Buck - Boost converter with Ir2110

[Bruno.do]

We are currently designing a Buck-Boost Converter circuit and using the IR2110 gate driver to properly drive the MOSFET by increasing the Vgs. The step-down (buck) operation works well — for example, at 20% duty cycle, we get around -2.4V output, which is expected.

However, when we try to boost the voltage by increasing the duty cycle, we face a problem. The duty cycle only goes up to around 68%, and beyond that, it doesn’t increase further. At this 68% duty cycle, the output voltage reaches -18.8V, but it doesn’t boost any higher no matter what we do.

We are wondering if this issue is caused by:

• Incorrect component selection (e.g., MOSFET, diode, etc.)
• A potential mistake in the circuit design
• Or if it’s simply a limitation of the system characteristics with the current setup.

Here are some additional details:

• The diodes used in the circuit are all 1N5819.
• The Vcc for the IR2110 is set to 15V (not 12V).
• The MOSFET in use is IRF640.
• Our target buck-boost range is from -2V to -25V output.

Could anyone suggest possible reasons why the boosting stops at -18.8V and how we might resolve this issue?

Any advice on improving the performance or identifying whether it’s a component, design, or inherent system limitation would be greatly appreciated.

Thank you!

 

[kd502]

Don't draw a MOSFET as a switch when you want to understand how it works. MOSFET is controlled by voltage between gate and source. You can turn on MOSFET by setting Vg=0V and changing source voltage below -5V. That is why this circuit can't work and it is impossible to do it with IR2110

You used high side driver wrong Vs is not connected anywhere.

Try with P channel MOSFET

 

[24vtingle]

Tes, or use Cuk converter, especially as you only have low power.
Cuk gives you low side drive and currnt sense.

Or use a mini boost IC to give you a high side 15V supply, (or rather a 5v supply "sat" on top of the 10v battery) and then use that to connect to your IRRxxxx and use a 5V drive NFET in the high side.

 

[Easy peasy]

The cap is the wrong way round in the schematic - what is the voltage rating of the parts you are using ? ( esp D2 - this sees Vin + Vout ) ) - if you exceed PIV for the diode - there can be a clamping action causing your issues !

--- Updated Apr 25, 2025 ---

Also - you inductor may be saturating ( your freq may be too low ) - or its Rdc may be too high for your load - post details of the L.

--- Updated Apr 25, 2025 ---

Also - your gate drive topology is not too flash - assuming N fet. You could put an N fet in the bottom, left of the L and drive directly - source on the left.

 

[Easy peasy]

-25 volts into 200 ohms is 125mA, 3.125 watts, since your L is 1mH, it is likely saturating at >=125mA, running from 10V you may need to be in CCM to get sufficient power transfer, power = 0.5 L. I^2 Freq for DCM, for CCM it is ( Ipk^2 - Imin^2 )

Vout/Vin = D/(1-D ) for the buck - boost

Using substitution and simplifying of the following 2 formulae: P = 0.5 L Ipk^2 freq, and: Vin/L = Ipk / Ton

We get a single solution for Ipk and Ton, here: 0.559 amps and 55.9uS, the off time will be 22.39uS, D/(1-D) = 2.5 required

D = 0.714 min for the required -25 Vout into 200 ohm. Higher to allow for losses.

the period = 78.29uS and the freq is therefore 17.155 kHz, the ave current in the L = 0.2795 amp

This will give the required power out, at lower loads the D needs to be reduced for voltage regulation. Make sure the L can handle the 560mA needed.

good luck !

 

[Calm down]

I feel that there is a problem with the usage of your MOS transistor. IRF640 is an N-channel semiconductor, and its G pin voltage must be at least one VTH value higher than the S pin, and it must last until the end of this driving pulse to fully conduct. In addition, the problem of reverse power supply to the battery through parasitic diodes of MOS during inductor discharge also needs to be avoided.

