to start with, I am very new to the hardware side so don't beat me up too much.

For a project I am doing I need to build a couple power rails, one 3.3v and another 24v. The power supply will be from a 3.7v 1A LiPo battery.

I started on the 3.3v first, I already had a 3.3v fixed voltage regulator (L78L33A), so I figured I could probably just use this, in the datasheet showed a schematic but didn't have the zener diode and resister values and I don't yet know how to calculate these so I figured I would just play around with different values to see what I could come up with.

I set my benchtop supply to 3.7v, 1A for testing and after playing around a bit I got a nice 3.3v output, still don't understand it but eventually I will learn.

Then, I hooked up the battery and did not get the same 3.3v output, instead it was much lower and when I probed right at \$Vin\$ with the battery in, I would get ~2.5v, however, when I probe the battery without it hooked up to anything I got ~3.9v.

I'm wondering why the results are different and if this is normal to have the different results? If this is, then should I have instead tested using on the battery since that is what it will ultimately be running off of?

Here is the schematic with the values I ended up with (I will start a new question about how to figure out Vz/Vx/R1).

enter image description here


Update 01

Based on the answers and comments, clearly this is def wrong for what I'm trying to do, I was reading the schematic for a boost circuit, will take the suggestions and try this again.

  • \$\begingroup\$ Please show the schematic of your actual circuit. "I probed right at 𝑉𝑖𝑛 with the battery in, I would get ~2.5v, however, when I probe the battery without it hooked up to anything I got ~3.9v." - what is the voltage on the battery when you measure 2.5V at Vin? \$\endgroup\$ Apr 13, 2020 at 18:52
  • 1
    \$\begingroup\$ Just curious, which datasheet said to use zeners on a 3.3V regulator to get 3.3V output? That can't be right. \$\endgroup\$
    – Justme
    Apr 13, 2020 at 18:55
  • \$\begingroup\$ The 78L33 is a 3.3 voltage regulator - no need to add a Zener and resistor around it - but it does need capacitors on the input and output. The drawing says the output voltage will be the Zener voltage PLUS the regulator voltage - that's about 8 volts. Fortunately, a linear regualtor can't boost the voltage, so you probably didn't damage anything. \$\endgroup\$ Apr 13, 2020 at 19:46
  • \$\begingroup\$ This is the datasheet: st.com/resource/en/datasheet/l78l.pdf, Figure 14, which now that I'm looking at it again, is a "boost" circuit, not sure what I was thinking. \$\endgroup\$ Apr 13, 2020 at 21:31

2 Answers 2


A linear regulator will only give you what you feed into it, at best, in the case of the L78L33, a volt or two less depending on the loading.

If you want the 7.6V output shown in the schematic, then you should feed it with about 10V.

For 3.3V output, the datasheet shows an input of at least 5.3VDC, at 0 to 40mA out. If you add the 4.3V zener you should give it about 10V.

For your 24V rail you need to make or buy a switching DC-DC converter, for the 3.3V supply you can consider an LDO regulator of better performance dropout-voltage-wise than the ancient L78L33 (which is also pretty power-hungry for battery use).

  • \$\begingroup\$ How do you know the input needs to be at least 5.3v? I was trying to figure this out but don't know what I'm looking for. \$\endgroup\$ Apr 13, 2020 at 21:34
  • \$\begingroup\$ The dropout voltage is given as 2V which means you should give it at least 3.3 + 2V = 5.3V. Also i.imgur.com/JFmRfFw.png \$\endgroup\$ Apr 13, 2020 at 22:12

That regulator is already a 3.3V regulator. You must not use any zeners with it. Zeners are only meant to make the regulator output voltage higher than the regulator would natively output.

The only problem is, it won't work when input is only 3.7V. It needs at least 5.3V on it's input to work properly.

Perhaps you would like to change the component to a LDO regulator like LM1117.


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