I have an XL6009 powered circuit that draws about 2 Amps at steady state. Actually, the ON/OFF switch is connected near the source input and before the DC/DC converter but, in this way and by design, the 2 Amps current must flow through the switch.

I would like, instead, to use the ENABLE pin on the XL6009. Connecting the switch to the EN pin would drive the circuit ON/OFF without requiring the same amount of current flow through it.

The problem is that I've noticed that EN pin left floating or driven to Vdd enables the circuit and to power off it should be driven to ground.

How, using simple components (like a resistor) use a simple ON/OFF switch to drive the EN pin on the XL6009?


simulate this circuit – Schematic created using CircuitLab

Thanks in advance for all your suggestions.

  • \$\begingroup\$ What is the topology created with XL6009 - boost, buck-boost or inverting? Draw a schematic. It is essential, because in boost disabling the regulator would not stop the current flowing to the load. \$\endgroup\$ May 30, 2017 at 20:19
  • \$\begingroup\$ Thanks @TodorSimeonov, I've added a schematic of the actual circuit. Input is 5V, output should be between 5 and 10V (around). EN pin now is disconnected (floating = ON). \$\endgroup\$
    – PeppeA82
    May 31, 2017 at 5:57
  • \$\begingroup\$ Doesn't EN float high when disconnected? Therefore you would need to connect EN to Vin via a resistor, so you can ground EN via switch to turn the output off? I have not tried this, just relying on the datasheet. \$\endgroup\$ Nov 26, 2017 at 3:39

1 Answer 1


This is a typical boost topology. You can't stop the current to load unless you break the circuit with a relay or MOSFET switch. Enable wont work. When IC is disabled it will not switch and boost voltage, but the input voltage - minus diode forward voltage will still appear at the load. If you look carefully at your schematic you can see that current flows from Vin through L1, D1 and goes to the load no matter if the SW pin of the IC is being switched.

For example if you have 5V input and 12V output, disabling the IC will stop it from switching and voltage at the load will not be 12V, but it will be 5V-0.4V = 4.6V.

The only way to stop it is using a mechanical or semiconductor switch in the current path (input positive or output positive).

UPDATE: Changing the topology if you will use it only for boost will compromise its efficiency. However depending on your output power and load requirements you could afford a lower efficiency by means of simplicity.

This is a schematic of a buck-boost module which is very popular. If you need a couple of pieces only for your project you can buy it ready assembled.

enter image description here

As I said, efficiency is lower than pure boost (like 70-75% compared to 80-85%) but using this topology you can take advantage of the "enable" pin. You can tie it with a 10k resistor to GND and place the switch between IN+ and EN. When the switch is turned on there will be +Vcc on EN an XL6009 will operate. When your switch is turned off the IC will stop and transfer capacitor C1 will prevent current flowing from input to output.

Here is a picrue of a popular module:

XL6009 buck boost module

The large SMD capacitor between the inductors and schottky diode is this blocking capcaitor.

  • \$\begingroup\$ Thanks for the clean suggestion. Could you suggest me a possible approach, either by changing the DC/DC or with the method you suggested? \$\endgroup\$
    – PeppeA82
    May 31, 2017 at 7:50
  • \$\begingroup\$ @PeppeA82 see my update. \$\endgroup\$ May 31, 2017 at 9:07

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