 

[Akanimo]

Hi,

Show your actual schematic so the problem can be tailored. For instance, your schematic is not showing the VS pin of the IR2110 being connected to the source of the MOSFET.

 

[Easy peasy]

@Akanimo - that connection cannot / should not be made as it will take the Vs pin 25V below ground - which the IC cannot tolerate - poof !

--- Updated Apr 25, 2025 ---

Using a P fet as the mosfet would perhaps be better, GD is then 0v ( on ) to 10V ( off ) - so the logic of the gate drive would need to be inverted, such that 72% ON time is 72% low, source on the left.

 

[Bruno.do]

[Message translated to English by moderator]

The VS Pin is connected to the MOSFET source. Otherwise, there seems to be a problem with the schematic. Seriously.

However, the circuit works to some extent. At 50% duty cycle, the output voltage reaches -10V, and both buck and boost operation operate in the range of about -3V to -19V.

I am still wondering about one issue.

When using the IR2110 in a high-side configuration with a single MOSFET, why does the output voltage stop increasing when the duty cycle reaches about 70%? It seems that the boost voltage is impossible beyond that point.

Note:
The inductor is rated for 3A.
The capacitor is rated for 50V.
I would appreciate any insight into why this limitation occurs and how to fix it.

 

[BradtheRad]

My own simulation shows you're operating near the limit of boost at 68% duty cycle. You must fully turn on the transistor in order to achieve 25 volts output. As stated in a reply it needs to be a P-device. Notice you have to apply bias in inverted manner from N-device.

The ON-resistance must be less than 1Ω. You may find a bjt PNP is easier to drive that low than a mosfet.

I used a 7k switching frequency in order to obtain 25V from a 1mH coil. If I raised the frequency (in simulation) then I could not reach 25V amplitude.

 

[Bruno.do]

So is there no way to solve this problem when using N-channel MOSFET?

When the switching frequency was changed to 7kHz, the duty cycle could reach 72% and the output voltage increased to -19.6V.

 

[kd502]

So is there no way to solve this problem when using N-channel MOSFET?

It is possible but requires -25V to turn off the MOSFET and Vdc1+7V to turn it on

 

[BradtheRad]

So is there no way to solve this problem when using N-channel MOSFET?

When the switching frequency was changed to 7kHz, the duty cycle could reach 72% and the output voltage increased to -19.6V.

Alternate arrangement of buck-boost converter. NPN at low side.
Load referenced to supply rail.

 

[Easy peasy]

Here is a better ckt:

and here are the results: ( exactly as calc above post #5, for timing etc ) 12.773kHz, Ton = 55.9uS, D=0.714 - note current scale.

note the high peak currents in the choke, 870mApk at start up, 720mApk at steady state full power, 440mA ave at -25V

note: if your source cannot provide these currents - you will never get full power at the correct D.

 

[kd502]

@Bruno.do​

Before you design anything you have to read datasheet and application notes.
From DT-97-3:

International Rectifier control ICs are guaranteed
completely immune to Vs undershoot of at least 5V,
measured with respect to COM. If undershoot ex-
ceeds this level, the high side output will tempo-
rarily latch in its current state. Provided Vs remains
within absolute maximum limits the IC will not suf-
fer damage, however the high-side output buffer will
not respond to input transitions while undershoot
persists beyond 5V.

When manufacturer says it should not work at all, you can't complain that is doesn't work as you expected.

 

[KlausST]

So is there no way to solve this problem when using N-channel MOSFET?

Why not using a dedicated switcher IC?
Or a flyback configuration?

Did you do research for your application?
I don´t see anything special, thus I guess it already is designed and used many many times.

I miss a clear definition of your working parameters. They are spread all over the thread ... some information is here others are there...

Klaus

 

[Akanimo]

Where are you getting the 15 V Vcc from?

 

[Easy peasy]

Where are you getting the 15 V Vcc from?

In this case from inside the sim.