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I'm trying to debug a incorrect output voltage from a boost regulator (TPS61220). The schematic is below. Pointed colors represent the measuring points in graphs underneath:

enter image description here

The graphs show the circuit behavior. enter image description here

The graph below is the input voltage and the L voltage captured with an oscilloscope: enter image description here

According to the datasheet, the ratio of R1 and R2 should result in an output voltage of cca. 2.8V. However the output (green) is never higher than the input and is rather unstable. The current coming from the source is a constant 15mA even though there is nothing connected to the output.

The prototype board was assembled on a breadboard.

The power source is expected to be unstable, vary in voltage between 0.8-2.5.

Questions:

  • How would one approach debugging this circuit?
  • Should the current be observed on the induct and it makes no sense to check the voltage (green)?
  • Would 15mA of current without a consumer imply an assembly problem?
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    \$\begingroup\$ Why do you have your feedback divider (R1/R2) connected to Vin?!? How is the IC supposed to know what its output is if you're not giving it any feedback from that output? \$\endgroup\$ – brhans Sep 21 '16 at 16:45
  • \$\begingroup\$ That looks like a good answer to me @brhans \$\endgroup\$ – Andy aka Sep 21 '16 at 16:50
  • \$\begingroup\$ also, what kind of supply collapses to 1.5V under a 50mA load? You may have unrealistic expectations of a too-small battery - it is trying, Vout > Vin from 21 to about 30s (but only just) \$\endgroup\$ – Brian Drummond Sep 21 '16 at 17:00
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    \$\begingroup\$ A "detail" like that ought to be in the question. What you're looking for is much more like an MPPT tracker, and you cannot expect it to produce a stable output voltage unless you can control the current demanded by the load. (Also the datasheet link is now broken). First question on the revised circuit is, why aren't you seeing teh oscillations? \$\endgroup\$ – Brian Drummond Sep 22 '16 at 10:57
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    \$\begingroup\$ @winny Updated the question with the graph from the oscilloscope. It also shows the input voltage which is the voltage on EN. \$\endgroup\$ – TheMeaningfulEngineer Sep 22 '16 at 14:51
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In order to regulate the output voltage, the IC needs to know what that voltage is.
Some ICs have an internal connection to this point in the circuit and usually an internal fixed voltage divider network which is compared to an internal reference.

If the IC is intended to have an adjustable output voltage (like yours), the usual method is to have the user/designer incorporate their own voltage divider, allowing a range of output voltages to be selected.
You choose the resistor values to achieve a division ratio such that when your desired output voltage is achieved, the value produced by the voltage divider matches the value of the IC's internal reference.
This divider of yours is connected to the feedback pin on the IC.

BUT - you have connected your voltage divider to your input supply voltage instead of the output - so your IC has no idea what the output voltage is.
Its not getting any feedback, so it can't regulate.

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  • \$\begingroup\$ Thanks for the answer brhans. I made a mistake in the drawings when posting the question. Updated the image and graphs. \$\endgroup\$ – TheMeaningfulEngineer Sep 22 '16 at 9:05
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Your VFB voltage is at 20mV, while the datasheet specifically shows it should be at 500mV. That explains why you don't get the expected output voltage. As to why the VFB is so low, you need to investigate that more.

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