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question regarding amperage

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KevinW

Member
I want to operate a 110 volt ac block heater and a 12 volt DC wall adapter that controls a led light which indicates the block heater is working.
If I wire these in series where the block heater fails the light also fails, will the 4 amp rating for the block heater have to match a 4 amp rating for the wall adapter?
I would think so but wanted to check.
Thanks.
 

dknguyen

Well-Known Member
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I don't think that will work the way you expect it to since both are expecting 110VAC but won't get that because they are in series.
 

KevinW

Member
Failed block heater Ron.
I operate a relay at the garage from inside the house to warm the engine on my tractor in the winter , I'd like to know that the block heater is functioning and not just receiving power.
 

dknguyen

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Failed block heater Ron.
I operate a relay at the garage from inside the house to warm the engine on my tractor in the winter , I'd like to know that the block heater is functioning and not just receiving power.
Find a light that runs off very low voltage AC and put that in series with the heater. Then it will only hog a small amount of voltage to run itself and the heater will take all the rest. If both light and heater expect 120VAC, they will both try and grab half leaving enough for neither to run.

But you have to find a light (or use a bunch of lights in parallel) that can handle the full current of the heater since it will also pass through he light and a chain is only as strong as it's weakest link. That probably rules out LEDs.

Do tractor block heaters break down that frequently? Because going parallel would be a lot more simple and straightforward.
 
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ronsimpson

Well-Known Member
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Here is a "current transformer" circuit.
Picture shows 110V---switch---4A block heater.
Break open the wire and insert a transformer. (in this case backwards)
Use any 110 or 220 V to low voltage transformer that is rated at 4A or more on the secondary.
In this example I wired the transformer for 220 to 12V. (will also take 110 to 6V).
The transformer will have 6 volts across the secondary causing 120 volts across the double primary.
Putting a 120V 0.2A 20 watt incandescent bulb on the secondary.
1535824140222.png
Another way to think about it:
120V 0.2A bulb on the primary looks like a 6V 4A bulb on the secondary.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
I recently had occasion to look at a 'Radon Fan Warning Light', it was from a local school who were complaining it didn't light up, so they didn't know if the fan was working or not. In the event there was nothing wrong with it, except they didn't understand it's function - it actually lights up if the fan stops drawing power - as it used a SPDT relay I rewired it to work as they wished.

It was a nice simple device, a resistor in series with the load, and a triac connected across it, which triggered the relay. It would work just as well in this application, but obviously with a lower value resistor.
 

Les Jones

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I would suggest finding a small toroidal transformer with a 9 or 12 volt winding. Wind about 5 turns on it with wire suitable for mains voltage and the 4 amps of your heater. These 5 turns will be connected in series with the heater. Insulate the ends of the primary winding as these will not be used but they will have quite a high voltage between them.) Connect a resistor of about 18 ohms across the 9 or 12 volt winding. Use about a 5 watt rated resistor. with the 5 turns connected in series with the heater measure the voltage across the resistor. You are aiming to get about 5 volts across this resistor. You will probably have to adjust this resiator value and / or change the number of turns that you wound on the transformer. When you get a voltage between 4 and 8 volts connect the input of a small bridge rectifier across the resistor and connect an LED in series with about a 220 ohm resistor to the output of the bridge. An alternative method is to have two groups of four diodes (At least 5 amp rating.) in series connected in parallel but with the opposite polarity. Each group of 4 diodes will conduct in one direction so with the forward voltage drops you will have about a 2,8 volts square wave between the ends. Connect two LEDs in parallel but opposite polarity. Connect these via a current limiting resistor (About 56 ohms.) to the ends of the groups of diodes. (An alternative to using 8 individual diodes would be to use two bridge rectifiers (At least 5 amp rating.) On each bridge rectifier short the + and - terminals together then connect the AC inputs to the bridge rectifiers in series. This may be easier than mounting 8 diodes.)

