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Hi all,
Because my old power supply released the magic smoke, I decided to build another one, but this time in needed to be digital.
So after searching the internet I came across the OPA549 power amplifier and looked ad the datasheet and found a perfect example see picture 11.
With the OPA549 I also use a MCP4922 DAC to provide the voltage and current adjustment. But I have problem. The voltage can be controlled very good im kind of surprised to how well it controls the output voltage. But for the current limiting it does not work at all, and Im not sure where to look.
The voltage that is provided at the V+ (10,11) is about 33.6VDC, the V- (5,6,7) are connected to 0VDC. The output of the OPA549 is connected to two diodes as per figure 8 including the carbon 10 Ohm resistor.
The OPA549 has a gain of 10. the voltage and current setpoint is provided by a voltage that comes from the DAC MCP4922.
Now for example:
Vout is set to 13.8VDC (1.38VDC -> dac output)
ILIM Is set to 3A ore (3.33VDC -> dac output)
Output load = Car light 12V 55W.
The resulting current would about 4A (55 / 13.8 = 3.98)
what I measure with my ampere meter is that the current peaks ad 3.018A and then continues to drop to about 1.745A. which is not what I expected. What I do expect is for the current to stay at 3A and drop to another value.
But what is even stranger is when I set the current limit to 5A which is above the calculated current draw, it only allows 2A to flow. you can even here the OPA549 make some sort of sound (Some sort of buzzing) which would indicate that current limiting is active.
Another scenario is when the output voltage of the dac is set to 0.016V which is (0V) the you would expect a current draw of about 4A but it does not you can still here the current limiting active.
I don't know what going on here maube you can help me out with this problem.
If you need any schematics please let me know.
Regards Jordy
Jordy,
You do not say anything about using a heat sink for OPA549 and without it 3A output current will clearly push the part into a thermal shutdown resulting in lower average output current as the part goes in and comes out of the shutdown condition – see below. Check the REF pin with scope for thermal shutdown status - LOW in shutdown mode - see below.
With (V+) of 33.6V and Vout of 13.8V, the voltage drop across the internal output transistor is about 20V and this results in the power dissipation inside the OPA549 package of: P=20V*3A=~60W. Without the heat sink, the typical junction to ambient thermal resistance, θJA, is 30C/W (see above) resulting is the temperature rise of ΔT=P*θJA=60W*30C/W=1800C! Of course the chip will never heat up to 1800C because at T=~160C the thermal shutdown protection circuit turns on temporarily disabling the output stage – see above.
Please verify that you use a heat sink to lower the OPA549 package thermal resistance value θJA =~2C/W or less necessary to prevent running into the problem of over-heating the chip; the thermnal shutdown temporarly disables the output at T=~160C until the chip cools down to T=~140C.
