dc/dc converterpower supply
I'm analyzing a power supply design. A Current mode PWM controller is used to step down the i/p 45-57 V/350 mA to 5 V/3 A. The design uses this part, http://datasheets.maximintegrated.com/en/ds/MAX5974E-MAX5974F.pdf
I read the datasheet but I couldn't understand, how exactly the stepping down is taking place? and how the o/p current is determined?
You should think of DC to DC converters as producing an output power that is about 90% of the input power. Therefore, if output is 20V at 0.3 amps, the power is 6 watts and the input power will be about 6.6 watts. From an 11 volt supply this means a current of 600 mA.
A "DC-DC controller" to my way of thinking is a buck or boost chip that requires an external MOSFET or two and, depending on topology can be as power efficient as 95%. Same rules apply; power in = power out * 1.05.
Each series block of DC conversion will reduce efficiency so if you boost to 24 volts then buck to 12 volts you might be looking at an overall efficiency of about 80% to 85%.
Some DC convertors are flyback types and these incorporate isolation between input and output. Typical of these are the modules from Traco Power or XP (to name but 2 suppliers).
If you have access to the feedback resistor then be very careful how you wire the switch up because cable lead lengths and added capacitance and inductance can make the feedback unstable and the DC converter sits there as an oscillator or maybe a bit noisier than the spec suggests.
If you are going to use a step-up, boost converter for the +30V, then it is very simple and economical to create the negative of that using switched-capacitor given the current requirement is low.
The schematics below shows the typical boost output stage in L1, M1, D, C1.
C2, D2, C3, D3 are the addition to generate the negative voltage. The negative voltage tracks the positive voltage minus one additional diode drop. If you don't want the diode drop difference, make D1 into 2 diodes in series (that would waste 0.7V in the +30V output).
simulate this circuit – Schematic created using CircuitLab
Best Answer
Start with the numbers you quote - 45-57V (lets average it to 51V) @ 350mA - this is a power into the circuit of 17.85W. Power out is 5V * 3A = 15W.
This tells me that the circuit has a power efficiency of about 84%. So, if it were a black-box and you got out 5V at 3A but needed to put into the black box nearly 18W, would this make it clearer where the 3A comes from?
The stepping down is taking place in the transformer. I'm going to continue using the term "step-down" for the rest of this explanation even though strictly speaking it's the primary inductance and secondary inductance that do the business of energy transfer - see the circuit below: -
Transformer (T1) has a primary coil and secondary coil and, energy is gained in the primary coil by briefly connecting it across the supply, Vs by the FET (circled in red) and a resistor labelled Rcs.
The primary energy is in the form of magnetic flux and when the "red" FET switches open circuit the magnetic flux energy has to go somewhere. This is where the "blue" fet comes in. The transformer secondary (being closely coupled to the primary) can take this flux, convert it to current and push this current into the 4 x capacitors on the output.
But it can only do this if the blue FET conducts after the red FET stops conducting.
Because T1 is "step-down" (i.e. secondary winding has fewer turns than the primary winding) the ampere-turns generated by the primary gets turned into bigger amperes for fewer turns in the secondary. Again I ask the purists to forgive me on my terminology.
I'm not going to go into the fine detail of the chip other than to say the pulses that turn the FETs on and off determine how much energy is stored in the magnetic field so that the output voltage remains regulated across a range of acceptable load currents.