Skip to main content

I have a question about the meaning of mA for a power supply

 

I am going to build a statistic grass applicator from plans available on-line.  

 

One plan calls for a 12vdc 100mA power supply, while another plan calls for a 500mA power supply. As I have no understanding understanding the mA rating I was wondering what the practical difference is between using a 500mA vs a 300mA or 100mA?

 

BTW - both plans are using the same negative ion generator. The only difference is the rating.

 

Thanks,

Ed

Original Post

Replies sorted oldest to newest

Hi ma stands for mili amp, so 500ma is equal to 1/2 AMP so you buy a 1000 ma power supply which would be capable of 1 AMP which would be good for what your saying assuming your load does not exceed 1 amp. Do you understand.

1000 mili amp = 1 Amp that is the relationship

500 mili amp would = 1/2 amp

750 mili amp would = 3/4 of an amp

1000 mili amp would = 1 amp

 

Hope this helps

 

Alan

Alan,

 

Thank you for the explanation.  If I understand your explanation correctly than the difference between the ratings is only relevant as it relates to amperage draw. As in the case of the static grass applicator both power supplies will be just as effective in dispensing the grass. It is just that the higher rated one can pull more amperage.  

 

As you can see my knowledge of electrical components is limited. Using a negative ion generator is new to me and I always like to have an understanding of an unfamiliar component's rating before I jump in.

 

Thank you for your help.

Ed

 

 

I went to the Noch website to find information on their Gras Master.  It would appear you are building a super high intensity static applicator.   Note the specs., Technical Data.

 

Here is some of the information.

60131 Gras-Master

 

Technical Data:

Input: 9V DC. battery.

Max. Operational Ampere: 10mA

Output: 15 kV DC    15,000 volts DC. 

Scenic Express detail shows 20,000 volts. 

Maximum Current: 1,5mA  .0015 amps

Power Supply: 9V Alkaline Battery

Battery Life Expectancy: max. 60 hours

In use operation indicated by illuminated red LED

On/Off Switch

 

The NOCH Gras-Master has been awarded several prizes and is the perfect tool for applying static grass on a model landscape.

Thanks to the electrostatic technique the grass fibres get electrostatically charged and will land in a vertical position on the scenery.

Easy to use:

Brush the NOCH Grass Glue, item no. 61130, to the selected area that you want to grass. Fill the white container on the Gras-Master 2/3 full with the NOCH Grass. Use the fine sieve for the short NOCH Static Grass fibres or the larger sieve for the longer grass fibres (Wild Grass, Field Grass and the Master Grass Blends). Attach the alligator clip to the landscape in the area you are going to grass. This will polarize the device. If you are unable to attach the clip, then we recommend sticking a small nail into the surface area and proceed to attach the clip. Switch on the Gras-Master and gently shake the device over the area covered with glue. Allow the glue to dry and brush off or vacuum the loose grass. You may use the loose grass again. When you complete your task, always switch the Gras-Master off.

Nozzle:

Use the included Nozzle to flock small areas e. g. you can apply long grass to even small sections. The Nozzle is also ideal for flocking areas that are difficult to reach. Two sieves are included with the Nozzle. Use the sieve with many holes (2) for the long grass fibres and the sieve with few holes (1) for the short grass fibres. Put the sieve on the Nozzle (number on the top) and screw the Nozzle on the Gras-Master. Fill the container only 1/3 full when using the Nozzle.

Contents:

Alligator Clip with Cable, 1 x Small Sieve, 1 x Large Sieve, Nozzle with two Sieves, 9 Volt Alkaline Battery (pre-installed), and 1 x Bag of NOCH Static Grass

 

.Summer Meadow. Item No. 08310.

Function:

The electrical circuit within the Gras-Master converts the 9V battery current into a stable high voltage reaching 15,000 V DC. The coating at the bottom of the white container is a special material that creates a negative pole carried by the grass fibres. The positive pole travels through the cable from the top of the device and releases through the alligator clip. As the fibres exit the sieve they are pulled mid-air through an electrical field allowing the grass to land in a vertical position on the

surface area.

To achieve an effective electrical field you should use glue that conducts electricity. The NOCH Grass Glue (no. 61130) is capable of performing this task.

 

Safety Instructions:

The Gras-Master is designed modern safety systems, however, follow these precautions.

