We are powering a circuit with a 12 Volt battery. We have a RF remote controlled Relay to turn the system on. When the system turns on the current draw is 15 Amps and then settles to 6.5 amps. The voltage drop is about 2 volts +- on start up too.(measured using a multimeter). Our problem is the RF circuit is bothered by this voltage drop and starts to act funny and not do what it is supposed to do when pressing the remote buttons. If I power the RF unit with an additional battery everything works fine. My thought is if I can put a cap on the power supply side of the RF unit it will act to smooth out the voltage issue. Can someone please let me know if I am thinking correctly and how I would figure out the correct value Capacitor. This is a small setup so I need to be able to put the cap in tight quarters. I look forward to any suggestions. Thank you in advance.
Electrical – Need Capacitor for 12 volt Circuit
capacitor
Related Solutions
you have to understand one the thing for eg,if you connect a capacitor across a 9V PP3 battery or some other battery (not fruit or vegetable battery) the capacitor would get immediately charged to the battery voltage (but it would take a certain time because of the presence of internal resistance would reduce the flow of charge Q; C=Q/V) and the current through the capacitor would get immediately reduced.
Click here and simulate the circuit by giving different R values like a)0.1 ohm b)1 ohm c)100 ohm d)1K and observe the volatge and current of capacitor for different values of resistor.
Where to start?
First, your HV unit is not a transformer. It is a high-voltage module. It puts out a pulsing high voltage.
So, the high voltage transformer was made to make arcs, and as such does not allow ... a no-load situation.
You are correct. Your capacitor will store charge and will destroy your module. Furthermore, you have taken the input power specification, (and it would be nice if you would share with us where you got the numbers) of 18 watts at 4.5 volts, and then calculated an output power of 18 watts/20 kv, giving .9 mA. This is wrong. Input power does not equal output power.
You have also failed to notice the part of the description that reads
Work: Input voltage 1.5V ~ 3V can work for about 1 minute, the input voltage exceeds 3V continuous work does not allow more than 30 seconds
In other words, you will not be able to run your unit for more than 30 seconds at a time. And it doesn't say how long you need to let the unit recover from 30 seconds operation, either.
Having failed to understand your HV unit, you have equally failed to understand your water capacitor. To begin with, ultrapure water is does not have an infinite resistivity. It is, in fact, 18 Mohm-cm. For your described capacitor, this amounts to ~ 2 Mohm. At 20 kV, that will require 10 mA which you cannot provide. You mention a "coating on the plates" which you believe will prevent current flow, but you do not describe it, and I hope you will forgive my suspicion that it may not work as you plan.
You speak of your desire not to break down the dielectric (the water) and then talk about how pulsing or AC voltages will avoid this. You seem to be unaware that the breakdown potential for pure water is ~3 MV/m. Since the spacing for your capacitor is on the order of 12.5 mm, the breakdown voltage will be ~ 37 kV, or twice your voltage, and you have no need to worry.
Given the dielectric constant of water (80), the effective capacitance of your capacitor will be ~ 100 pF. Since your module has a diode in its output, you are correct in thinking that you will get just a DC output, since there is no major discharge path. You cannot discharge your capacitor from the input side. Any such discharge mechanism would have to be synchronized with the HV pulses, and would have to withstand 20 kV. Such a circuit will be neither simple nor cheap.
Your selection of series resistor is very strange. Your calculation for voltage drop make it clear that you have used the input figure of 4.5 amps, and this has absolutely no application to the output.
So, overall, I have no way give you advice, except to suggest that you do something else. I don't see any instance where you seem to actually understand how your proposed circuit would work, or why, and that is not a good starting point.
Best Answer
Either try a big capacitor across the RF receiver being fed through a big resistor which will more or less isolate it from the sudden transients when the circuit is triggered, or (the more sophisticated, but more reliable solution), use a boost converter to feed the RF section so that the battery voltage is irrelevant, the boost converter will always supply a voltage above the battery voltage (and it'll be a fixed, regulated voltage too) so that the RF section gets a steady supply voltage under all conditions.