Problem with MAX756 Boost converter

Question

I'm using MAX756 as a DC-DC boost converter to provide 3.3V from of a single 1.5 AA battery. I've implemented the circuit on page 1 of datasheet with 1N5819 instead of 1N5817 and 144uF caps between Vin and Gnd instead of 150uF. Also Vin comes from bench power supply.

I've tested three 22uH inductors with different current capabilities (all above the required 1.2A due to datasheet (page 6).

~SHDN and 3/~5 is connected to out. LBO is left open.
The problem is the Vout is 3.3V in no-load. But as soon as I insert the 220Ohm resistors, Vout drops and in Iout = 60mA, the Vout drops below 2V.

Here is the picture of the setup: lower rail in Vin and upper rail is Vout

I've shown the other two inductors that I tested.
Anyone knows where the problem might be? Thanks in advance.
P.S.
Glenn W9IQ says:

Take a bench supply capable of 2 amps output and set it to 7 volts. Put a 4.7 ohm, 10 watt non inductive resistor in series with your inductor and put this circuit across the supply. Monitor the inductor voltage with your scope. You should see a nice exponential, asymptotic curve that heads toward zero when power is applied. This will happen within 1 us of applying power. If the beginning of the curve has a different slope, your inductor is in saturation at less then 1.5 amps. But your problem is more likely the ESR of your inductor and capacitor. Use the right parts - it makes a difference! –

I've paralleled 5 , 2 Watts 22 Ohms resistor (so 4.4 Ohm in theory and 4.7 due to DMM).
Here is the circuit:

And here is the scope's result immediately after applying the power :

Answer 1

{Added diagrams of SinglePoint VDD and SinglePoint GND}

Circuit works with no load and with 220 ohm (12mA) Iload, with 3.3 volts output.

Consider inductive bounce on the breadboard. Assume 2" of loop (in Gnd or elsewhere), or 50nH. Assume 0.1Amp in the IC's internal FET; the IC timing circuits have a minimum ON time, and thus some minimum current; we'll use 0.1amp.

Those internal FETs can be very fast, so assume 10 nanosecond to turn OFF or ON.

Result? Vbounce = L * dI/dT = 50nH * 0.1a /10nS = 0.5 volt bounce.

Nothing is well-controlled when GND or the GND rail is bouncing around with 0.5 volt.

Suggestion: build your circuit atop a sheet of copper (one-side copperclad).

Here is one physical setup that will generate this 0.5 volt upset

^{simulate this circuit – Schematic created using CircuitLab}

By the way, this is not the output swing, but inductive upset within (inside) the GND wiring between the IC pins. (for this example)

Answer 2

The app note shows that the minimum input voltage is 2 volts but you are attempting to power it with 1.5 volts.

The selection of the inductor and capacitor is quite critical to the operation. Look at page 6 for recommended suppliers and part families.

Your AA cell has an internal resistance and limited energy capacity. You should monitor the input voltage to your switcher to make sure the cell is holding up. Take note that the best efficiency for this boost converter is about 80% with proper component selection.

Building a switcher supply on a protoboard is never a good idea. With the high currents and high frequencies involved, the protoboard introduces too many strays. Dead bug / Manhattan style construction on a solid ground plane is a better approach.

Answer 3

The output voltage on the MAX757 is set by a voltage divider between ground and the output which is connected to the feedback input (pin 2), the formula to calculate the required resistors is:

VOUT = (1.25) [(R2 + R1) / R2]

To get 5V I have used 30K for R1 and 10K for R2, for 3.3V you could use 18K for R1 and 11K for R2.

If you can find it an alternative part is the MAX756 which is a bit easier if you only need an output of 5V or 3.3V as the voltage can be set by connecting pin 2 low for 5V or high for 3.3V which negates the need for the voltage divider arrangement.

https://www.electroschematics.com/wp-content/uploads/2010/11/portable-5-volts-power-box.png