I built a simple circuit to produce -5V from regulated +5V but it doesn't work unless the voltage is at least +6.5V as input.

  • The LT1054C shoud work ok from 3.5V but I measure 0V at output with my multimeter when Vin is +5V.
  • If I raise Vin to +6.5V, the Vout is -6.5V and my circuit works normally. Seems that 6V is the minimum to get the negative voltage output.
  • from the datasheet, the switching frequency of the DC DC step down LM2596 is 150 KHz while the LT1054C can be configured between 15 and 35 kHz. So it impossible to sync both. I think the issue could be here maybe?
  • I use tantalum capacitor (marked "Ta"). Other are Al standard cap. I followed the LT1054 datasheet. I tried Cin=Cout=100uF (tantalum) too.

(I bypassed the problem by using 7805 and 7905 regulators in each corresponding inputs but it's ugly)

Why can I get directly the -5V from the +5V?

LT1054 datasheet

enter image description here

The buck regulator I use (12VDC to +5/+6.5VDC): enter image description here

enter image description here

I will be using 12V unregulated input and regulation to -5V.

It is still not working. I measure Vout = +0.4V with Vin=+12V. It's curious that Vref=+7V while it should be +2.5V (internal reference). I tried a second chip (LT1054C N8 by Linear Tech.), same thing. Both chips work fine as simple inverters (no resistors, Vout=-Vin if Vin>+6.5V)

I tried a LM336-2.5V reference as external reference instead of using the reference pin, same thing.

Breadboard circuit: [link]https://drive.google.com/open?id=1Rgna1OPtDUJwWGj8AU4rmA7glmsUp6G5

enter image description here

  • \$\begingroup\$ No need to synchronize anything! You should only be able to see the DC component from the first regulator, save for a very small ripple. Where is your output capacitor on the buck? \$\endgroup\$
    – winny
    Apr 25, 2019 at 21:53
  • \$\begingroup\$ How did you apply 6.5V to it? If you used an external power supply for that? Did you use that supply for testing on 5V as well? \$\endgroup\$
    – Huisman
    Apr 25, 2019 at 21:54
  • \$\begingroup\$ the buck is a DC DC step down, it is supplied by a DC12V PSU. I don't have output capacitor, should I put some big one? I use this kind of DC DC step down [link] mpja.com/images/30148-large.jpg \$\endgroup\$
    – Vinlar
    Apr 25, 2019 at 22:07
  • \$\begingroup\$ I added a 1000uF cap at the output of the buck, that didn't help. @winny \$\endgroup\$
    – Vinlar
    Apr 25, 2019 at 22:17

2 Answers 2


It looks like you are missing the R1/R2 resistor divider network used to set the desired output voltage for the voltage converter. From the LT1054 datasheet section (http://www.ti.com/lit/ds/symlink/lt1054.pdf):

The error amplifier of the LT1054 drives the PNP switch to control the voltage across the input capacitor (CIN), which determines the output voltage. When the reference and error amplifier of the LT1054 are used, an external resistive divider is all that is needed to set the regulated output voltage. shows the basic regulator configuration and the formula for calculating the appropriate resistor values. R1 should be 20 kΩ or greater because the reference current is limited to ±100 μA. R2 should be in the range of 100 kΩ to 300 kΩ.

The typical application circuit at the top of section 8.2 shows where resistors R1 and R2 should be placed:

enter image description here

Edit 1

Followup on your other thought, the electrical characteristics section of the datasheet specifically calls out using tantalum capacitors for Cin and Cout, so I think you're fine using that type rather than Al.

All regulation specifications are for a device connected as a positive-to-negative converter/regulator with R1 = 20 kΩ, R2 = 102.5 kΩ, external capacitor CIN = 10 μF (tantalum), external capacitor COUT = 100 μF (tantalum) and C1 = 0.002 μF

Edit 2

Two other ideas:

  1. The datasheet makes specific mention that when using the LT1054 as an unregualted inverter, equal nominal values should be used for Cin and Cout for best results (Capcitor Selection https://www.analog.com/media/en/technical-documentation/data-sheets/1054lfh.pdf)

  2. I suspect that when configured in an unregulated circuit the LT1054 can't effectively invert voltages lower than 7V. There are other application circuits like the positive doubler which accept voltages as low as the minimum Vin of 3.5V, but all of the electrical characteristics for -5V regulated output are measured with Vin from 7-12V. If you use the resistor divider values for -5V regulated output and a Vin of 7V or higher, do you see a regulated -5V out?

  • \$\begingroup\$ I tried this circuit and it doesn't work too. Check this datasheet [analog.com/media/en/technical-documentation/data-sheets/… on page 10. I need a basic unregulated voltage inverter, so the resistors aren't needed. \$\endgroup\$
    – Vinlar
    Apr 25, 2019 at 22:24
  • \$\begingroup\$ As mentioned in first post, I do use tantalum capacitors for Cin and Cout. \$\endgroup\$
    – Vinlar
    Apr 25, 2019 at 22:25
  • \$\begingroup\$ Have you measured the amount of current drawn from the output. I was suggesting using the feedback pin because the voltage losses can actually be very high (1.1V at 100 mA, enough to potentially explain your symptom) and the feedback resistor divider allows the chip to compensate for output current even when just being used as an inverter. \$\endgroup\$ Apr 25, 2019 at 22:44
  • \$\begingroup\$ I didn't measure the current drawn, I don't think my multimeter draws 100mA. For Vin = +5V, I read Vout = +0.5V AND For Vin = +6.5V, I read Vout = -6.5V No load is connected at the output. I only use the multimeter to check voltage. \$\endgroup\$
    – Vinlar
    Apr 25, 2019 at 22:50
  • \$\begingroup\$ Okay, second question, when you tried using the resistor divider values for -5V regulated output with a 6.5V input, was the output a correct -5V in that case? I'm noticing that all of the electrical characterisitcs for both the TI and AD versions are measured with Vcc from 7 - 12V. It might be that while 3.5V is a valid supply voltage for some applications (like the voltage doubler in fig. 20), that as an unregulated voltage inverter only 7V+ can be used. \$\endgroup\$ Apr 25, 2019 at 22:57

Seems ICs were damaged, I changed them and I was able to produce -5V from +5V without issue using 2x 100uF tantalum capacitors. So, I didn't use the regulator circuit with resistors.


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