# LT3680 switcher output doesn't have regulation under no load

I have a switcher circuit based on LT3680. I implemented the typical circuit shown in the datasheet in a well laid out 4-layer PCB. However, I find that the input voltage is present at the output at no load condition. When I add load of 100ohms and below, I get 5.05 constant output. The typical circuit is meant for 5V, 3.5A load. My loads draw only a few milli amps. So could that be because if the inductor saturating or something. The switcher has feedback and as far as I understand should be able to regulate the output whatsoever. Regulation kicks in only when I add 100 ohm or below. Nothing with 10K, haven't tried with anything in between since I don't have a wide range of resistors available with me. Please have a look at the circuit used. Datasheet can be found at:

http://cds.linear.com/docs/en/datasheet/3680fb.pdf

• So could that be because if the inductor saturating or something Inductors only saturate when the current becomes too high, so no, that's not it. – Bimpelrekkie May 30 '16 at 10:41

The problem is very likely to be loop compensation.

This is a current mode architecture, so the output pole changes with load, but the output zero does not.

The output pole is at $\frac {1} {2 \pi\ R_o C_o}$; rearranging as $R_o = \frac {V_o} {I_o}$ we get:

$f_p = \frac {I_o} {2 \pi V_o C_o}$

At very low load currents, the output pole is at a very low frequency and when added to the dominant pole at the $V_c$ pin, the arrangement could easily be unstable.

Positive feedback could indeed make Vo = Vi.

To cure this, you could add a capacitor (need to calculate it) from $V_c$ to ground and add a small (100pF perhaps) capacitor across R46

Another point to note is that for low load currents, the size of the inductor is very small (as load current increases the required inductance value decreases).

The inductor in the application circuit has been chosen for about 1A of ripple current; if you have only a few mA of load, then I calculate the inductor should be of the order 120μH assuming a maximum load of 100mA (for a ripple current of 40mA).

The datasheet gives guidance on this

$L = \left (\frac {V_O + V_D} {F_{sw} \cdot \Delta I_L} \right) \cdot \left ( 1 - \frac {V_O + V_D} {V_{in(max)}} \right)$

See page 10 of the datasheet.

• Hi. Thanks for the reply. I guess it's the inductor value. The typical application circuit seems to be designed for 3.5A current. – sdk Jun 6 '16 at 4:22
• I fixed the issue by replacing the inductor. Redesigned the circuit for 1.5A max current and replace the inductor. That fixed the issue. – sdk Jun 26 '16 at 4:47
• I am glad you have a solution :) – Peter Smith Jun 26 '16 at 17:00