Les.
 

crutschow

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Note that the current transformer design in post #7 can generate a very high voltage across the 120V winding if left open circuited, so be sure to always have the light bulb load (or a short) on the winding when the heater is connected.
 

Diver300

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I would suggest finding a small toroidal transformer with a 9 or 12 volt winding. Wind about 5 turns on it with wire suitable for mains voltage and the 4 amps of your heater. These 5 turns will be connected in series with the heater. Insulate the ends of the primary winding as these will not be used but they will have quite a high voltage between them.) Connect a resistor of about 18 ohms across the 9 or 12 volt winding. Use about a 5 watt rated resistor. with the 5 turns connected in series with the heater measure the voltage across the resistor. You are aiming to get about 5 volts across this resistor. You will probably have to adjust this resiator value and / or change the number of turns that you wound on the transformer. When you get a voltage between 4 and 8 volts connect the input of a small bridge rectifier across the resistor and connect an LED in series with about a 220 ohm resistor to the output of the bridge. An alternative method is to have two groups of four diodes (At least 5 amp rating.) in series connected in parallel but with the opposite polarity. Each group of 4 diodes will conduct in one direction so with the forward voltage drops you will have about a 2,8 volts square wave between the ends. Connect two LEDs in parallel but opposite polarity. Connect these via a current limiting resistor (About 56 ohms.) to the ends of the groups of diodes. (An alternative to using 8 individual diodes would be to use two bridge rectifiers (At least 5 amp rating.) On each bridge rectifier short the + and - terminals together then connect the AC inputs to the bridge rectifiers in series. This may be easier than mounting 8 diodes.)

Les.
I made a current transformer for a heater to monitor when the thermostat cut in.

I used an old conventional (non-toroidal) transformer, and I cut away and removed the secondary. I put a bridge rectifier and LED across the primary, and a single turn of the heater supply wire (live only) through the space where the secondary had been. The insulation was still on the single turn of wire. The LED came on when the heater took current.

I estimated that I had around 2000 turns on what had been the primary. The mains is 230 V here, and it was a 3 kW heater taking around 13 A, so that gave me around 6 mA though the LED

My suggestion, if using a toroidal transformer, is to wire to the primary to a bridge rectifier and an LED, or two back-to-back LEDs in parallel. A single pass of the 4 A wire though the middle of the transformer should be enough.

You don't need to worry about the voltages. A small toroidal transformer will have around 500 - 1000 turns on a 120 V primary. When you pass the heater supply wire through the middle, that is one turn, and with the primary shorted, the current will be proportional to the heater current, but reduced by the turns ratio. So a 4 A input, with 500 turns on the output (that was the primary) will give about 8 mA on the secondary. If there are more or less turns on the output (that was the primary) you will get a smaller or larger current.

Do make sure that you have a load connected to the output that can take current in both directions.

A transformer and LED arrangement like that will not cause any noticeable voltage drop, and nothing will get hot.

I don't think that the arrangement using the 6 or 9 V winding, (that was the secondary) will work as well. There are far fewer turns on the 6 or 9 V winding, so you will get more current than you want for an LED. There may also be a problem that you need a bigger magnetic field to get enough voltage for the LED, which would mean that you need more turns of the heater wire, so you could end up with far more current than you want in the 6 or 9 V winding, and then a big resistor to bypass or limit the current.
 

KevinW

Member
Thanks to all that responded, I'm going to give it some thought now before I attempt to wire this up.
Lots of input here to consider, thanks again.
 

ronsimpson

Well-Known Member
Most Helpful Member
Making a "current transformer" from scratch is hard and unpredictable.
Here is a data sheet of a real CT.
Many CTs can only output a fraction of a volt. Not enough to light a LED. In this case don't use a small CT like from CoilCraft.com
This part (AX-0500) can output near 3 volts.
Many CTs have a turn ratio of 1000:1. So your 4A will output 4mA which is maybe too small for a LED.
The AX-0500 will output 8mA.

