Help with LT3489 boost circuit

Nov 01, 2017 11 Replies

We've got a 3V6 (battery) to 14V boost circuit using a LT3489.



It works fine except the idle current consumed by the circuit with no load is high @ 15mA.



We've pretty much followed the data sheet in terms of both the schematic and the PCB layout.


15mA seems quite a high idle current.

Can anyone point us in the right direction with this?


Data sheet:

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Our schematic:

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Thanks.


Is it switching with no load? The data sheet shows it to be 4mA max when not switching.

Rule out a leaky output capacitor first.

Cheers

Clive

Yep, we swapped the cap and have tried electrolytic as well as MLCC caps and it doesn't change the current draw.

Yes, with no load it pulls around 15mA.

We put the circuit into LTSim and it seems to work fine with a quiescent current of 2mA.

Can you try measuring with a different meter or adding some more input filtering? You don't have a lot of that, and some DMMs are known to behave oddly when there's a lot of ripple present. Or in the presence of AC fields in general.

E.g.:

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-- john, KE5FX

In that case, swap the chip. The internal switching transistor may be leaky. I don't know what internal protection these have, but operating without the diode (say a momentary open circuit) could produce a very high voltage on the transistor.

Cheers

Clive

Purchase their demo board and see what it does

Have you looked at the switch node? You may have some rather large oscillations, causing more inductor heating (or EMI). Layout is really important in switchers.

I understand. But is it switching at that moment? If so, what is the duty factor. You may be wasting all your energy in the switching devices due to the very low on time vs switching speed.

Try some several small loads to get the switcher off 0 duty cycle and see what happens.

I think you are right.

Re-reading the data sheet it says the quiescent current is 2mA with the device active (shutdown pin high). It has a note 'not switching' for this parameter.

We found if the input voltage is taken above the programmed output voltage the quiescent current it indeed 2mA. So we think the circuit is probably working as designed, we just mis-interpreted the data sheet.

Thanks

If we have it

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