High voltage output from a isolated DC-DC convertor module

[cowana]

In a project, I'm using a TRACOPOWER TMA0505S DC-DC converter to produce an isolated 5v supply from a USB connection.

Module datasheeet:
http://www.farnell.com/datasheets/1763529.pdf

The TMA0505S model runs from 5v, and produces a single 5v output.

However, in my circuit, I'm getting 5.662v on the output - higher than I expect.  The input to the DC-DC converter is 5.17v, which is within the quoted 5v ±10%. I'm powering a microcontroller and LCD display (and backlight), pulling between 20mA and 50mA. The DC-DC convertor has all the required capacitors.

I emailed the manufacturer to see if I had got a faulty unit, and they replied that the TMA series are unregulated converters, with a main purpose of isolation of a constant input voltage. They linked me to the following part of the spec, showing it performing as expected:



When I selected the part, I was considering this part of the spec - I expected the output to be 5v ±3%.



Assuming the manufacturer is correct and the device is performing in spec, where did I go wrong? Is the Voltage set accuracy not how accurate the output voltage is? I had assumed the load variation was talking about the output dropping with increasing load.

Any insight appreciated!

[mij59]

Hi,

Very basic data sheet, at the bottom of page 2 is a note that the output voltage is a full load.
Take a look at the  application note.

[DanielS]

The good old accuracy vs precision thing.

The accuracy spec on this thing likely refers to deviation from a given output value once the operating point has been determined - you are getting 5.662V with your current setup and as long as your input voltage and output current remain the same, you can expect output voltage to remain within 3% of that.

An unregulated supply is simply a high-frequency transformer. Its output voltage is proportional to its input with some allowance for wiring, switching and rectifier losses. In this case, the 1.2% output change per 1% input change implies a transformer turn ratio of 1:1.2 so at 5V input, the transformer has 6V output minus 0..35V for its Schottky diodes, which yields your 5.6V output at low load. I you increase the load, the output can droop by 10% and you get about 5.1V. Also notice how the load regulation/variation spec starts at 20% load minimum? So you need to be pulling at least 40mA to even be within a range where you have any sort of regulation spec.

If you want 5V +/- 3%, you need a regulated supply.

[cowana]

Very basic data sheet, at the bottom of page 2 is a note that the output voltage is a full load.

And that's the bit I'd missed - now it all makes sense. I'd assumed the converter was doing some regulation from their nice-sounding figures - but you are very right that that is just in the 'best case' full load situation.

An unregulated supply is simply a high-frequency transformer. Its output voltage is proportional to its input with some allowance for wiring, switching and rectifier losses. In this case, the 1.2% output change per 1% input change implies a transformer turn ratio of 1:1.2 so at 5V input, the transformer has 6V output minus 0..35V for its Schottky diodes, which yields your 5.6V output at low load. I you increase the load, the output can droop by 10% and you get about 5.1V. Also notice how the load regulation/variation spec starts at 20% load minimum? So you need to be pulling at least 40mA to even be within a range where you have any sort of regulation spec.

I had noticed in my first (unloaded) powerup that the output was even higher - I had put that down to the internal capacitance charging up, and assumed it would come back down with a small load on it. Great explanation of the internal workings .

The accuracy spec on this thing likely refers to deviation from a given output value once the operating point has been determined - you are getting 5.662V with your current setup and as long as your input voltage and output current remain the same, you can expect output voltage to remain within 3% of that.
...
If you want 5V +/- 3%, you need a regulated supply.

Perfect - as I can be sure the output will stay above 5.35v, I can put a nice 5v LDO linear regulator (350mV dropout) on the output, and get the voltage I need.

I'm glad my circuit is doing exactly what it should be - when devices are faulty they have a habit of failing further in later life.

Cheers,
Andy

[Christopher]

I had this problem -- 6-7V on a 5V output DC-DC converter. Nice how the datasheet only states what happens at full load and not at a low load 

I put a series current limit resistor with a zener shunt & a big ish parallel cap on the output as I usually only power a few analog chips or a small PIC.

[mariush]

Cowana, for up to 100mA, you may want to check out ADUM5000 isolated dc-dc convert IC :

http://uk.farnell.com/analog-devices/adum5000arwz/ic-isolated-dc-dc-conv-180mhz-soic/dp/2102522
http://www.digikey.com/product-search/en?FV=fff40042%2Cfff800df&k=adum5000&mnonly=0&newproducts=0&ColumnSort=-1000009&page=1&stock=0&pbfree=0&rohs=0&quantity=&ptm=0&fid=0&pageSize=500

It's a big more expensive if you buy just a few but at quantity, it's probably cheaper than that larger, heavier converter. It's also more regulated, claims 200mV peak to peak noise and maximum 5.4v output so it may save you from using extra parts to stabilize the output voltage.

The other downside (besides the price) is that it's only about 35% efficient at full load, compared to about 70-80% with that tracopower module.

[Zero999]

If you only need to draw 50mA, try the connecting the LM2936z-5 to the output - it has a drop out voltage of under 200mV.
http://www.ti.com/lit/ds/symlink/lm2936.pdf