Hi,

I was hoping to convert my 1V DC square wave into an AC wave. I was using the square wave to pulse an inductor coil but now I would rather put an AC signal through the inductor.

First here is my working square wave circuit:


And now here is my attempt at AC coupling by putting a capacitor in series before the inductor but clearly I am missing something here:


I certainly don't have to use a polarized electrolytic capacitor but it was the only high farad capacitor I had on hand (1000uF for a low cutoff frequency). I would certainly like to know if a polarized cap is acceptable for this circuit anyways. Most importantly though, what am I missing to make an AC signal flow through the inductor?

 

What are you trying to achieve, making a buck boost circuit or type of flyback voltage booster, or just want to pulse a coil with 1V dc on /off?

 

After the first few pulses the top of the capacitor will be charged to the 1 volt power supply level. The mosfet can pull it up to +1 volts but there is nothing to pull it down. You would need a resistor between the mosfet source and ground.

Les.

 

You can try arranging the circuit like this to demonstrate simple AC coupling:


After a few hundred milliseconds for the 1 uf cpacitor and the 100k resistor to settle down your waveform should look like this. I assumed 1 kHz, you can scale according to your chosen frequency.

 

See

 

Ok thanks Les I thought there should be another path to ground, just wasn't exactly sure.

Trying to fully understand DickCappels and Bordodynov's schematics, both have V1 and V2. Looks like in both, V1 is feeding the gates of transistors/mosfets and V2 is the supplied power to the emitter/drain. If I were to guess, would V1 control the voltage level and V2 control the frequency of the resulting AC wave?

I still don't understand how the DC ground also functions as AC ground. If my DC is 1VDC where would the -0.5V come from?

 

I still don't understand how the DC ground also functions as AC ground. If my DC is 1VDC where would the -0.5V come from?

There's nothing unique about "ground" it's just a reference point for voltage measurements, AC or DC.

When the 1V-0V signal goes through a capacitor it is level shifted to +0.5V to -0.5V.
This occurs because a capacitor's average DC current must always be zero.
So for the capacitor output to have a 1Vpp signal it must be 0.5V for half the time and -0.5V for the other half so that the average DC current is zero.
That's why you see the shift in level at the start of the simulation in post #5, the output starting at 0V-1V and settling into -0.5V to +0.5V.

Make sense?