Changing voltage 10VAC to 5VDC

AlbuquTurkey · #1 February 7th, 2005, 03:46 PM

I would like to be able to take the 0 to 10VAC that comes from the CT that Rupp likes and convert that to 0 to 5 vdc that SECU16 likes.
How can you do this?
If possible (and cheep) it could be an inexpensive power monitoring system.

Barry Gordon · #2 February 7th, 2005, 04:09 PM

Easy. Just pickup at Radio Shack the cheapest 5 volt wall wart. That puts out DC. Plug it or wire it into your 110 V source. Get the lowest power (amperage) unit you can find, the lower the power, the less expensive. I do it with a little AC 5 volt unit and feed it into an analog input on my digital / analog input board (already in place). If your good with electronics a bridge rectifier, and a resistor voltage divider will do the same thing.

AlbuquTurkey · #3 February 7th, 2005, 08:29 PM

I like the wall wart idea. But is there one that is linear from input of 0 to 10 (ten) Volt AC input with a 0 to 5 volt DC side out put. Even if it is not linear but is consistant with the variation through the range.

I realize that it is not a wall wart it will have to be a transformer with input 0-10 AC and output 0-5 VDC. If anyone has an idea of where to find this cheep I would like to know.

rocco · #4 February 8th, 2005, 01:00 AM

I assume that the CT you are referring to is one of the CT30xx current transformers that have been batted around. If so, the burden resistor sets the current-to-voltage ratio, and correct selection of that resistor can give you 0 - 5 volts.

If you post which CT you want to use (if you have a preference), and/or how much current you want to measure, we can come up with exactly what you need.

Mario from Spain · #5 February 8th, 2005, 03:38 AM

I'm in exactly the same boat.

I ordered two CR3110-3000 CTs from CRMagnetics.

I plan to use one to measure a 25 Amp circuit and one 15 Amp circuit. Both circuits are one phase (just phase and neutral) 220V 50 Hz.

rocco · #6 February 8th, 2005, 10:54 PM

Mario:
The equation for the burden resistor value, R, is:
R = 3100 * V / I where:
V is the output voltage you would like, and
I is the maximum current that you would then give you that voltage, V.

For your 15 Amp circuit, you would use a 1000 Ohm resistor (any wattage you want, it only dissipates 25 milliwatts at full current), which will give you 4.8 volts when you have 15 Amps.

For your 25 Amp circuit, you would use a 620 Ohm resistor (again, as small as you want, this one will dissipate 40 milliwatts), which will give you 5 volts when you draw 25 Amps.

Hope this helps.

Mario from Spain · #7 February 9th, 2005, 02:28 AM

Thank you very much, rocco.

I'll run now to buy the resistors.

siliconmelt · #8 February 9th, 2005, 07:50 AM

Quote:

Originally Posted by AlbuquTurkey

I would like to be able to take the 0 to 10VAC that comes from the CT that Rupp likes and convert that to 0 to 5 vdc that SECU16 likes.

I see that suggestions have been given to put a current limiting resistor inline to drop the voltage from 10vac to 5vac. Will the secu16 accept this as an analog dc signal? Seems to me, a bridge rectifier as mentioned already is needed to filter the ac to dc right?

AlbuquTurkey · #9 February 9th, 2005, 08:02 AM

I am still concerned that this is not going to change AC current to DC current.

I was looking at doing the whole house. I have not looked at the panel yet to verify but I believe that it is a 300 amp service at 220V. I would be attaching to each of the 120V legs for the load reading. That would be two at 150 amps.

The CT I would be using is the CR3110 - 3000. In the first paragraph in the writeup on the CT says that it can create upto 10Vac.
http://www.crmagnetics.com/pdf/3110.pdf

The Ocelot is a DC unit and excepts upto 5 V. I understand the V=IR/3100 but I don't see the AC to DC. Does it matter?

rocco · #10 February 9th, 2005, 03:47 PM

Quote:

Originally Posted by AlbuquTurkey

I am still concerned that this is not going to change AC current to DC current. . . Does it matter?

I completely overlooked the fact that to Ocelot's input is DC only, if it really is. I don't have an Ocelot, and have never seen a schematic of their analog front end, so I am going to be conservative here.

A full-bridge rectifier, using schottky diodes, between the CR3110 and the burden resistor should do the trick. Since the voltage is developed across the resistor, that is where you want to measure it. If the bridge was after the resistor, you would be measuring the voltage at the resistor MINUS two diode drops. With the diodes first, the CR3110 will be generating a current though them, not a voltage, and their effect will be less significant. To minimize the effect further, use schottky diodes for their low voltage drop. I would also put a small-value capacitor, around 1 nF, in parallel with the resistor to deal with the ripple.

