MIC29302 regulator with pass transistors not working

[radeohedca]

I need about 15 Amps at 10.5VDC for an RF amplifier project so I chose to use a low dropout MIC29302 adjustable 3A regulator so that I could run it off a 12v car battery.  I have the output of the regulator connected to four TIP3055 pass transistors in an emitter follower arrangement with 0.15 ohm resistors on each emitter.  The regulator and pass transistors are mounted on a big heatsink. I am sampling some of the output voltage at the feedback input of the MIC29302 regulator through a voltage divider pot.  If I set the pot to the desired voltage then start loading the output it begins to drop and eventually levels off about 1.5v lower than the set voltage.  It seems the MIC29302 regulator needs to see about 300~400mA of load in order to regulate but that seems excessive (and a waste of power & heat) considering the 29302 datasheet says it should only need 10mA load. In order to get my required ~10.5v I need to set the unloaded output to almost 12v but I fear the over-voltage may damage my power amplifier when initially connected.
Is there a better way to do this ?
Thanks.

[mawyatt]

The base of the 2N3055 will need to be ~11.5 volts to supply 10.5 volts at the load. With only 12 volts available for the regulator, likely not enough headroom voltage.

Best,

[radeohedca]

I don't think that is the issue since it acts the same even with 13 or 14V at the input.

[mawyatt]

Then check for oscillations, with the added delay of the 2N3055 and load this might be a problem. LDOs are sometimes very sensitive to the effective load capacitance at the output, and the capacitance at the LDO output and delay with the 2N3055 and whatever capacitance on the load may be enough to cause instability.

Best,

[radeohedca]

I previously scoped it and there were no oscillations but I did find the issue....   comes down to RTFM ! 

The MIC29302 regulator datasheet states that the Adjust input needs to see about 10mA of current flow from it's own output (not after the pass transistors) in order to properly regulate. I thought I could just sample the output after the pass transistors and bring that back to the voltage divider at the Adjust input since that is the true output to the load after the diode drop thru the pass transistors but it needs to sample it at the output of the regulator. I put a 15 ohm resistor between the regulator output and the adjust pot, still sampling the output of the pass transistors as well.  I am now able to pull 15A at 10.5V with very little change in the voltage at full load.
 

[mawyatt]

Think your original concept should work with the feedback from the output, but the regulator may need a minimum load current to work properly. Keep the feedback from the output but add a shunt resistor of ~500 (10v/20ma) ohms to ground from the regulator 29302 output, that should satisfy the minimum load current requirement but still provide load regulation with feedback from overall regulator output.

Best,

[magic]

Put 47Ω from reg output straight to the load?

Nevermind, that's exactly what you have done
It's the right solution, as it keeps the regulator's output biased with 10mA that it needs and delivers that current to the load so it isn't wasted.
You still need to ensure that the load never falls below 10mA, for example by making feedback network be 1kΩ.

[mawyatt]

The issue with applying a resistor directly from the 29302 output to the load is than now the load current must not go under the minimum load current for the 29302 or the 29302 loses ability to control the output voltage. According to the data sheet this minimum current is 7ma.

Whether this minimum load current is a concern or not, depends. I would not want to use an unloaded supply where the output drifts up and is uncontrolled, thus charging the regulator output capacitors to the maximum supply input voltage. Once the load is connected it sees this higher than regulated voltage from the output caps until they discharge to the proper regulated level. Same situation for a load that is remotely controlled, the voltage would drift up and subject the load to a higher than regulated voltage until the load is enable. Either case could damage load circuits from over voltage.

Since adding an additional resistor is in question I would recommend adding this resistor directly from ground to the 29302 output to sink the minimum current at minimum expected output voltage and avoid any potential higher than expected voltage conditions at the overall regulator output, loaded or unloaded.

Best,

[mawyatt]

Put 47Ω from reg output straight to the load?

Nevermind, that's exactly what you have done
It's the right solution, as it keeps the regulator's output biased with 10mA that it needs and delivers that current to the load so it isn't wasted.
You still need to ensure that the load never falls below 10mA, for example by making feedback network be 1kΩ.

I see you've edited your response while I was typing then got distracted, had to bring in the groceries

Since the OP is discussing 15 amps load current, don't think 10ma will matter. Also if you make the feedback network look like 1K this will certainly work, but then again you are "wasting" the 10ma. In fact you could do away with the added resistor and just use the feedback network as the minimum load sink.

Best,

[magic]

Hmm, fair enough. But it works in those cases where the "real" load can be guaranteed to pull enough current to bias the regulator.

you could do away with the added resistor and just use the feedback network as the minimum load sink

Nope, that you can't do because the feedback divider is taken from 2N3055 emitters so the minimum load required increases β times.

[mawyatt]

Hmm, fair enough. But it works in those cases where the "real" load can be guaranteed to pull enough current to bias the regulator.

you could do away with the added resistor and just use the feedback network as the minimum load sink

Nope, that you can't do because the feedback divider is taken from 2N3055 emitters so the minimum load required increases β times.

Agree, that would take more current, likely ~100ma.

The best overall solution is what I said originally with the shunt resistor of ~500 ohms from the 29302 output to ground which should provide enough current with margin over the total expected output voltage range and temperature, and with regulated output loaded or unloaded.

Best,

[radeohedca]

Thanks for your input guys... I seem to be on the right track now.
I've been browsing some past posts and there seems to be a good group of knowledgeable peeps on here.  Great resource !

Hugh
VA3TO