# Non-inverting Buck boost

#### declan2693

Joined Feb 2, 2018
37
Hi guys,

I have simulated a buck boost with a positive input I get a negative output correctly. Now I want to create a non-inverting output and from what i've seen the circuit doesn't change? How do I create a positive output with a positive input

#### Alec_t

Joined Sep 17, 2013
13,443
Is your circuit going to remain a mystery? Post the schematic and the simulation file here and there's a chance we might be able to help

#### Dodgydave

Joined Jun 22, 2012
10,830
Just a minute ,, I'll get my crystal ball out....

Joined Feb 20, 2016
4,290
You need to Buck the trend of thread starters who ask questions without supplying the relevent info to Boost you chances of getting some help here.

#### BobTPH

Joined Jun 5, 2013
6,527
A boost converter followed by a buck converter, both in the standard topology, will output a positive voltage referenced to ground.

Bob

#### declan2693

Joined Feb 2, 2018
37
All I want to know is how to create a non-inverting buck-boost converter

#### AnalogKid

Joined Aug 1, 2013
10,308
Then all you need to do is tell us the input and output voltage ranges so we can determine if you know what you are talking about.

Then all you need to do is go to the Linear Technology web site and pick out a chip.

ak

#### declan2693

Joined Feb 2, 2018
37
Then all you need to do is tell us the input and output voltage ranges so we can determine if you know what you are talking about.

Then all you need to do is go to the Linear Technology web site and pick out a chip.

ak
My input range is 2.9V to 4.2V and output I want +5V on a buck-boost converter. I want to simulate a non inverting buck-boost.

#### tsan

Joined Sep 6, 2014
130
My input range is 2.9V to 4.2V and output I want +5V on a buck-boost converter.
Boost converter is enough on this case, where input is always lower than the output voltage

#### AnalogKid

Joined Aug 1, 2013
10,308
My input range is 2.9V to 4.2V and output I want +5V on a buck-boost converter. I want to simulate a non inverting buck-boost.
Since the output voltage always is greater than any possible input voltage, why do you think you need a buck-boost converter? In a buck converter, the output always is *less* than the input.

ak

#### declan2693

Joined Feb 2, 2018
37
Since the output voltage always is greater than any possible input voltage, why do you think you need a buck-boost converter? In a buck converter, the output always is *less* than the input.

ak
Yes, I just want to simulate a "Non inverting Buck-Boost" boosting first and later do results for the buck side. I have a PSIM circuit of the Inverting Buck Boost. I just dont know how to make it Non inverting which is what I am asking.

#### AnalogKid

Joined Aug 1, 2013
10,308
For a single-inductor circuit you need a full H-bridge switching circuit. For a standard schematic power flow (left to right):

Buck: inductor left side is switched. Right side goes to the output filter.

Boost (flyback): inductor right side is switched and goes to the output filter through a rectifier. Left side is tied to the input power source.

http://www.linear.com/product/LT8705A
http://www.linear.com/product/LTC3785

They also have devices with the MOSFET switches built-in, but since your current requirement still is a secret I can't recommend anything.

ak

Last edited:

Joined Feb 20, 2016
4,290

#### ebp

Joined Feb 8, 2018
2,332
The short answer is that you can't make a non-inverting single-switch, single-inductor buck-boost converter. The inverting buck-boost can work because there is never any direct path between the input and the output. All of the energy is stored by the inductor during the switch ON time and delivered to the load during the switch OFF time.

If you turn the diode around:
• As soon as the switch is turned on the output is connected directly to the input with no means of limiting the current, so the output can't ever be lower than the input while the switch is ON. At very low power you could get away with this, relying on the output capacitor as the sole energy storage element, but it just isn't workable at any significant power because of the huge magnitude of current.
• When the switch is turned on the inductor begins to charge, the current rising according to δi/δt = V/L - the equation that rules all behavior in switch mode power supplies. When the switch turns off, the inductor "tries" to keep the current flowing at exactly the same magnitude and in the same direction as it was the instant before the switch opened. Current would try to flow from the "top" end (as normally drawn in a schematic) to ground - but the diode prevents current from flowing from the output side into the inductor and the switch prevents it from flowing from the input side. There is simply no path for the current. But that stored energy must go somewhere. And it will, but it won't be pretty. The top end of the inductor will go to some very large negative (with respect to ground) voltage and either cause the switch or the diode to break down, at which point it is all over. You could put a diode across the inductor, as you might for a relay coil, so the current has somewhere to go, but then all you are doing is shifting around charge you can't use. The inverting buck-boost allows the energy stored in the inductor to be delivered to the output capacitor and the load.
You can simulate this. Turn the diode in the inverting circuit around. Pulse the switch on briefly and do a transient analysis, watching voltage and current from the time you turn the switch off until some time after you turn it off. If you use an ideal switch (no resistance when ON) you may need to put a small resistor in series with the capacitor to keep the current from being infinite at switch turn-on.

