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Topic: 16VDC to 12V DC (Read 526 times) previous topic - next topic

anchitdaga

Oct 21, 2019, 01:51 pm Last Edit: Oct 21, 2019, 01:55 pm by anchitdaga
Hi,

I have a Steam Generator installed at my place for using steam shower.

Old Water Solenoid Valves have no rating mentioned on the motor. I had to check them through Volt Meter. Output from machine is 15.8V & Old Solenoid Valve amp 540mA. This system is about 10 Years old and Solenoid Valves are rustic and have to be replaced.

New Solednoid Valves available and purchased from Market are 12V and 500mA.

I also read in another forums that installing a resistor of 1.2ohms would solve the step down problem

Please advise what changes or additions should I do to existing setup to make them compatible.

I am just a regular guy with no expertise in electronics. Request for your help in simple and layman words.

Idahowalker

You could use a 7812.

Robin2

#2
Oct 21, 2019, 02:27 pm Last Edit: Oct 21, 2019, 02:27 pm by Robin2
New Solednoid Valves available and purchased from Market are 12V and 500mA.

I also read in another forums that installing a resistor of 1.2ohms would solve the step down problem
Using Ohms law V = R * I then the resistance of the 12v coil would be 24 ohms ( 12 = 24 * 0.5)

Following the same logic to get 500mA with 16v would require a resistance of 32 Ohms (16 = 32 * 0.5)

That suggests that you would need a resistor of 32 - 24 = 8 ohms if you want to use the 16v supply with the new valves.

The voltage drop across the 8 ohm resistor will be 4v so the energy dissipated will be 4 * 0.5 = 2 watts. That suggests to me that you should choose a 5-watt resistor. It will probably get quite hot so make sure it is suitably ventilated.

(Hope my maths is correct)

...R
Two or three hours spent thinking and reading documentation solves most programming problems.

JCA34F

Quote
Output from machine is 15.8V
AC or DC? 8Ω resistor may not work with AC.

MarkT

One thing a resistor is fine for is working the same with AC as DC!

AC voltages are quoted as rms, specifically so all the relations:

V = IR
P = V^2 / R
P = I^2 R

all work the same for resistive loads.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

JCA34F

#5
Oct 22, 2019, 05:01 am Last Edit: Oct 22, 2019, 05:03 am by JCA34F
But what about the ac inrush? The R may drop enough V that the solenoid plunger never bottoms completely, (sealing the magnetic path) and buzzes 'til it smokes. Been there, done that.  :)

Robin2

#6
Oct 22, 2019, 10:09 am Last Edit: Oct 22, 2019, 10:10 am by Robin2
But what about the ac inrush? The R may drop enough V that the solenoid plunger never bottoms completely, (sealing the magnetic path) and buzzes 'til it smokes. Been there, done that.  :)
Interesting. I know very little about AC electrics and my immediate reaction to your Reply #3 was that more resistance might be required, but this Reply suggests (to my ignorant mind) that a lower value would be needed.

How should the OP calculate the resistor value if it is an AC supply?

...R
Two or three hours spent thinking and reading documentation solves most programming problems.

anchitdaga

Interesting. I know very little about AC electrics and my immediate reaction to your Reply #3 was that more resistance might be required, but this Reply suggests (to my ignorant mind) that a lower value would be needed.

How should the OP calculate the resistor value if it is an AC supply?

...R
Hi Robin,

Please advise If I am correct i.e. to make the above system work I should install a 8ohm resistor with 5 watt capacity ??

Regards

Robin2

Hi Robin,

Please advise If I am correct i.e. to make the above system work I should install a 8ohm resistor with 5 watt capacity ??

Regards
I believe that's what I said in Reply #2 - have you checked my maths?

And is your system using AC or DC?

...R
Two or three hours spent thinking and reading documentation solves most programming problems.

jremington

#9
Nov 04, 2019, 05:49 pm Last Edit: Nov 04, 2019, 11:37 pm by jremington
Quote
But what about the ac inrush?
No "inrush" with a inductor, AC or DC.

DC charging curve:

JCA34F

#10
Nov 05, 2019, 05:51 am Last Edit: Nov 05, 2019, 06:11 am by JCA34F
Check the steady state AC current on a solenoid with the plunger seated, then check it again with the plunger either blocked, so it can't pull in, or removed completely. Do the second test quickly, without the iron, the coil will overheat and burn out fast.
What is the inductive reactance with no plunger in the coil? WITH the magnet iron plunger seated in the coil?
I don't have an AC solenoid or relay around the house to perform a test.

Robin2

Isn't this getting a little esoteric for the OPs needs?

He has a device that is works fine with a 16v power supply and a 16v solenoid. All he needs to know is how to make the same 16v system work with a 12v solenoid.

Won't an extra resistor be adequate?

...R
Two or three hours spent thinking and reading documentation solves most programming problems.

anchitdaga

I believe that's what I said in Reply #2 - have you checked my maths?

And is your system using AC or DC?

...R
I have absolutely no clue about electronics hence the maths. I am afraid I will have to blindly follow your advice. I am 100 percent sure supply to Valves is DC and not AC.

Supply to Steam Generator is AC and I believe the circuit/converter inside would covert it to DC before supplying to Valves.

MarkT

Definitely use a resistor, make sure it has adequate power rating.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

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