1535850823492.png
You run one wire (not both) through the CT.
Normally you would put a resistor on the output of the CT like in the picture.
Because you don't care to "measure" the current but to just turn on a LED then forget the resistor and put two red LEDs across the output.
Do not use white LEDs because they need 3 volts.
You must use two matching LEDs. (same type and color)
1535851645862.png
The load of 4 amps into the CT will output 8mA into the LEDs.
https://www.digikey.com/product-detail/en/talema-group-llc/AX-0500/1295-1132-ND/4172197
DigiKey has 3000. I have made this circuit.
 

Reloadron

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I would just buy one of these. You can build and try different approaches but I have seen these in the US (maybe on Amazon) for about 10 to 12 USD. I used them with heating elements to detect burned out elements. Simple, easy and non intrusive. The RED LED turns on at 0.75 Amp so it should do fine for you.

Ron
 

KevinW

Member
The led I was planning to use is an smd5050, (deck light).
I can mount it under the eve of the garage and see it from the house.
I'll know if the heater is working when I initially turn it on from thirty feet away.
It generally stays on for about two hours.
12vdc
https://www.tweaking4all.com/wp-content/uploads/2014/01/5050LED.pdf

What I have been using is a light that indicates power going to the block heater but the heater burnt out, the light came on for a couple of hours and the engine wasn't getting any heat.
 

alec_t

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I used an old conventional (non-toroidal) transformer, and I cut away and removed the secondary. I put a bridge rectifier and LED across the primary, and a single turn of the heater supply wire (live only) through the space where the secondary had been. The insulation was still on the single turn of wire.
I did something similar: except that my 'transformer' was just a small shaded-pole motor. No unwinding necessary. Plenty of space for a turn or two of insulated mains wire.
 

Reloadron

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The led I was planning to use is an smd5050, (deck light).
I can mount it under the eve of the garage and see it from the house.
I'll know if the heater is working when I initially turn it on from thirty feet away.
It generally stays on for about two hours.
12vdc
https://www.tweaking4all.com/wp-content/uploads/2014/01/5050LED.pdf

What I have been using is a light that indicates power going to the block heater but the heater burnt out, the light came on for a couple of hours and the engine wasn't getting any heat.
The LED you linked to, the 50/50 is a RGB (Red, Green, Blue) tri color LED. Any reason you plan a tri color? Doing it the way you propose will require a separate power for the LED. Here is an example of what I suggested in post #14. The AC hot is simply passed through the loop. The LED (which comes in a choice of colors) is line powered. They include "snap-in" hardware so mounting requires a single hole. The LED lead wires can be extended.

LED Indicator.png

Pictured with AWG 12-2 ROMEX cable. Figure for under $15 USD it's done. I use them to detect open heating elements in large systems. I can easily see the LED at 50 feet even in bright lighting. Obviously simply detecting voltage to your heating element won't work for the reasons you cited.

Ron
 
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ronsimpson

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The led I was planning to use is an smd5050,
It will be hard to use this LED. It is made up of three LEDs that are not the same.
Red voltage is from 1.8 to 2.7.
Green voltage is from 2.8 to 3.6.
Blue voltage is from 2.8 to 3.6.
It would be better to get two white 3 volt LEDs.
I think you will need two LEDs. One for the positive and one for the negative voltage. OR You could add a full wave bridge so both + and - current goes through one LED. The 4 diodes will need 1.4 volts and a white LED will need something in the 3 to 3.6 volt range. (total is 5 volts)

You will need a turn ratio that gets you the right current. 4A/10mA= 400:1
AND You will need a transformer that will not saturate at 5 volts or what ever the LED voltage is.
 

KevinW

Member
I opted for the tri colour light for visibility only.
A white light might be difficult to see on a bright sunny or snowy day.
I won't see the light directly because it will be flush with the soffit.
 

ronsimpson

Well-Known Member
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I opted for the tri colour light for visibility only.
If you wire all three colors together in parallel only the red will light.
----edited----
You can run the three colors in series so the same current goes through all three colors. The results is 4.8 to 9.9 volt white LED.
 
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