If anyone wants to see a schematic, let me know.

toscal · #11 February 9th, 2005, 04:31 PM

Try this link here http://www.ee.washington.edu/circuit...32/theory.html
Ok I know its for 12V AC to 10Vdc but in the text there are the formulas to change the resistance for the desired Output voltage.

WayneW · #12 February 9th, 2005, 05:21 PM

Quote:

Originally Posted by rocco

The equation for the burden resistor value, R, is:
R = 3100 * V / I where:
V is the output voltage you would like, and
I is the maximum current that you would then give you that voltage, V.

Where does that 3100 come from? I found the same formula on the CR-3110 spec sheet, but the spec sheet also states it only has 3000 turns on the secondary. I realize the difference between 3000 and 3100 is pretty small and may be a typo, but I wanted to make sure I wasn't making a bad assumption.

Mario from Spain · #13 February 9th, 2005, 06:03 PM

Quote:

Originally Posted by rocco

If anyone wants to see a schematic, let me know.

Don't know if I need and schematic but I don't understand? Can you explain me how is the bridge after or before the resistor? Or maybe you mind the resistor before or after the bridge?

Barry Gordon · #14 February 9th, 2005, 06:08 PM

Pardon my eyes. When I read the original post I read 110VAC, hence my reply.

AlbuquTurkey · #15 February 10th, 2005, 11:39 AM

The schmatic that toscal linked is great. Now I can build one. But is there one on the market?

Michael McSharry · #16 February 10th, 2005, 01:24 PM

When I do 5VDC voltage regulation I use a 7805 regulator. The input can be pretty sloppy and it will clean it up to a nice 5VDC output. They sell for under $1 and maybe a little more at Radio Shack. The circuit can be a xfmr, full wave bridge and cap as was posted above or a half wave using a single diode and cap, and the 7805. This puts the parts count at under 5 pieces and can easily be easily soldered with point to point wiring.

Other fixed voltage regulators are available in the same family and variable ones also exist if you want to add a variable resistor to tune-in a specific voltage.

rocco · #17 February 10th, 2005, 10:39 PM

For those trying to do this: YOU DO NOT WANT A VOLTAGE REGULATOR !

A voltage regulator will take the voltage that comes in and regulate it down to a FIXED voltage. Then the Ocelot will read the same voltage whenever the incoming voltage is higher than (by some amount, between .04 and 2 volts) the setpoint of the voltage regulator. This won't help if you are trying to make a measurement.

It's clear where the misunderstanding came from: "Changing voltage 10VAC to 5VDC" doesn't convey the fact that the voltage varies from 0 to 10 volts. But in reality, the CR3110 doesn't put out 10 volts. It can put out UP TO 10 volts, depending on the burden resistor. You simply set that resistor to a value that will give you your desired voltage, 5 volts in this case.

Wayne:
The 3100 comes from the equation on the bottom of their data sheet. Their equation defines voltage as a function of resistance and current. I simply reworked it to give resistance as a function of voltage and current.

Mario:
Sorry if I wasn't clear of the placement of the bridge. Here is another attempt.
the output of the current transformer (a proportional current) goes into a full-wave bridge rectifier, thereby rectifying the AC current signal to DC. The output of the rectifier is a DC current and goes to a small filter capacitor to help smooth the remaining ripple, and then to the resistor, that gets a voltage developed across it proportional to the current, and finally to the Ocelot to measure that voltage.

I will try to do a schematic before the weekend.

Michael McSharry · #18 February 10th, 2005, 10:49 PM

Rocco, you are correct I should have read the top of the tread to see what was actually being attempted.

rocco · #19 February 11th, 2005, 02:04 AM

hi

Mario from Spain · #20 February 11th, 2005, 05:50 AM

I think when you convert a AC to DC by a full wave rectifier bridge you have to multiply by 1.41 (or something like this I'm not sure about this conversion factor) so the maximum voltage at the burden resistor will be 5x1.41 = 7.05 Volts if rectifying it full wave.

To have 5 Volts DC we need at the input of the rectifier 5/1.41=3.54 Volts.

So we need to calculate a burden resistor that gives us 3.54 Volts AC and at the voltage and current we want.

Quote:

Originally Posted by rocco
The equation for the burden resistor value, R, is:
R = 3100 * V / I where:
V is the output voltage you would like, and
I is the maximum current that you would then give you that voltage, V.

For example:
For 220 Volts 15 Amps: R = 3100 * 3.54 / 15 = 731.6 ohms.
Looking for a standard resistor like 680 ohms.
Deriving from same formula V = R * I / 3100 so V = 680 * 15 / 3100 = 3.29 Volts

Then, by using a 680 ohms burdern resistor we will have at the Secu16 input 3.29 * 1.41 = 4,64 Volts DC (so we have about 10% security margin)

I studied electronics 22 years ago, so: Is this at all correct?