would be best. It will only pass 0.1mA, with a bias voltage of 200mV, with a junction temperature of 25oC.https://www.vishay.com/docs/85508/bat54.pdfA 100mA fast blow fuse, is also a must but it might not react in time to avoid destroying the diode, then the meter. A small fuse will be fast, but not have such a high breaking capacity, so it will arc over, at extremely high overloads. A larger fuse, with a higher breaking capacity will be slower and therefore might not protect the meter and diode in time.

author: bsfeechannel
author_rank: Super Contributor
timestamp: September 14, 2018, 04:36:47 am
content: I have the impression that galvanometers are capable of withstanding an overload voltage or current of two or three times the full scale for a couple of seconds. What makes me conclude this is the fact that the old analog multimeters used germanium--and modern ones, silicon--diodes for protection.Also, the thinnest enameled wire, 60 AWG, has a diameter of 8µm, half the diameter of the thinnest human hair, and is capable of conducting a direct current of 150µA; far beyond the full scale of a regular 50µA meter.But, of course I can be wrong.What I also see for protection, specially in multimeters, is a resistor placed in series with the galvanometer like that:It is there to protect the diodes I guess. That will raise the burden voltage up a little, but, if something has to burn, it better be a resistor.Anyway, galvanometers are a lost art. They are still being used in the industry but information and expertise about them is difficult to find.

author: JS
author_rank: Frequent Contributor
timestamp: September 14, 2018, 05:42:39 am
content: Yes, they don't have much trouble withstanding overloads. I have one in the factory that's been working for years with four 30s overloads per day, about 2X the full scale.The shunt also can take some abuse, so in many cases they don't need much protection, if fool proof is required the expected abuse must be specified to define a protection, HRC fuses might be required, depending on the conditions. Two schottky diodes could be an option, uf fast and beefy ones are required (expecting very high fast transients). If slow small abuse is expected a PTC might be enough, but the burden voltage is increased, which in some cases is not a problem, like in mains powered devices, where even loosing a few volts could be ok.JS

author: Zero999
author_rank: Super Contributor
timestamp: September 14, 2018, 08:01:05 am
content: Quote from: JS on September 14, 2018, 05:42:39 amThe problem with beefy diodes is they will conduct more current, at lower bias voltages. For example, the SB320, a 3A Schottky rectifier will be passing 100mA, with a bias voltage of 200mV. A higher voltage rating diode, such as the SB3100 will have a higher forward voltage, therefore conduct less current at 200mV, but it will allow more current through the shunt, when overloaded, so it's trade-off.https://docs-emea.rs-online.com/webdocs/12cf/0900766b812cffcc.pdfAs mentioned before, HRC fuses will be lower, than smaller fast blow ones, hence another trade-off. I suppose another option is to use both an HRC fuse, and a small fast blow fuse in series.The one good thing about PTC resistors is they're resettable, but they're very slow and tent to lower their tripping current and increase conduction losses, after a few cycles. Another option is a magnetic circuit breaker, but I've not seen any in really low current values.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 14, 2018, 01:14:03 pm
content: The usual way is a pair of diodes. Depending on the meter Si or Schottky type diodes are better suited. With a low burden Schottky diodes are more suitable - with a higher burden normal Silicon diodes might be more suitable.Usually the analog meters are not that accurate anyway - so a little extra current through the diodes are less of a problem than with higher resolution digital ones. For gross over-current there are fuses. If the diodes a strong enough the fuse current might be relatively high.

author: schmitt trigger
author_rank: Super Contributor
timestamp: September 14, 2018, 01:53:09 pm
content: If this is a valuable legacy component, that you must protectat any cost...... Don't flame me. What I propose will sound as heresy but:Why don't you use a front end buffer opamp? They can withstand much higher differential voltages, such that you could use some large power silicon diodes across the shunt. And you can limit the opamp's output to whatever you want.

author: Wolfgang
author_rank: Super Contributor
timestamp: September 14, 2018, 09:48:57 pm
content: Lets see ...- regular burden voltage is 200mV- lets put a hefty 3A diode in parallel, say, an 1N5408. At 200mV this diode should have no measurable current (Silicon standard is ca. 0.65V)- Now something bad happens and we have 3A across the diode. The datasheet says now we have 1V across the meter.This is a 5x overload, the needle will jump to the right corner, but it will certainly NOT die. Lets proove this:Regular max is 50mA * 200mV = 10mW. 5x Overload = 250mW, most of this dissipated in the shunt. No meter dies because of that.

author: gbaddeley
author_rank: Regular Contributor
timestamp: September 15, 2018, 01:29:34 am
content: Quote from: 001 on September 13, 2018, 07:57:14 pmWhat is the FSD current that you want to measure?Is it 50mA FSD with the shunt, or without the shunt?What is the maximum possible overload voltage across the meter, or current through it?

author: ogden
author_rank: Super Contributor
timestamp: September 15, 2018, 06:07:53 am
content: Quote from: Hero999 on September 14, 2018, 08:01:05 amLet's not forget initial specs: 50ma, burden 200mV. It means that shunt is 4 Ohms. My math may stand corrected, but IMO even for small 0.125W resistor 0.7V is needed to reach it's nominal dissipation so high voltage diodes are very OK here.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 15, 2018, 07:11:03 am
content: The meter movements are usually rather robust. They are usually less than 100 µA and 0.5 V, thus less than 50 µW nominal power. Even at 10 times overload this would be only 100 times the power and thus 5 mW.The shunt can actually be a problem for the higher current ranges, like 1 A. The burden with analog meter is often rather small (e.g. 50 mV to 100 mV). If the diodes limit the voltage to something like 1 V, thus would be a 10 fold overload and this might not trip the fuse very fast. 10 W might be too much for the shunt - though it depends, as shunts should be normally operated with low self heating and thus quite some headroom. The highest current range is usually best protected by a fuse.Which type of diode to choose depends on the meter: if the meter is low burden, it can be worth choosing low forward drop diodes (e.g. schottky). With increasing temperature the diode drop also tends to go down and after gross overload some time for cooling might be needed to get back to rated performance anyway.