Despite NE555N datasheet says output (pin 3) voltage is only about 2 volts lower than Vcc, I have only +6.3v on pin 3. Is it a digital multimeter misreading?
A ordinary digital multimeter will average the pulse train on the gate of the FET into what looks like a DC level.
The MOSFET used there is IRF3205, but currents beyond 6-7A force it to heat up, and 10A (with 100-110 Watts on secondary) even more (heatsink is ready to boil water)
An NE555 is a poor choice for driving a MOSFET (even logic-level) as it cannot sink and source much current. Put a monolithic MOSFET driver IC between the 555 and the MOSFET and your power dissipation will improve.
Does MOSFET overheat due to undervoltage on gate (hence not fully opened) or cracked transformer core? Do I need some other enhancements for this circuit?
Get an oscilloscope (even a relatively old one) and probe the switching nodes. It's the only way to really know what's going on in the circuit.
Given its size, that is likely to be an audio coupling transformer, rather than an AC mains transformer - assuming its core is separate metal laminations.
(If it has a ferrite core, then it may be designed for high frequency use, in a PSU, but it looks like metal laminations).
Audio coupling transformers usually fall into 2 classes : wide bandwidth for pro audio use (ideally 20Hz to 20kHz) or telephony, usually 300 to 3400Hz. The measurements you have made so far suggest the latter.
The transformer's performance is mainly determined by three quantities (and their interaction with the circuitry around it)
- Turns ratio (determines voltage gain)
- Primary inductance (determines LF performance)
- Leakage inductance (determines HF performance)
It's also worth knowing three more quantities:
- Primary resistance (measured at DC, presumably 1 ohm from your question)
- Secondary resistance (measured at DC)
- Leakage capacitance (can be indirectly determined)
Turns ratio is close to the best voltage ratio you measure, probably 3.6:1.
The ratio of secondary to primary resistances will be approximately the turns ratio squared (not exact, it depends on the nearest wire gauge) so your secondary R is probably about 13 (10 to 16) ohms.
Primary inductance is measured across the primary (duh!) with the secondary open circuit. If you can't measure inductance directly, you can connect a known capacitor in parallel and find the resonant frequency.
Now the primary inductance is effectively in parallel with the load (its purpose is to magnetise the core, generating the flux that communicates signal to the secondary). So it, combined with the source impedance of your signal generator, (actually that source impedance plus the 1 ohm primary resistance) form a series R-L circuit - which is the high pass filter you have observed. (600 ohms is a traditional source impedance in audio work, your signal generator may be different, maybe 50 ohms if it covers radio frequencies)
simulate this circuit – Schematic created using CircuitLab
You can improve the LF performance by driving the transformer from a low source impedance (e.g. an audio power amplifier) reducing R1.
You can't eliminate the primary resistance (R2) for perfect LF performance, but I've seen transformers driven from negative source impedances to partially cancel it. (Not a common trick : as you can probably guess, unstable when the transformer is replaced with a better one!)
Now the leakage inductance is measured in the same way, but with the secondary shorted. Smaller is better; it determines the high frequency performance of the transformer.
simulate this circuit
The leakage inductance forms a series L-R circuit with the load, which is R3 / turns SQUARED. So to get good high frequency response, keep R3 high.
But there is a practical limit on this : the leakage capacitance C1 resonates with leakage inductance, limiting the high frequency response, and providing a frequency response peak. Use this peak to determine the capacitance. It can be controlled (damped) by reducing the value of R3 or a separate Zobel network
Best Answer
It's a bit of a misnomer to call the flyback converter magnetics "a transformer" because it is being used as an inductor with a load so the inductance is chosen to support the low DCR conduction losses and same for the secondary.
Normally the inductance ratio is adjusted to give around 50% duty cycle for the desired load or less at minimum losses.
Basic math
The energy storage density has a limit to transfer about 100~200W max before the topology becomes less efficient than a forward converting transformer topology where the transformer no longer has to store the energy being transferred as it is coupled by the mutual coupling with a much lower impedance path.