I've created a 12 VAC -> 24 VDC boost power supply based on the TPS55340 regulator. I used a reference design from TI's WEBENCH tool, with some selected part substitutions for parts availability, coupled with a pretty standard AC->DC rectifier rated for 3 A. The load is a series/parallel array of LEDs (\$V_f = 2.7-3.7\text{ V}\$). Each series of 6 LEDs should draw 20 mA and total 240 mA.

The problem I'm having is that the voltage as measured between the +24V NET and GND is approximately 17 V, which only gives approximately 2.5 mA to each series of LEDs and makes them very dim. The datasheet (linked above) says that the output voltage is selected via the divider attached to the FB circuit and computed via $$V_{\text{out}} = 1.229\text{ V} \times \left(\frac{R_{\text{FBT}}}{R_{\text{FBB}}} + 1\right)$$

The standard example - which I happen to be using - is 187k/10k for 24 V. I did pull those two resistors off a board and measure them to verify.

Any pointers to send me on the right direction here? I'm really new to hardware design but this seems like a pretty straightforward boost circuit that I've screwed up somehow.

Critical Parts

Entire BOM - Google Sheet

TI's Reference design TI WEBENCH Design

My Circuit as Built enter image description here

Layout Top layout top

Layout Bottom layout bottom

  • 2
    \$\begingroup\$ With switchers it is important to see the layout and/or a picture, also parts selection, especially the inductor saturation current is important. Can you put links to datasheet for inductor and information about input and output caps? Also 220µF after rectifier bridge is too low, it has to store energy when rectifier does not conduct. Can you confirm the value of this cap? \$\endgroup\$
    – bobflux
    Sep 22, 2021 at 13:14
  • \$\begingroup\$ Your schematic appears to indicate the AC live and AC neutral are connected to your switching chip via a bridge rectifier. AC live and neutral are usually in the hundreds of volts and, for anyone who has a degree of caution, dropped to a lower voltage (and one that is isolated) via a transformer. \$\endgroup\$
    – Andy aka
    Sep 22, 2021 at 13:23
  • \$\begingroup\$ Measure the voltage on the feedback pin. If it's low, disconnect the load, does it go up? If so, check specs on the inductor first. \$\endgroup\$
    – Aaron
    Sep 22, 2021 at 13:24
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    \$\begingroup\$ Betting on input capacitor being too small, test by probing Vin with a scope. \$\endgroup\$
    – bobflux
    Sep 22, 2021 at 14:32
  • 2
    \$\begingroup\$ At 24V 240mA out, it should pull 480mA from 12V. dv/dt=i/C so with a 220µF cap, input voltage drops to zero between rectifier pulses. Capacitor should be increased to 2200-4700µF. \$\endgroup\$
    – bobflux
    Sep 22, 2021 at 15:16

1 Answer 1


Here's a simulation with 220µF post-rectifier cap.

The current source sims a boost regulator. Since it tries to output a constant 240mA at 24V, it will input roughly constant power (neglecting efficiency) so its input current is proportional to 1/Vin. So I've used a function source, with "v(vcc)>3" to set current to zero when voltage is below 3V (I'll assume the boost just shuts down, also this avoids a division by zero) and "0.45*12/V(VCC)" to simulate the boost input current. I realized after taking the screenshot that it should be 0.5A instead of 0.45A but whatever.

enter image description here

As you can see the input cap is too small, so as it discharges, the boost draws even more current (green plot) until it sucks all the energy out of the cap and shuts down.

Switching to 1600µF looks a lot better, the capacitor stores enough energy to last until the next AC half-period where it gets recharged.

enter image description here

When the cap voltage doesn't drop too much (just a few volts) the boost input current is constant enough that you can use the back of the envelope approximation of:

\$ dv/dt = i/C \$

\$ C = \frac{ i.dt }{ dv } \$

So if you want dv = 2V drop between periods ; dt is say 8ms because even with a 50Hz AC period the rectifier should conduct 20% of the time, and i=0.5A, then this gives a required capacitance of 2000µF. 1600µF is close enough, it should work fine. 220µF is definitely not enough though.

  • \$\begingroup\$ Yessir. Thanks for the thoughtful and thorough replies & answer. \$\endgroup\$
    – pseabury
    Sep 22, 2021 at 20:20
  • \$\begingroup\$ Which software were you using to simulate this? \$\endgroup\$
    – pseabury
    Sep 24, 2021 at 18:47
  • 1
    \$\begingroup\$ @pseabury spectrum-soft.com/download/download.shtm \$\endgroup\$
    – bobflux
    Sep 24, 2021 at 18:50

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