The capacity of the battery is listed as 1000 mAh (milliamp-hours). So any combination of of current flow and time multiplied together should yield 1000.

That is not strictly correct. The battery capacity is probably rated on a 20 hour draw. If you draw 10 times faster (2 hours) the capacity must be substantially derated. Off the top of my head I would guess the capacity may drop to 75% or less; thus only a 1.5 hour run-time.

 

I agree that some batteries may behave that way, but there are many different battery technologies, and they don't all behave the same way. We are dealing here with rough approximations, to get an idea of the practicality of using a 3.7V battery to produce 110 VAC at 400 Hz. Without putting too fine a point on the matter I think everyone would agree that it is impractical to power such a panel with a battery. If a fast current draw reduces the available capacity it only reinforces the basic objection to using that battery for that application. Also it is a better approximation than having no idea at all what you are proposing.

 

please does anyone have an idea on how to build a resonant circuit inverter with 3.7V input rather?

I told you in #10 what to start with.

Since the circuit you need could kill you, and because I perceive your skills to be low (no offense, just a guess), and finally because you can buy a suitable solution for <$10, I'm reluctant to say much more.

 

Maybe something like this would work.

http://www.bing.com/images/search?q...2DE86E0F046610D5A6032503BA756&selectedIndex=0

 

OP here, so I am wondering if this would work =>

If I use a 555 to produce a square wave, then run it through a circuit with the panel paralleled with a 10uF capacitor and then connected to an inductor with 5H and a resistor with 95 ohm resistor, the panel would receive roughly 100v AC. Would something like that work?

 

This would be exceedingly unlikely. How do you get to 110VAC at 400 Hz.?
Show us a schematic. Words just don't make it with electronics.

 

The specific datasheet mentions 400 Hz., but I don't think there is anything magic about that frequency. A higher frequency could mean smaller components.

A 555 produces a square wave. I was thinking that a Wein Bridge oscillator might be a better fit for this application.

Since the EL panel looks like a capacitor, a higher freuequency would mean more current drawn since the impedance is lower. Stick with the rated 400 Hz.

Bob

 

So is this a possibility? Problem is I don't know how to generate that signal using a 3.7v battery.
C2 is the capacitance of the panel.

 

You may be able to produce the voltage you want, but there is no load to draw any current. If you change the load from reactive in parallel with a high impedance scope probe to something which draws actual power the voltage will collapse.

110VAC / .6A ≈ 183 Ω

Put such a resistance across the "panel" capacitor and see what happens.

The answer to your other question is to make a Wein Bridge oscillator and AC couple the resulting 1V pk-pk (0.707 VRMS) signal into your resonant circuit.

 

1. Is it possible to have the Wein-bridge oscillator directly in series with the resonant circuit?
2. If I need to couple the oscillator, how do I determine how big of a capacitor to use?

I put a 183 ohm resistor in series with the 45nf capacitor and the voltage did collapse down to ~16 volts

 

The series coupling capacitor should have a "low" impedance at the frequency of interest aka 400 Hz. In this case "low" might be regarded as a few ohms.

It is almost always a mistake to try "extracting" power from a resonant circuit. If you need power, it should come from the power supply.

 

So if I don't use a resonant circuit at all, I would just be powering the panel directly from the oscillator? Which I suppose would bring me back to my original problem of creating an inverter to power my panel. My restraint is that a 3.7V 1000mAh battery is used.

 

Take a look at something like this:

http://www.micrel.com/_PDF/mic4826.pdf

 

If I were to build a Wien Bridge oscillator though, I would need a positive and a negative voltage supply to drive the op-amp. How can I do this with only 1 battery? Or possibly 2 batteries?

 

You create a virtual ground at Vcc/2. You need to select an opamp that will work from a single supply. When you AC couple the output the DC component will be removed and you have the 1 V P-P signal that you had in your simulation.

 

Okay, I will have to look into that, Do you know of any op-amps that work from a single supply? Also, I put the resistance value of the panel that was calculated before in series with the capacitance of the panel and I was still able to achieve 110+ VAC across the load. All I had to do was adjust the resistance that was connected in series with the inductor.

 

Take a look at something like this:

http://www.micrel.com/_PDF/mic4826.pdf

Nice find, hot off the presses.

 

Something like this perhaps?
Winding the 3-coil transformer would be the trickiest part.

 

I will keep the micrel inverter in mind should it all fail. I am trying to design a circuit from scratch, that's why I am not jumping to parts that already exist. I don't really have a tight budget, so it is okay if the final product that I create is more expensive than an off the shelf product. It's more for the learning experience.

At the moment, I am looking to find an op-amp that uses only one power source and figure out a way to create that virtual ground.

 

I will keep the micrel inverter in mind should it all fail. I am trying to design a circuit from scratch, that's why I am not jumping to parts that already exist. I don't really have a tight budget, so it is okay if the final product that I create is more expensive than an off the shelf product. It's more for the learning experience.

At the moment, I am looking to find an op-amp that uses only one power source and figure out a way to create that virtual ground.

If you take the single supply opamp and make an oscillator out of it and then AC couple through a capacitor you don't actually need to do anything else. The capacitor will remove the DC component from the output.

You only need to make a virtual ground if you want to use a dual supply opamp with a single power supply voltage.

http://en.wikipedia.org/wiki/Virtual_ground