12V DC/DC for amplifier. Current limiting?

[Pierre]

Hello all.
I am (also) trying to design a 12V to +/-55V dc/dc converter for amplifier use, with a power level of around 700W.
The prototype is up and running quite ok (although efficiency is only around 86% due to excessive transformer leakage inductance, I have to reduce it).
The thing is that I would like to add a current limit protection. But with the currents involved (around 60A), it is not practical to use current sense resistors without loosing lots of watts in them.
Another possibility is to use Rds(on) of the mosfets, blanking out the voltage drop between source and drain when the mosfet in particular is off. However, that's too dependent on temperature.

I am familiar with coilcraft current sensors, having used them with success in the past. But they are specified for 30A "only".
I was thinking on measuring the current to the 12V input to the transformer center tap. My idea is to split that connection in two thick and identical wires of the same length (2-3cm), passing only one of them through the transformer. That should theoretically produce a measurement of HALF of the total current, assuming that the other wires carries almost exaclty the same current. Do you think that this idea is practical/useful/crazy?

I don't need 0.1% precision in the measurement, but something useful for protection with an error of, say, 5-10%.

The same for the input fuses. I plan to use 2 x 32A automotive fuses in parallel at the input. Good or bad idea?

Thanks for your suggestions!

[Ouroboros]

It is very easy to make your own current transformer. All you need is a suitable ferrite toroid with (say) 500 turns of this wire for the secondary, and the primary is simply the feed supply wire passing through the centre. You then load the secondary with a resistor. Obviously the ferrite (not dust-iron) must not saturate, but this shouldn't be a problem with a single-turn primary.

[Eva]

Consider measuring the voltage drop across a portion of the +12V PCB track or plane. You can use a fet input op-amp with a common-mode input voltage range including the positive rail, like TL071, to amplify and condition this small voltage drop...

[Pierre]

Ouroboros,
With 1-single turn primary, the flux is higher than with more turns, so it is the worst possible situation (B is inversely proportional to the number of primary turns).

Eva,
Very nice idea, but wouldn't it be difficult to measure such a small voltage drop (for 1milliohm track, 60mV max) in the presence of switching noise, etc, cleanly?
I get you when you refer to FET-input opams, such as TL071, 81, etc, for their extremely low offset voltage and input current, right?

I will try to do some tests with different load measuring with an oscilloscope and multimeter to see if there is really something to measure.

Best regards,
Pierre

[Ouroboros]

It is normal in a current transformer for the primary to be a single turn. The resistive loading on the secondary, togther with the large step-up ratio means that the impedance presented by the transformer to the primary turn is negligable.

A current transformer is a special case, and it needs to be thought about differently from a normal transformer.

[Eva]

Originally posted by Pierre
Ouroboros,

Eva,
Very nice idea, but wouldn't it be difficult to measure such a small voltage drop (for 1milliohm track, 60mV max) in the presence of switching noise, etc, cleanly?

Interference pickup is not that bad when you amplify the signal locally with very small PCB loops.

I get you when you refer to FET-input opams, such as TL071, 81, etc, for their extremely low offset voltage and input current, right?

No, I mean that these fet op-amps can sense voltages near or slightly above the positive supply rail, while most op-amps can't. On the other hand, conventional bipolar "single-supply" op-amps can only sense voltages near the negative rail (ground).

I will try to do some tests with different load measuring with an oscilloscope and multimeter to see if there is really something to measure.

Best regards,
Pierre

Test leads will pick up a lot of switching spikes and ringing. As I previously mentioned, the signal should be amplified locally before using it. It's not the same to carry a 0-60mV signal than a 0-6V signal.

[Electrone]

I like to detect the voltage drop on the power input wire using a couple PNP transistors or a PNP transistor and a diode. Then there isn't a need for a separate power supply for an IC chip. The extra diode or transistor increases the sensitivity and provides temperature compensation. I wrap a sense wire around the power input wire.

[luka]

Hi Electrone

I like your idea, do you have any pic of that for easyer imagination?

[Electrone]

Hi Luka. Thanks.

Here a rendition which doesn't have a resistor placed in series with the compensation diode junction. '7' is at the battery end of the power input wire obtained through the sense wire wrapped around it. '6' is at the power supply end of power input wire. The values of the components depend on the other circuitry. The end of the capacitor shown as being connected to '7' may be better connected to ground in many cases. Then you would more likely need that extra resistor. Another option is to make the extra resistor the one controlling when the circuit begins to operate instead of relying on the Resistance value of the input wire alone, which may be in the miliohm range, by the way. The resistor in series with the emitter of the active transistor may not always need to be in there.

[d2134]

The same for the input fuses. I plan to use 2 x 32A automotive fuses in parallel at the input. Good or bad idea?

Not a good ideea. Consider using a single 63A fuse.