Help on Constructing a Circuit with 205VDC to 12VDC

[gabrielsky]

Hi Everyone,

I am trying of constructing a circuit but i do not have much electrical knowledge to put me through this. I hope someone here could provide some advice for me.

Here it is:

I have a circuit that is powered up by 230VAC (main).
I am trying to tap a voltage (205vdc) somewhere from the middle of this circuit. I'm not sure how they converted into 205vdc but i know this voltage currently drive a dc motor.
My question is, can i tap this voltage out in parallel to drive another soleniod valve of 12vdc? I'm sure i need to reduce the voltage to the device of 12vdc.
Does it means that i need a resistor to reduce this?

I have checked the desired specs of the solenoid valve:
Power: 0.4W
voltage: 12vdc

Therefore is the calculation below correct?
load Current: Watt/volt = 0.4/12 = 0.03A

which means to say that resistor should be volt/current
=(205vdc-12vdc)/0.03A = 6.4k ohms

 

[berkeman]

Hi Everyone,

I am trying of constructing a circuit but i do not have much electrical knowledge to put me through this. I hope someone here could provide some advice for me.

Here it is:

I have a circuit that is powered up by 230VAC (main).
I am trying to tap a voltage (205vdc) somewhere from the middle of this circuit. I'm not sure how they converted into 205vdc but i know this voltage currently drive a dc motor.
My question is, can i tap this voltage out in parallel to drive another soleniod valve of 12vdc? I'm sure i need to reduce the voltage to the device of 12vdc.
Does it means that i need a resistor to reduce this?

I have checked the desired specs of the solenoid valve:
Power: 0.4W
voltage: 12vdc

Therefore is the calculation below correct?
load Current: Watt/volt = 0.4/12 = 0.03A

which means to say that resistor should be volt/current
=(205vdc-12vdc)/0.03A = 6.4k ohms

Welcome to the PF.

What exactly are you trying to do? To be honest, we get a bit worried when a new poster comes along without much knowledge of AC Mains circuits and safety, and asks for advice on how to hook something up. What is the application, what is the machine, and why are you trying to do this with so little training? (not trying to be insulting here, just trying to assess whether I can keep this thread open under the PF rules...)

 

[vk6kro]

You can get 12 V DC power supplies. These are in the form of a small box (a "wall wart" or a "plug pack" depending on where you live) that plugs into the power outlet on the wall and they have two wires coming out that have 12 V DC between them.

This is a lot safer than somehow trying to generate 12 volts directly from the mains.

 

[gabrielsky]

Hi,

Alright..This is a UV curing machine which runs by 230VAC main.
when it is power on, the uv lamp runs continuously and UV light coming out is controlled by a dc motor.
when a switch is pressed, a timer and dc motor is activated as long as the timer is set.
thus when the time is up, all voltage is cut off.

Now why am i trying to hook this solenoid up is because, I wanted to drive a actuator using this solenoid valve. This actuator will fix on to some jig that holds on to the UV light guide. So that when the switch is pressed, (timer),(dc motor) and (solenoid valve) is activated and gets cut off when time is up.

Here the sequence:
switch on --> timer activated, dc motor activated (for UV light to exit & shine), solenoid valve activated (actuator goes down to position the UV light on the desired point) --> timer stop ---> all voltages cut off --> dc motor close to block UV light from exiting) --> solenoid valve (drive actuator up to bring the UV light away from the point).

I am sorry if it brings about trouble to you guys. I understand your worries.
Thanks anyway!

 

[vk6kro]

OK, if you understand the safety aspects, you can run the solenoid off 205 volts DC, assuming it is able to supply an extra 33 mA.

You put a large resistor in series with the solenoid.

This resistor would have 205 - 12 volts across it. That is 193 volts.

It has 0.03333 amps flowing in it. So, by Ohm's Law, the resistor must be 193 volts / 0.0333 amps or 5790 ohms.
And you need to know how much power it would dissipate.
Power = 193 volts * 0.033 amps = 6.43 watts.

Sadly you cannot buy 5790 ohms resistors, but you could use a 5600 ohm resistor but it would need to be rated at 10 watts. It might be better to put two 12 K resistors in parallel to get 6000 ohms. These could then be 5 watt resistors and this may work out cheaper than a single 10 watt resistor.

 

[gabrielsky]

Hi vk6kro,

Cool. thanks for the advice.
But i have another questions.

When i use a voltmeter to measure across the existing dc motor running on 205vdc, it measure a 205V on DC mode. But when i switch the voltmeter to AC mode, it also shows a value of 100.5V...

from the voltmeter, its telling me that the voltage there is 205V in DC and 100.5V in AC.
is there any concern on the AC part?
by the way, why is there an AC voltage found across the dc motor?

 

[vk6kro]

It depends on how the meter is wired internally.

If the meter just switches a diode in series with the input, it will work on DC, but only with the leads one way around. It will use a different multiplier resistor, too, so you can't believe the AC reading. If it reads on DC it certainly is DC while AC or DC can produce a reading on the AC scale.

It is also possible that the 205 volts DC is not pure DC but pulsing DC. This will pass through a capacitor if the meter uses one to split off AC, so you will get a reading.

Just use the meter on the DC ranges.

 

[gabrielsky]

Perfect... I got your explanation clear.. I will try it out.
Thanks ya!