Keep this item away from Children. Never physically touch or place a foreign object inside the white container when the device is ON or in operation.

The formation of  sparks is possible during operation. Keep away the Gras-Master from flammable liquids, gas or any combustible materials during operations and storage. Make sure the item is stored in a dry area and protect it from humidity, moisture and wetness. Never use or turn on the device, if you notice moisture or wetness in or on the Gras-Master.

Remove the battery if you place the Gras-Master in storage. Do not throw your used batteries with your household waste. Dispose them according to your local environmental rules. Check your local government waste regulations, if you are not aware of the environmental rules in your area. Electrical and electronic devices do not belong in the trash!

Cleaning Tips:

Make sure the Gras-Master is turned off and battery is removed. Never change or stress the black conductive material in the white grass holder, in the Nozzle or in the two sieves 1 and 2. Do not use solvents, liquid cleaners or water to clean your Gras-Master, or it's parts! Simply use a dry towel.

 

Keep these instructions for reference.

 

Somewhere in your build instruction there should be an indication of the high voltage developed for the static charge.  This voltage and the conductivity of the glue, would play a big part in how well the device works. IMO

I never had a problem using Elmer's white glue.

 Best wishes with your project

Mike CT.

 

Last edited by Mike CT

A Watt is a Watt.  VA is just an alternative.  AC gets trickier because of the sine wave form and a Power Factor is also used.  For DC, pure resistive loads.  1 W = 1Volt-Amp (VA).

 

A transformer has a Voltage and a Current rating.  In your case the call out is for a transformer that produces a steady 12VDC and capable of at least 100ma in one case and 500 ma in the other.

 

If you tried to run the 500ma set up on the 100ma transformer you would be trying to pull more current then the device is rated for and it would overheat and fail.

 

If you chose a 1000ma device  you have a reserve of capacity and will run below the rating.  But you not as efficient either.  So it is a balance.

 

Think Train transformers, it is similar.  G

Originally Posted by GGG:

 

A transformer has a Voltage and a Current rating.  In your case the call out is for a transformer that produces a steady 12VDC and capable of at least 100ma in one case and 500 ma in the other.

 

If you tried to run the 500ma set up on the 100ma transformer you would be trying to pull more current then the device is rated for and it would overheat and fail.

 

If you chose a 1000ma device  you have a reserve of capacity and will run below the rating.  But you not as efficient either.  So it is a balance.

 

Think Train transformers, it is similar.  G

Just to add what G says and this applies mainly to unregulated wall warts. The nominal voltage is only close at rated current. Most wall warts are unregulated. They put out as much as 1/3 more voltage below their rated current. Probably not much of in issue for a static grass circuit but can be for other voltage sensitive devices like Miller Signs DC-DC convertors.

Use a 12V 500ma wall wart for something that requires 10ma and it will likely never get below 15 volts.

The alternative is make sure you have a regulated wall wart. They will be marked such. In that case the voltage will stay close to the rated voltage regardless of current draw.

I have been using a 12VDC wall wart unit for more than 10 years to supply my Tortoise track switch motors. Seldom are more than two motors operating simeltaneously and of course the operation is intermittent.  

 

Although the Analog DC panel meter registers  a steady 12 volts the meter has a 2.5% potential variance. However my more accurate shop meter reflects the same steady 12 volts.

My backup for wart transformer failure is an old HO Power Pak but haven't needed it so far.

Originally Posted by Dewey Trogdon:

I have been using a 12VDC wall wart unit for more than 10 years to supply my Tortoise track switch motors. Seldom are more than two motors operating simeltaneously and of course the operation is intermittent.  

 

Although the Analog DC panel meter registers  a steady 12 volts the meter has a 2.5% potential variance. However my more accurate shop meter reflects the same steady 12 volts.

My backup for wart transformer failure is an old HO Power Pak but haven't needed it so far.

Dewey, FYI Tortoise motors draw current all the time. More at stall but still small like 16ma. They are designed for a very low stall current and won't burn out even when not turning.

 

Pete

I just wanted to thank everyone for taking the time to help and for providing so much great information.  It was all very helpful and I am a lot more knowledgeable now about this thanks to you all.  

 

I  drove over to Radio Shack this morning looking to buy the needed parts with no expectation they would have all the parts.  They were having a 25% off sale on all their electronic components.  They had all the parts I needed.  They also had a box of new unused power packs from various devices.  I picked up 2 12vdc 300mA power supplies for $2 each. My negative ion generator arrived for Oatly yesterday.  Some days are just flat out more fun than others! 

