I'm new to the electronics world. I've been trying to build a switching voltage regulator circuit using the LM2574N-5G (datasheet) IC. I'm trying to switch 9V (from a 9V battery) to around 5V. I've tried to implement the circuit that the datasheet suggests, which looks like this:

Notice the red box I superimposed on the schematic. The manufacturer suggests a resistance of 3.1Kohms for R2 to achieve an output of 5V. I went ahead and used 2Kohms for R2 to try and get a voltage between 3.3V and 5.0V.

Here's my attempt to implement this circuit:

I connected two LEDs in series as a load at the output, and they immediately burnt up. I quickly unplugged the battery and checked to see if the battery or any of the other components were hot - they weren't.

After removing the LEDs, and with no load on the supposed "regulated" output, I then measured the voltage between the output and ground, and I measured 7.40V! I swapped R2 with some different value resistors and measured the voltage each time. Here is what I found:

R2      - Voltage
22 ohm  - 5.01v
200 ohm - 5.24v
510 ohm - 5.61v
1k ohm  - 6.20v
2k ohm  - 7.40v

Nothing was running hot during these measurements. Am I missing something obvious or critical? How am I meant to interpret this data? I wouldn't be surprised if my problem can be attributed to a silly mistake on my part - I have no intuition for this sort of thing. Thanks for any input.

  • \$\begingroup\$ Your test load was just two LEDs in series, no resistors? \$\endgroup\$
    – dylanweber
    Sep 8, 2019 at 17:07
  • \$\begingroup\$ @dylanweber Yes...? \$\endgroup\$
    – Paul M.
    Sep 8, 2019 at 17:09
  • \$\begingroup\$ What LEDs did you use and what was their forward voltage? \$\endgroup\$
    – dylanweber
    Sep 8, 2019 at 17:10
  • \$\begingroup\$ I used these: cdn-reichelt.de/documents/datenblatt/C500/… \$\endgroup\$
    – Paul M.
    Sep 8, 2019 at 17:17
  • \$\begingroup\$ So it says in the data sheet that the forward voltage for those LEDS is between 1.8 and 2.4 volts, so putting two in series would have effectively pulled the output down to about 4.2 volts, shorting the supply. \$\endgroup\$
    – dylanweber
    Sep 8, 2019 at 17:19

4 Answers 4


You have committed two errors:

First: The data sheet shows a direct connection from the supply output voltage to the feedback. You put a resistor in series. The data sheet hints that the feedback pin draws current -- there are resistors in there. That means that the feedback pin will see a lower voltage than the output, and the only thing the chip "knows" is to regulate the feedback to 5V. So it is natural to expect that putting a resistor in series will get you a higher output voltage.

Second: LED's, or at least bare LEDs, are junction diodes. This means that their current increases exponentially with increasing voltage. Diodes vary, and LEDs more so, but the current will double for every 10 or 20mV of increase in voltage. This means that over the range of currents where you can see light and the LED isn't burning up, the LED will "try" to be a constant-voltage load.

There are hobby projects out there that connect LEDs up straight to dry cells. This works because a dry cell (particularly a cheap one) has some internal resistance, and a voltage that'll naturally light up a red LED.

However, if you connect an LED to a stiff voltage source you'll most likely get no light, or you'll burn up the LED. For reliable operation you must put a current-limiting resistor in series with it.

So, congratulations, you've made an LED burner-upper.

Get rid of the resistor in the feedback path. You can't reliably light two of your LEDs in series off of 5V without fancy circuitry, so wire each one in parallel with a suitable resistor. "Suitable", in your case, is something that puts 30mA through the LED when it is dropping 1.8V: $$R = \mathrm{\frac{5V - 1.8V}{30mA}} = 106\Omega$$ Round that up to 110 ohms and you'll get plenty of bright light.


So here are the two problems I see with your design and approach:

  1. The LEDs that you attached were red LEDs with forward voltages of approximately 2.0 volts. When put in series, you create a configuration of diodes that will pull down the output to ~4.0V, which will cause a large amount of current to be sunk from your switching regulator. To fix this problem, you would need a current limiting resistor which you can calculate from Ohm's law.
  2. The open circuit output voltage is not indicative of the performance of the regulator in real conditions. The datasheet for the regulator, specifically the 5V regulator section, specifies that the testing conditions they chose were between 100mA and 500mA. Now it should still work at low currents, but the datasheet also specifies that the value of \$R_2\$ should be \$3.1k\Omega\$ with the 5V version of the device, so the behavior of the output at such a low switching frequency with no load and only electrolytic capacitors is undefined and might be "unpredictable."

The manufacturer suggests a resistance of 3.1Kohms for R2 to achieve an output of 5V.

No. R2 is a resistor inside of the chip!

You're using LM2574N-5G, which already has R1 and R2 built-in, and has a fixed 5V output. You don't need nor want any external resistors - you've already got 5V coming out.

To have control over the output voltage, you'd need to use the adjustable version LM2574M-ADJ, and then you'd use an external R1 and R2. But if all you want is 5V, then using the adjustable version isn't necessary.


It appears that you have a 5V version of the device, so a direct connection from output to feedback will provide that. The table in the red box is confusing, where fixed voltage versions use the direct feedback connection, whereas the adjustable versions use an external voltage divider composed of external resistors for R1 and R2..


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