It would need two 1.5A secondaries. If you could find 3V secondaries then a 9VA transformer would do, but 6V is going to be more common, so would require a 18VA transformer.
It would need a drive circuit that could drive 2Ω, identical the the circuit that would be required without the transformer.

 

OK, then. Let me try to give an actually helpful answer.

I'd wind the transformer with a center tap in the primary. Connect the center tap to B+ and ground the other two legs with transistor switches (say, N-channel MOSFETs). This way, you are driving the gates of ground-referenced transistors which is easier than a totem-pole. Use any microcontroller with a two-channel PWM to drive the MOSFET gates. You'll probably need a low-pass L/C filter on the secondary. As others have discussed, the turns ratio will depend on the required output voltage to push 5W into the load. The higher the switching frequency, the lower the magnetizing inductance can be and the smaller the output filter components. You also probably want to be out of the audio range so I'd start with a 20 KHz PWM carrier.

If you power this thing with a single Li-Ion cell, say an 18650, you can find a microcontroller that will operate directly from B+ without requiring an LDO. An ATTINY4 would be fine.

You can find MOSFETs with low voltage gate drive and 5 watts isn't much power so you might not even need an active gate driver. ON Semi MCH6436 would be a reasonable choice.

You'll almost certainly want to wind your own transformer. I like EFD cores for DIY transfomers. Cores and PC-mount bobbins are readily available. I did a quick napkin calc with an EFD20/10/7 and came up with 1/2 volt per turn so your primary would need 16 turns (8 on each side of the tap). (EDIT: this is a lot of turns. How about 50 KHz and 6 turns).

For the output filter, 470 uH and 470 nF gives you a 10 KHz cutoff. This filter is going to be pretty non-critical because the carrier frequency is so far from the base band.

 

You'll almost certainly want to wind your own transformer. I like EFD cores for DIY transfomers. Cores and PC-mount bobbins are readily available. I did a quick napkin calc with an EFD20/10/7 and came up with 1/2 volt per turn so your primary would need 16 turns (8 on each side of the tap).

I get 313 turns. Are you making the mistake of thinking that the transformer operates at the PWM carrier frequency?

 

I get 313 turns. Are you making the mistake of thinking that the transformer operates at the PWM carrier frequency?

Indeed I am. I was thinking that the transformer was passing the PWM waveform but missing the fact that this waveform isn't DC balanced so the flux is going to walk up and down at a 625 Hz rate. Yeah, need a bigger core for this.

 

Typical sine wave inverters use a high frequency (but DC balanced) PWM waveform to drive a step-up transformer for HV DC. Then they turn the HVDC into low-frequency sinusoidal AC with a bridge, usually using IGBTs. This saves having a low-frequency transformer at the cost of having to power the secondary circuitry somehow.

 

Indeed I am. I was thinking that the transformer was passing the PWM waveform but missing the fact that this waveform isn't DC balanced so the flux is going to walk up and down at a 625 Hz rate. Yeah, need a bigger core for this.

A 1.5VA or 3VA mains transformer bobbin and Silicon iron laminations would do. Because it's 625Hz the flux is a lot lower than a mains transformer, so 5W would be easily achieved. Using a split-bobbin would let you use the leakage inductance as the L in the LC filter, so all that would be needed to remove the carrier would be a capacitor across the output.
I haven't any specs for the Ae of that size core, nor the core loss to be able to calculate the optimum flux density.

 

Typical sine wave inverters use a high frequency (but DC balanced) PWM waveform to drive a step-up transformer for HV DC. Then they turn the HVDC into low-frequency sinusoidal AC with a bridge, usually using IGBTs. This saves having a low-frequency transformer at the cost of having to power the secondary circuitry somehow.

The more professional ones use a 50Hz transformer driven with PWM, wound for a certain leakage inductance which forms the L in the LC filter.

 

How about modifying a cheap sine-wave inverter to raise the output frequency? If you started with the topology I mentioned (high-frequency DC/DC converter followed by a modulator bridge), you wouldn't have to mess with the transformer calculations or drive circuitry. You probably wouldn't have to change the filter and, if you did, it would just be to put in a smaller capacitor. Just change the PWM to the IGBTs and change it before the gate drive/level shifting stuff. You can find a 50W inverter for under $20.

 

The OP has an amplifier application. "The sine wave will be a signal generated by a circuit before transformer driver"

 

A 1.5VA or 3VA mains transformer bobbin and Silicon iron laminations would do. Because it's 625Hz the flux is a lot lower than a mains transformer, so 5W would be easily achieved. Using a split-bobbin would let you use the leakage inductance as the L in the LC filter, so all that would be needed to remove the carrier would be a capacitor across the output.
I haven't any specs for the Ae of that size core, nor the core loss to be able to calculate the optimum flux density.