 

Thanks,

Ed

Originally Posted by Ed Walsh:

rt - Your explanation does clear it up for me.  For the grass applicator application, the higher rating would result in a stronger static charge to the grass?

 

thanks,

Ed

I don't know how these work, but yes the grass applicator would have more current available with the 500mA device, if the applicator can use up to that amount of current.

 

Other posters here have probably given better explanations than I have and provided more info as well. I am just a hobbyist/dabbler in electronics, unlike some of the pther posters that really know their stuff.

 

 

Originally Posted by BigBoy4014:

"The 500ma device will supply more current than the 100mA device. The 500mA device will supply more power in watts (also VA in DC)."

 

If my memory is correct, the VA refers to power in AC voltage/current only, where as Watts refers to power in DC voltage/currents

It's a direct conversion with DC. Volts x Amps = Watts and VA or 1 watt = 1 VA. AC is a little trickier because of the power factors which have to be considered, the true RMS and all that stuff. I don't know the numbers, but there is a multiplier that needs to be used in the calculations, it is not a direct conversion like DC.

 

Others like GGG and GRJ explain it better above. They probably also know how to do the calculations properly. I do not without looking it all up, then I may still have trouble. About all I know is that it is there. 

 

 

To continue the discussion, PF (power Factor) is very important to know how much actual power vs apparent power is being used. PF= W/VA. It is good practice to use VA for AC and W for DC to let customers know that there is some heat loss and power loss due to a less than "1" PF. There are power factor corrected power systems that approach unity(1) and they are very efficient in creating minimum heat while providing max output power at a given input power.

 

For the purpose of our hobby, it is not a big concern regarding PF or heat/power loss unless one is using maximum currents out of a 12A AC plug.

"The 500ma device will supply more current than the 100mA device. The 500mA device will supply more power in watts (also VA in DC)."

 

Technically, this statement is wrong.  The device will supply only as much current as the load "requests" or needs at the rated voltage.  Anything left unused is a cushion for any increase in demand.  In the statement, it should have "can" in place of the two "will"s.  In your case, the current supplied by the larger unit will be nearly identical to the current supplied by the smaller unit, except for small differences in the voltage sags of each device under load, but the larger device will have more cushion.

Originally Posted by Dale Manquen:

"The 500ma device will supply more current than the 100mA device. The 500mA device will supply more power in watts (also VA in DC)."

 

Technically, this statement is wrong.  The device will supply only as much current as the load "requests" or needs at the rated voltage.  Anything left unused is a cushion for any increase in demand.  In the statement, it should have "can" in place of the two "will"s.  In your case, the current supplied by the larger unit will be nearly identical to the current supplied by the smaller unit, except for small differences in the voltage sags of each device under load, but the larger device will have more cushion.

Yes, I agree, 'can' would have been a much better choice of wording. I tried to clear up the part about the device using only what it required in my second post after Ed asked about it. The OP also said he had very little understanding of these things, so I was trying to keep it simple.

Originally Posted by Dale Manquen:

"The 500ma device will supply more current than the 100mA device. The 500mA device will supply more power in watts (also VA in DC)."

 

Technically, this statement is wrong.  The device will supply only as much current as the load "requests" or needs at the rated voltage.  Anything left unused is a cushion for any increase in demand.  In the statement, it should have "can" in place of the two "will"s.  In your case, the current supplied by the larger unit will be nearly identical to the current supplied by the smaller unit, except for small differences in the voltage sags of each device under load, but the larger device will have more cushion.

Actually technically it is not wrong.  The 500ma can and will supply more current then the 100ma when called for, but some of this is taken out of context and I think folks understood what BB 4014 was stating.

 

I think if you go back to the original post, there were 2 different devices being looked at.

 

One required 500ma transformer and the other a 100ma transformer.

 

So in the application that requires 500ma, it might be a fair assessment that it needs more than 100ma of current.

 

We are now in overkill mode.   G

Add Reply

Post

OGR Publishing, Inc., 1310 Eastside Centre Ct, Suite 6, Mountain Home, AR 72653
800-980-OGRR (6477)
www.ogaugerr.com

×
×
×
×
Link copied to your clipboard.
×
×