# 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. – Tony Stewart Sunnyskyguy EE75 Sep 29 at 19:20
• Are you using the voltage divider as the output, or to modify the feedback signal? – JRE Sep 29 at 19:24
• VOUT= VREF(1 + R1/R2) . note correction where VREF= 1.23V – Tony Stewart Sunnyskyguy EE75 Sep 29 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 at 19:44
• Ok. That's not good. Consider your load, and what that does to your voltage divider ratio. – JRE Sep 29 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 at 20:28

Firstly, I would recommend not using a voltage divider to get your different output voltages. You should look into using stacked zener diodes with a current limiting resistor on the top of them. This would only be good too if the different voltages are going into a high impedance load as well. If you you don't have a high impedance load with your current configuration or with the zener configuration, then whatever your output voltages are connected to will as affect your resistor divider. This will also affect the Vout of your buck boost. The other thing you may want to consider is that your inductor may be going into saturation by loading it with the voltage divider. You should probably use something with a PID controlled output and look into the saturation issue as well. Boost converters can be very finicky when it comes to boosting a voltage on the scale you're trying to do it at and then having different loads on the output, especially without feedback going into the controller or switcher. The zeners will help the outputs stay the same as long as your Vout stays above the combined output voltages of the zeners.

• Going back to what I said about saturation and the current you're using. I'd bet that you are pulling your 220uH inductor into saturation. Think about using a smaller inductor. Also, what frequency is this switching at? – Dave M. Sep 29 at 20:48
• Also, almost 1 amp at 50 volts can be a serious shock hazard. Be careful please. – Dave M. Sep 29 at 20:59
• I have the Boost converter hooked up to a transistor that is controlled by a 555. The 555 is oscillating at 280hz and 82%. I will play with smaller inductors too, thanks for the help. – JSHS Sep 30 at 0:42
• I understand you're in highschool and using a 555 is cheap, but 555s have a huge variance with temperature. You may want to try and use a cheap ucontroller instead. It will also give you the capability of programming a PID controller and having a continuous control of the output. – Dave M. Sep 30 at 1:03
• @DaveM.: You should not need a 555 or other clock source. The 2588 has its own oscillator, and will run at 100kHz if you don't mess with it. If you apply a clock to 1 (/On, Off) then you are simply turning the 2588 on and off at your clock rate - 280 times a second according to your comment. If you apply the clock to pin 6 (Sync) then each pulse will cause the 2588 to "stutter step" as it tries to synchronize to your slower rate. Don't do that. Just let it run at its own rate. – JRE Sep 30 at 4:46