Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Current Measurement Problem

Status
Not open for further replies.

muashr

Member
Hi,

  1. I tried to measure the current of a dc source (with specifications: 5V/200mA) by connecting the ammeter's +ve with the +ve of dc source and ammeter's -ve with the -ve of the dc source. However, the current shown was less than the actual 200mA. The measured current was about 130mA.
  2. Across a source of 25V a series combination of a 10MOhm resistor and Ammeter is connected. A Voltmeter is also connected in parallel to the source. With this setup the Ammeter shows no current! The Multimeter used as Ammeter & Voltmeter can be seen **broken link removed**. Can anyone help in this regard?
  3. Can someone give me advice about measuring current especially with very less amount i.e. in uA?
 
An ammeter typically has a very low resistance. When you attach it directly across a voltage source, you are effectively shorting out the voltage source. In this case, the dc source may be going into a protection mode such as current limiting, or fold-back or crowbar mode. You really should not do a measurement this way unless you know exactly how the dc source will respond to a short circuit. If the dc source has built-in current limiting, then it would be ok to do it this way, but many dc sources do not have limiting and putting a short across their output may result in a lot more than the rated current flowing through the meter. In such cases, the fuse in the meter can open, or the meter may be damaged, or the dc source can even be damaged. I think that the best way to measure the current from the dc source is to put a load resistor on the source, in this case 25 ohms would be appropriate (using ohms law), and then place your ammeter in series with the resistor.

In the second case, using Ohm's law, I calculate that the amount of current to be expected through the 10Mohm resistor is .0000025 amps (2.5 microamps) which is likely too little for the meter to measure. Why not try a 1K ohm resistor instead?

Ohms' law is very useful in these cases. It says that in a resistive load, voltage across a resistance is equal to the current through it times the resistance. V=IR. simple algebra allows you to rearrange this in two other ways I=V/R and R=V/I so when you know two things you can always calculate the third.
 
The basic rule:
A voltmeter goes across (in parallel with) both terminals of the voltage source.
An ammeter goes in series with one terminal of the voltage source to the load.
Mix them up and you can get fireworks or an incorrect reading.
 
Yea, so. The service of your house may be rated at 200 A @ 120 VAC [simplification] (your 200 mA value) and if you only had one 10 W light bulb on, you won't measure 200 A. The service, or in your case, the power supply, has to be rated higher than the item being powered.
 
There is no such thing as "200mA output" unless the power supply limits the current to 200mA. It is probably 200mA maximum allowed output. Then it can supply its rated voltage or more when the current is zero, very low or as high as 200mA. If you try to draw more than 200mA (like your current meter that is like a short circuit) then something will burn up.

Ohm's Law says that with 5V and 200mA then a resistor load is 5/0.2= 25 ohms. So connect your current meter in series with a 25 ohms resistor and when connected to the 5V it will show 200mA.
 
Simple DC source with 5V/200mA output.

"Simple" DC source can fall into two categories: constant current source or constant voltage source.

I guess constant voltage source is "simpler". It'll try to maintain voltage regardless of current (5V up to 200mA). If you short it as you did, it won't be able to maintain the voltage any more and it'll either die, disconnect, or limit the output.

Constant current source will try to maintan current regardless of voltage (200mA up to 5V). If you short it, you get the rated current, but the voltage will be very small. Some current sources will not be able to operate at such a small voltage and will disconnect.
 
I have NO IDEA what everyone is doing here. The markings on a power supply refer to the limits of the supply.

A 5 V, 10,000 A power supply will read 150 mA if that is what the load draws at 5V.

Inserting ammeters CAN reduce the voltage to a hopefully insignificant value.
If the load draws more than what the supply can deliver, then the supply USUALLY protects itself. It may go into current limiting and supply the 200 mA at 1V say. I have one supply rated at 10A at (0-32V), and if you short the leads, the ammeter reads 10A and the analog voltmeter reads near zero. It's trying to supply 10 Amps to a very low load in milliohms or so and the voltage reduces accordingly.

A fuse or a "controlled break" is another protection mechanism. If the transformer fries within the enclosure, the riskof fire is minimal. Yea, the power supply is broke, but they told you on the label what it could do.
 
A simple cheap wall-wart produces up to 10VDC with no load and produces its rated 5VDC at 200mA with lots of ripple.
 
I have NO IDEA what everyone is doing here. The markings on a power supply refer to the limits of the supply.

KISS,

He didn't say it was a power supply. He said it was a power source. Could be a solar panel, a battery, a wind turbine or anything else he's experimenting with. I asked what it was, but he didn't really say.
 
the OP said:
current of a dc source (with specifications: 5V/200mA)

Northguy said:
He didn't say it was a power supply. He said it was a power source. Could be a solar panel, a battery, a wind turbine or anything else he's experimenting with. I asked what it was, but he didn't really say.

or could be ... a language barrier. I'll rule out a solar panel and a battery because the cell voltages don't compute.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top