I'm using a Traco TEC 3-0923 that converts 9vdc to 2 legs of +/-15vdc. It's rated for 100 mA per leg and 3W total. I have a guitar preamp that has 4 op amps, so this is applying +/-15 v to them, or 30v across the op amps. I have 1R resistors inline to measure the voltage, they aren't precision 1Rs, just standard so they probably are higher than 1R by a bit.

I measure total current draw to be 200 mA at 9.44v using a current meter. I have a 1R resistor on the input, and measure 0.195v across the 1R. So that makes sense. On each leg, I measure 0.038v across a 1R, or 38 mA, I can't use a current meter here so that's why I have 1R's.

200 mA at 9.44v is 1.9W, or 0.95W per side.
Divided by 30v is 32 mA per side. I measure 0.038v across the 1R on each leg, or 38 mA.

38 mA/side seems like I have roughly a safety factor of 3 on it's 100 mA/side capacity. But the 1.9W on the input looks like only about 50% headroom compared to the 3W rating. I don't understand that.

So I'm wondering if I need the 100mA/leg version, or if I could use the 2W - 67mA version, the Trace Tec 2-0923. It's $10 each instead of $15. If I look at current/leg, I have lots of room, but if I look at the Watt rating, I should stick with the 3W version.

 

Try using 10 milli ohm resistors .

General statement .
These things have a variable efficiency
Dependent on load .
How many of these are you making ?
Imho , if it's a few 100 off, go with the bigger part . Your time is worth more than you will save.
If your making thousand off ,
Then go with bigger part .. you will be able to get such a bulk discount , the extra price is peanuts. .
If your making the 10 of thousands , design your own supply.

 

Using a 0.01Ω resistor will make the voltage drop roughly 0.0003v. Won’t that make it harder to measure? I’m not sure the reason behind that.

 

Using a 0.01Ω resistor will make the voltage drop roughly 0.0003v. Won’t that make it harder to measure? I’m not sure the reason behind that.

Your right .
I'd normally be looking for a 10 to 100 mV max across a measuring resistor, so the resistor has minimum impact on the circuit working.
Remember the meter is not calibrated for the waveform you see on the input or output of the dcdc.
They are very spiky, so tend to under read . The input to the dcdc is probably most spiky, so probably miss read the most .

 

General statement .
These things have a variable efficiency
Dependent on load .
How many of these are you making ?
Imho , if it's a few 100 off, go with the bigger part . Your time is worth more than you will save.
If your making thousand off ,
Then go with bigger part .. you will be able to get such a bulk discount , the extra price is peanuts. .
If your making the 10 of thousands , design your own supply.

You're right. I'm making a handful of these, not 10 or 100 or 1000. So an extra $5 is best.

 

38mA @ 30V is1.14W
200mA @ 9.44V is 1.89W
Efficiency is 60%. Fair for a small flyback converter. It will probably manage better than at full power, because quite a few mA will be going to run the circuitry.
The rating will be for the output power. The input power will exceed the output power by 1/η.
Use the 2W, it is likely to be quieter than the 3W. The 3W will produce current pulses that are shorter with higher current.

 

Ah I see. Thanks for the tip!
So 38 mA is 57% of the rated 67 mA/side Also 1.14W about 57% of 2W. So that makes sense.