change level of positive square wave signal to negative

Question

A general question:

Lets say I have a square wave signal from 0 to 5 volts. What I would like to do is create a circuit that will convert this 0 to 5-volt signal to a -2.5V to +2.5V for example. ( remove the dc )

I do NOT want to have a negative voltage reference for the -2.5V in my circuit. What options do you think can create this kind of circuit? (for example, high pass filter maybe with Schmitt trigger? Other options? )
My square wave is at 200kHz and 50% duty cycle.

More info:

The square wave signal drives a transformer and after the transformer, a diode and a capacitor rectify the signal in order to create a DC output.

What I want to build, is a floating voltage that can be controlled by the frequency of my square wave. I need to connect this voltage in another circuit so that is why I use the transformer, for isolation and not having a ground reference on the secondary side.

The above question that I have is more general, in order to understand about level shifting.

In my case:
My AC waveform that I power on the primary side, can be a square wave or sine wave (I am not sure what is the best). The current that I need on the secondary side is about 1A. My transformer is a 1:0.8 (primary:secondary) and the voltage on the secondary side that I need, is about 10-12 volts and this voltage should be adjustable by the frequency or amplitude of the primary side's AC waveform.

Thanks a lot.

Answer 1

You apparently want to drive a transformer primary with a 5 V square wave, and what you have goes from 0 V to 5 V.

First, check whether this is even a problem at all. How much current will the transformer draw with 2.5 V DC input? Would 5 V DC input saturate the core or not. If the average DC current is acceptable and the core won't saturate, then you don't need to do anything.

Otherwise, put a capacitor in series.

However, you have bigger problems since you don't seem to understand transformers and you specs don't make sense:

My transformer is a 1:0.8 (primary:secondary) and the voltage on the secondary side that I need, is about 10-12 volts

This won't work. With a 5 V square wave in and a 1 to 0.8 ratio, you will get a 4 V square wave out.

I you want 10 V out, you need a transformer with more than 1:2 ratio. 1:2 will give you 10 V out unloaded. You need a bit more to compensate for the loss in the transformer, and the diode drop in rectifying the output.

Overall, you need to step back and figure out what you really want. Your question makes no sense as currently written.

Answer 2

Ok, it seems to me that you want to build some kind of power transfer circuit. Your input signal is not called PWM, because it has a fixed 50% duty cycle.

Concerning your question, you want to shift your original 0/5V (mean 2.5Vdc) signal to -2.5/+2.5 (mean 0Vdc) voltage level.

You can do this with a simple series capacitor. The output terminal of the capacitor is undefined concerning the DC operating point and thus can be tied to GND (or any other voltage level) with a (high impedance) resistor.

For driving the transformer with a low impedance driver, you might want to add a voltage follower stage.

Answer 3

How can you say PWM and not care about PW. Are you aware of flux walk thermal runaway?
– My Comment 18 hours ago

Generally, transformers for SMPS fall into 3 categories;

1) Flyback for <150W using Volt-sec [V-s] charge discharge products with deadtime T>L/R - equal Vs time product to prevent current integration towards core saturation. - Basic law V * dt = L*dI, Imax=V/R(loop)
2) Single half bridge primary center-tap to V+ with more linear transformer characteristics but still has dead time to prevent flux walk (most common on ATX supplies)
3) Dual half-bridge drivers to each supply rail to untapped primary with a dual half bridge on each primary end and deadtime as above. Ref When a core average current has a DC component, it integrates the B flux to one end or the other. Remanence or magnet B field memory after power off also stores energy like a cap* with voltage and saturation is the 10% drop in inductance before it becomes a flat line ( zero inductance) and the coil now is just a resistor low DCR causing thermal runaway in the driver and core. (failure)

This* is similar to a cap going over voltage into dielectric breakdown into a short circuit.