It is also necessary to know how long the dump is supposed to sustain 1.7amps and how low the voltage is allowed to go to be acceptable. Also it would be helpful to know what the load is. 1.7 amps at .23 volts implies an Ohmic load of .135 ohms but if the load is not purely resistive (and I suspect it is not) that can matter.JoeSalerno said:This is a pretty simple question, one that there is likely a simple answer to. So, I need to use a capacitor as a rapid dump battery, not a filter or anything. The required current for this application is 1.7 Amps, and only 0.23 volts is necessary. I'm not too sure as to what amount of farads a capacitor should have to supply this amount of power. If I'm missing anything, or you need more information, just let me know. Thanks in advance.
After looking at that article, the simple example circuit it shows is exactly what I'm trying to do. It's just a switch, capacitor, and the resistor (nichrome). The purpose of this circuit is to just heat the piece of nichrome up once very fast. I understand it will have to be recharged after every power drain. Looking at the equation given on the website, I realized a few things. I'm pulling my voltages and amperage from a nichrome calculator btw. I'll assume a wattage that correlates with a higher temperature as the starting voltage, and the wattage that correlates with the lower temperature I need as the post-drain voltage. So Vc=0.23 Vo=0.48 t=1 R=0.14 and I will solve for C.phinds said:Do you understand capacitive discharge through an ohmic load (which is what you have after all)? You seem to think it's like throwing a switch on a battery on and then off again. It doesn't work like that.
http://www.learningaboutelectronics.com/Articles/Capacitor-discharging.php
When selecting a capacitor size, you should consider the required capacitance value for your circuit, the maximum voltage it can handle, the frequency of the signal, and the type of capacitor (e.g. ceramic, electrolytic, etc.). You should also check the physical size of the capacitor to ensure it will fit in your circuit.
The required capacitance can be calculated by dividing the charge (in coulombs) by the voltage (in volts) in the circuit. You can also use the formula C = Q/V, where C is capacitance in farads, Q is charge in coulombs, and V is voltage in volts.
It is generally not recommended to use a capacitor with a higher capacitance than what is recommended for your circuit. This can cause issues with the circuit's performance and may lead to damage. It is best to stick to the recommended capacitance value.
If you use a capacitor with a lower capacitance than what is recommended, it may not provide enough charge storage for your circuit, leading to potential performance issues. It is best to use the recommended capacitance value or higher.
No, there is no one-size-fits-all solution when it comes to capacitor size. The best size will depend on the specific requirements and components of your circuit. It is important to carefully consider all factors before selecting a capacitor size for your circuit.