# Problem with MAX756 Boost converter

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 :

• Are you really expecting people to try figure out what is connected where? and imagine where it might be behind components they cannot see through? or are you going to provide a schematic? – PlasmaHH Jul 6 '17 at 11:12
• @PlasmaHH Sorry but I've already said that I've implemented the circuit on page 1 of datasheet. I'm adding the image ;) – Zeta.Investigator Jul 6 '17 at 11:14
• What is it you're testing, exactly? If it's to see whether the device will or won't work way out of spec with the recommended layout guidelines thrown out the window, I'd say you've succeeded. – Finbarr Jul 6 '17 at 12:18
• You may be able to, but as Glenn pointed out this particular application note shows a minimum input of 2V. And even that will assume a well laid out PCB rather than long loops of wire and push-fit connections that aren't great for handling current spikes. – Finbarr Jul 6 '17 at 12:39
• (Longtime Maxim apps eng here: this is why we started offering Evaluation Kits, switch-mode power supplies are very sensitive to pcb layout and component selection. Making it work on solderless breadboard is a lost cause.) – MarkU Jul 6 '17 at 20:03

{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.

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)

• nice! that's some serious calculations! But I'm not seeing 0.5V Vout swing. So you are suggesting I can not actually test this IC on a breadboard? How about perfboard and solder? – Zeta.Investigator Jul 6 '17 at 15:11
• If you do use perfboard and solder, make the GND a piece of braid or solderwick or copperfoil (small piece of single-sided if fine), and solder the 2 IC pins to this [bottom of cap on REF pin] , and solder the 2 big capacitors to this. – analogsystemsrf Jul 6 '17 at 15:51
• Added diagram for SinglePoint VDD, as well as SinglePoint GND. – analogsystemsrf Jul 7 '17 at 14:38

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.

• I'm using bench power supply to simulate an AA. Anyway I've also tested Vin = 2V and the problem persists. All these inductors meet the required 1.2A saturation current. The IC works for no-load and 220 Ohm load but higher Iout result in failure – Zeta.Investigator Jul 6 '17 at 11:29
• What are the part numbers that you used for L and C? – Glenn W9IQ Jul 6 '17 at 11:32
• Unless you have tested these components to ensure they meet the ESR and other requirements, you can assume that this is where your problem is. – Glenn W9IQ Jul 6 '17 at 11:48
• Before you go there, what is the series resistance? Your cap is looks like an electrolytic when it should be a tantalum. – Glenn W9IQ Jul 6 '17 at 11:56
• 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! – Glenn W9IQ Jul 6 '17 at 13:31

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.