Ignore the wrong position of the IC when I took a photo ;D

I have based my design from here: https://www.pcbway.com/blog/technology/DC_to_DC_Boost_Converter_using_UC3843.html

UC3842 Datasheet: https://www.mouser.com/datasheet/2/149/uc3843-309322.pdf

The problem is the output (15.7 V) is just below the input voltage (16 V), only change by the diode drop (0.3V). The out pin of the IC also do not have any pulse.

This is my first time doing analog stuff but I know the basic operation of a boost converter. I don't understand exactly how the chip works. I keep it at 16 V as it is the UVLO on. My goal is to have the output be greater than the input (16 V) and my life is complete. I have checked: every connection, values and no shorts, voltage coming to the IC, etc... My components are limited only to what I salvaged, that's why I have different values. But I have 10 UC3842 I bought online.

What do you think is the problem?

Is the shunt resistor in series with MOSFET too high? Is the MOSFET model incompatible? Is the capacitance critical at Isense pin? Are these UC3842 chips dead? (I've swapped them, same results) I was surprised there is a shunt in the high-current MOSFET (efficiency?), does it need to be high power resistor like I used? Is the board type inappropriate for my application?

Bonus question: Do current mode controlers do calculus? They seem to react with slope of current vs time from what I've read. Can't understand it tho.

What I have: Lab PSU Analog scope 60 MHz 2 DMMs 1st year EE

I'm already 4 days tried everything in my mental capacity. Any advice/tips or theory what's wrong? THANK YOU SO MUCH! My not working design:

My basis:

The actual circuit:

  • 1
    \$\begingroup\$ Welcome! Forget making any switch mode converter on a perfboard with long leads. You need to design a PCB for it with ground plane. \$\endgroup\$
    – winny
    Commented Jun 22, 2023 at 20:44
  • 2
    \$\begingroup\$ Section 11 of the datasheet contains layout guidelines. You managed to violate all of them. \$\endgroup\$
    – CL.
    Commented Jun 22, 2023 at 20:46
  • 1
    \$\begingroup\$ Alas, you can't actuate a MOSFET in a power converter with so long leads, no way. The components placement must be as compact as possible, with a MOSFET close to the DRV pin of the controller and a loop involving the MOSFET + shunt + inductor and input capacitance as short as possible. If you don't respect these rules, it won't work. In absence of voltage on CS pin of the UC384x, it should issue pulses if \$V_{cc}\$ is beyond its UVLO, the CMP pin is high and the oscillator at the Rt/Ct pin is alive. Don't lose faith in power electronics, "it's a long way to the top if you wanna R'n'R!" : ) \$\endgroup\$ Commented Jun 22, 2023 at 21:04
  • \$\begingroup\$ @winny yeah so I need to overhaul this. Why are long leads illegal with switch modes tho? Thank you for answer. \$\endgroup\$ Commented Jun 22, 2023 at 21:23
  • \$\begingroup\$ You have UC3842 in the question but -3 in the schematic; which is it? As you can see in the datasheet (UVLO start threshold), this makes all the difference on a low voltage design like this. \$\endgroup\$ Commented Jun 22, 2023 at 21:23

3 Answers 3


What do you think is the problem?

The problem is all of the wires between components. DC DC converters are very susceptible to problems with parasitics as series resistance and inductance can create issues with switching or feedback.

So what can you do about it? If you have a good digital multi meter that can measure milliamps and millivolts you can put the probes between two points of wire and measure the voltage drop on the wire.

Put the meter in millivolt mode, do this in AC and DC modes. The first thing to check would be ground. If you see a significant difference in voltage say more than a few mV, then that section of ground would need to be increase with a better conductor.

Also check the feedback resistors, you may need to shorten the wires.

The easiest way to fix it would be to change the feedback resistor dividor point.

  • \$\begingroup\$ Thank you, it seems that long wires are actually the problem here. But how can I test if the chip even tries to oscillate? There's no waveforms in RT/CT pin or OUT pin. But there's 5v at Vref pin. \$\endgroup\$ Commented Jun 22, 2023 at 22:34

Your under-voltage lockout can be as high as 17.5 V for the UC3842. Rather use the the UC3843 that is aimed at lower voltage applications.

Below is a working design (many in the field) that can produce 30 V at over 3 A, and was designed to operate from 11 V to over 16 V input. It has slope compensation to prevent sub-harmonic oscillations for duty cycles near 50% and over. The phase margin is around 65 degrees, with about 12 dB gain margin, which ensures good stability.

33V boost sc

waves sc

Slope compensation removed.

33v nsc diag

Sub-harmonic oscillation can be seen clearly.

33V nsc

Scope captures from an actual working prototype unit during stress testing with 12 V input.

(Vg and VI-sense)


With no slope compensation, sub-harmonic oscillations start to appear near 50% duty cycle.

sh osc

Load test.

Electronic load

FRA sweep test.

Gain phase proto


33v proto


If you want to create a better layout without needing to use software you can use a copper clad board with Manhattan style soldering using MeSquares.

The result would look something like this. https://makezine.com/article/technology/manhattan-style-circuit-in-a-copper-cladding-chassis/

MeSquares are small pads you can glue to the copper board. They are available here: https://www.qrpme.com/?p=product&id=MeSb


Also, yes you can do calculus using circuits, for example, here is how to make an integrator circuit: https://www.electronics-tutorials.ws/opamp/opamp_6.html


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