#### declan2693

Joined Feb 2, 2018
37
The short answer is that you can't make a non-inverting single-switch, single-inductor buck-boost converter. The inverting buck-boost can work because there is never any direct path between the input and the output. All of the energy is stored by the inductor during the switch ON time and delivered to the load during the switch OFF time.

If you turn the diode around:
• As soon as the switch is turned on the output is connected directly to the input with no means of limiting the current, so the output can't ever be lower than the input while the switch is ON. At very low power you could get away with this, relying on the output capacitor as the sole energy storage element, but it just isn't workable at any significant power because of the huge magnitude of current.
• When the switch is turned on the inductor begins to charge, the current rising according to δi/δt = V/L - the equation that rules all behavior in switch mode power supplies. When the switch turns off, the inductor "tries" to keep the current flowing at exactly the same magnitude and in the same direction as it was the instant before the switch opened. Current would try to flow from the "top" end (as normally drawn in a schematic) to ground - but the diode prevents current from flowing from the output side into the inductor and the switch prevents it from flowing from the input side. There is simply no path for the current. But that stored energy must go somewhere. And it will, but it won't be pretty. The top end of the inductor will go to some very large negative (with respect to ground) voltage and either cause the switch or the diode to break down, at which point it is all over. You could put a diode across the inductor, as you might for a relay coil, so the current has somewhere to go, but then all you are doing is shifting around charge you can't use. The inverting buck-boost allows the energy stored in the inductor to be delivered to the output capacitor and the load.
You can simulate this. Turn the diode in the inverting circuit around. Pulse the switch on briefly and do a transient analysis, watching voltage and current from the time you turn the switch off until some time after you turn it off. If you use an ideal switch (no resistance when ON) you may need to put a small resistor in series with the capacitor to keep the current from being infinite at switch turn-on.
This is exactly what I need. Thank you!

#### AnalogKid

Joined Aug 1, 2013
10,308
Something to consider is that the name "inverting buck-boost" is incorrect. In terms of the circuit topology, it is a boost converter. There is no "buck"-ing. For example, if the input is +5 V and the output is -3 V, that is an 8 V difference. Algebraically, the output is 3 V greater than the input, and that is a boost circuit. In fact, many regulator chips, such as some of the National Semiconductor "Simple Switcher" parts, show the inverting regulator as a minor variation of the boost circuit.

ak

#### ebp

Joined Feb 8, 2018
2,332
It is actually a flyback converter. At no time is there current flow directly from the input to the output, as happens in both the boost converter and the buck converter.

#### declan2693

Joined Feb 2, 2018
37
Update:

So Ive been working on simulation of the non inverting buck boost.
Trying to use the the boost side
Voltage in 4.2V
voltage out 5V
Duty cycle 0.54

I have the controls in circuit with the input voltage above the mosfet threshold. (Not its connected correctly)

Well from my attempts I have only been able to boost but way above 5V something like 9 with flucuations which is wrong.

Anyone spot whats wrong with it?

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#### djb

Joined May 17, 2008
31
If you are still online, you have to understand the buck converter alone, using a P channel mosfet (high side switch), and the boost converter alone, using N channel mosfet (low side switch).

In your diagram you have N channel for high side switching, something is relatively difficult to achieve, since usually we need +10V above source voltage to get a Vgs of 10V to power up the gate. That's why N channels are used usually for low side switching (connected to the ground)

For the high side switching you have 3 options. 1. use an integrated gate controller that support N channel high side switching, OR 2. use a P channel mosfet on high side, OR 3. make a voltage doubler with 555 timer or joule thief, to get the +10 Volts you need.

The boost circuit seems ok. Have in mind that controlling all that will not be so easy, to achieve the right feedback and also to control both mosfets at the same time. Post again if you decide to continue this project.