It would not be called anything because your specifications do not make sense.
You can specify a voltage or a current but not both for a source. If the voltage is specifed, the current will depend on what load is connected to it. If you specify the current, the voltage must automatically...
You use a PWM frequency of N times the output frequency desired. N=16 or 32 is probably good enough. Then you change the duty cycle of the PWM on each period with the value of the sine at that point in the waveform. The sine values can be stored in a table. You would use half the wave and...
Hi kitt, and welcome to the forum.
What is a "variable voltage IC?" Part number? Schematic?
What is needed to increase the current is a power supply capable of supplying the current. If the LEDs are to be controlled by an electronic signal, you will need a transistor to switch the...
The internal diodes are connected correctly.
But each chip must have a 120 Ohm resistor between its + and the power supply. And each chip's final - should be connected to ground. The chips are not connected to each other the way you have it.
Bob
Hi, Mr Haines, and welcome to the forum.
Tell us something about your best idea so we can get an idea of what you are talking about. If it has merit, you will get help here.
Bob
I am a PIC fan myself. Depending on what I need, I use any of the 12F, 16F, 18F, 24F or 33F series. For Atmel, there are, I believe, ATTiny And ATMega. These are all much more recent designs.
Bob
What are you using for the reference voltage for the ADC? If you are using the battery, you will always get a max reading when reading the battery voltage, no matter what it is. To measure battery voltage you need an independent voltage reference. If the chip has an internal 2.0V reference...
If you use a single supply amp, you will want to bias the + input at 1/2 the supply voltage and use a capacitor coupled (in and out) inverting amplifier configuration.
Bob
I always get confused about the meaning of p-p. It seems to be used as either the single sided peak of a symmetric waveform, or from the positive to the negative peak. Looking it up on Wiki, though, you are right, I have 10V p-p and you only need one op amp to amplify if you want 5V p-p which...
The only way to cause the phone to take less current is to lower the voltage. But this will most likely cause it to shut down as well. I would try for 4.5V to see if that helps.
Bob
That is actually pretty good performance. If you have a scope I would look at the waveform on the inductor to see if it is saturating. If so, a higher frequency would improve things, if not a lower fequency might give you more output power.
Bob
Why is the area of a right triangle only 1/2 base * height?
This is not an impertenent question. It is for eactly the same reason. When we charge a capacitor with a constant current, the voltage / time looks like a triangle. The energy put into the capacitor is the integral of the V * I...
Here is a circuit that does what you want. I produces a 5V p-p sine wave at 60KHz. The op amp used is a 5MHz GBP single supply op amp. Any op amp with similar performance should work.
You will probably have to adjust the resistors in the phase shift network to get the right frequency...
I misread your OP as Hz instead of KHz.
I tried simulating my idea with an LM324 and it was not capable of enough gain at 60KHz to output 5V p-p. You would need a faster rail to rail op amp to make it work. But it does look like it would work.
I cannot think of anything simpler. You...
To get a 1/2 W into 8 Ohms, you need:
Vrms**2 / R = 1/2
Vrms**2 = 4
Vrms = 2
But that is the RMS voltage, so you have to multiply by sqrt(2) to get the peak voltage, so
Vpeak = Vrms * sqrt(2) = 2.828V
In a perfect world you could get this by using a bridged amp at 3V, but not...