# Dividing output voltage from a buck boost converter

I am a student in high school and have been working with buck boost converters. I currently have a buck boost converter that steps up 5V to about 58V. I created 3 voltage dividers that should step the output voltage from the buck boost converter to 30V, 35V, and 40V. I then wired the voltage dividers to a 3 way DIP switch. This should allow you to choose which voltage to use (30V, 35V, or 40V via the switch.) However, the voltages I am getting as an output are nowhere near the predicted voltage (given $$\V_{out}=V_{in}(R_2/R_1+R_2)\$$.)

For example, I am getting an output of 12V with two 1k resistors. I then tried using a voltage dropping resistor to get the 3 voltages. This worked initially, with 10k ohms getting 30V, 15k ohms getting 35V and 23k ohms getting 40V. However, when I turned the converter off and back on, all of these voltages were different (i.e the 10k resistor was giving a $$\V_{out}\$$ of 40V instead of 30V.) Also, the converter does not give the same $$\V_{out}\$$ each time it is turned on. The output usually ranges from 52V to 62V but is mostly a steady .939A. The power supply is a 5v usb plugged into an iPhone charger port.

Here is the schematic to the buck boost converter.

Here is the link to the datasheet for the LM2588.

If anyone can offer an explanation to why $$\V_{out}\$$ is so inconsistent, or offer an alternative method for creating 30V, 35V, and 40V that would be fantastic. I am happy to provide any other information if necessary.

• These feedback mods. must be done with twisted pairs and relatively short leads to avoid interference. But your formula looks wrong. and test results incomplete. Sep 29 '19 at 19:20
• Are you using the voltage divider as the output, or to modify the feedback signal?
– JRE
Sep 29 '19 at 19:24
• VOUT= VREF(1 + R1/R2) . note correction where VREF= 1.23V Sep 29 '19 at 19:24
• Thanks Sunnyskyguy EE75 and JRE for the replies. I am using the voltage dividers as the output. The output from the Buck boost converter is going into the voltage dividers, and the switch is just choosing which voltage divider will be used at that time. The voltage dividers that I am mentioning are not part of the buck boost circuit or schematic.
– JSHS
Sep 29 '19 at 19:44
• Ok. That's not good. Consider your load, and what that does to your voltage divider ratio.
– JRE
Sep 29 '19 at 19:58

1. Switching power supplies and breadboards don't mix. Breadboards have parasitic capacitance between pin rows, and parasitic inductance in the rows - not to mention in the wires. The contacts are also often weak and don't hold well - the resistance varies just by moving the wires. The contacts also are not made to have amperes of current flowing through them. You really nead a PCB for a switching power supply. At the very least, you need a carefully soldered circuit built on a piece of perfboard to get a switching power supply to work reliably.

2. A voltage divider is a poor way to generate a lower voltage from a higher one. Your load changes the ratio of the divider. The correct way to get what you want is to switch the feedback resistors so that the regulator generates the output voltage you want.

This section of the datasheet shows how:

Pick a single value for $$\R_2\$$, then figure values for $$\R_1\$$ that give your desired $$\V_{out}\$$ and use a switch to select which $$\R_1\$$ is connected to set your output voltage.

• That makes sense, thanks! Ill go try it.
– JSHS
Sep 29 '19 at 20:28