Yeah, I concur. This is a good approach to getting the transformer.

 

The OP has an amplifier application. "The sine wave will be a signal generated by a circuit before transformer driver"

Ah. I missed that. Then what he needs is just a class B push/pull amplifier with the transformer Ian0 described. But I can't believe that it wouldn't be cheaper to buy an OTL audio amp. $7 on Amazon

 

We still don't know why it needs to be transformer isolated!

 

What sort of application is this, anyway??

Cable identification. There will be a pick up device at the other end to pick up the signal.

 

It would need two 1.5A secondaries. If you could find 3V secondaries then a 9VA transformer would do, but 6V is going to be more common, so would require a 18VA transformer.
It would need a drive circuit that could drive 2Ω, identical the the circuit that would be required without the transformer.

Do you mean I will need to design a driver that can drive 2ohms first, then find a suitable transformer?

 

Cable identification. There will be a pick up device at the other end to pick up the signal.

You need five watts for cable identification? And the impedance is 2 ohms? Must be a really unusual situation.

 

You need five watts for cable identification? And the impedance is 2 ohms? Must be a really unusual situation.

What would you suggest the number would be? The 5W was the spec I was given. The 2ohm was a number I made up so the discussion can continue.

 

What would you suggest the number would be? The 5W was the spec I was given. The 2ohm was a number I made up so the discussion can continue.

Well, the impedance of most cables is in the range of 50 to a couple hundred ohms. And if I were making a cable identifier, I think would actually want a constant current source. You start with the worst case length and type of cable, and how much signal you need to be able to detect it at the far end, work backward from there. Also this seems like the sort of thing that would be better

What would you suggest the number would be? The 5W was the spec I was given. The 2ohm was a number I made up so the discussion can continue.

Well most cables have impedances in the tens to hundreds of ohms. 50 to 150 ohms would probably cover a good bit of the cable out there.

As to signal strength, I guess it depends on how long the cable is and what you're using to pick it up at the far end. But I can't imagine needing even one watt. Personally, I think I would go with the current source which would self-limit if there was a short.

Also, this seems like the sort of thing that would be battery powered. If so then why would you need isolation? Isolation from what?

 

Well, the impedance of most cables is in the range of 50 to a couple hundred ohms. And if I were making a cable identifier, I think would actually want a constant current source. You start with the worst case length and type of cable, and how much signal you need to be able to detect it at the far end, work backward from there. Also this seems like the sort of thing that would be better


Well most cables have impedances in the tens to hundreds of ohms. 50 to 150 ohms would probably cover a good bit of the cable out there.

As to signal strength, I guess it depends on how long the cable is and what you're using to pick it up at the far end. But I can't imagine needing even one watt. Personally, I think I would go with the current source which would self-limit if there was a short.

Also, this seems like the sort of thing that would be battery powered. If so then why would you need isolation? Isolation from what?

Unless, maybe the cable is in- circuit? Maybe the impedance is low because it's already hooked up to something. And perhaps the transformer they want you to drive is a current transformer so you can inductively couple to the cable in question. If this is the case, I can see why you would need five Watts or even more. But I still think you're impedance number is probably too low

 

Unless, maybe the cable is in- circuit? Maybe the impedance is low because it's already hooked up to something. And perhaps the transformer they want you to drive is a current transformer so you can inductively couple to the cable in question. If this is the case, I can see why you would need five Watts or even more. But I still think you're impedance number is probably too low

I don’t claim to fully understand this, as I am just researching.

The transformer isolation is also a spec I was given. The reason I thinks it’s about 2ohms (or somewhat low resistive load) is the cable to be identified is shorted at the other end.

Good point about current transformer, how would that work if using a current transformer?

 

I was thinking it would work like an amprobe meter. You know the kind where you open the clamp put it around the wire and then let go of the handle to complete the magnetic circuit? But instead of measuring the current, you would inject the 625 Hertz current. The tricky part will be that you have to put the current transformer around a single conductor. You can't put it around a pair because you you need current in one conductor to go one way and the other conductor to go the opposite direction. For this to work the near end of the cable would need to be shorted and the far end could either be shorted or not.

A current source makes even more sense if the far end of the cable is shorted. If the near end is open then you would make electrical connections to the conductors. If the near end is also shorted you could use the current transformer trick I've been describing. You'll also need a similar kind of pickup coil to look for the signal.