# Buck-Boost Converter - Application

1. ## Buck-Boost Converter - Application

Hi there, I have real limited knowledge regarding electronics, hope you guys don't mind helping me here.

I've read about how does the flow of current in this circuit, the Buck-Boost Converter. But still don't quite understand. Do you guys mind explaining how do we control the circuit to work as a Buck and on the otherhand work as a Boost? Is there away of controlling? As in, having it switch to perform a buck converter and having it switch to perform as a boost whenever it is needed.

Apart from that, if this is a circuit where we couldn't control the Buck or Boost operation of it, do you guys mind suggesting what is the purpose of contructing such circuit?

It seems that, it makes more sense if one can control and determine whether he/she wants the circuit to operate under Buck or Boost :) Is this right? Sorry if I'm talking nonsense :D

I'm trying to a buck as well boost, depending on my input of the converter circuit, is there a way where I can implement this circuit or I have to develop two circuits: a buck converter and a boost converter separately.

Thank you very very much in advance...

2. ## Re: Buck-Boost Converter - Application

Originally Posted by devonsc
I've read about how does the flow of current in this circuit, the Buck-Boost Converter. But still don't quite understand. Do you guys mind explaining how do we control the circuit to work as a Buck and on the otherhand work as a Boost? Is there away of controlling? As in, having it switch to perform a buck converter and having it switch to perform as a boost whenever it is needed.
Based on the description in the attached figure, you can switch between buck and boost operation by controlling the duty cycle (D) of the PWM.

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3. ## Buck-Boost Converter - Component selection

Hi,

Thanks Nicleo.

Hope you don't mind guiding me again :)

a.) Regarding the component selection of the buck-boost converter, can I implement the same method I used for a boost DC/DC converter?

b.) In this buck-boost converter, does it mean that the positive pin of the capacitor is being grounded?

c.) Is it true that I will reduce even more ripples with another ceramic capacitor in parallel with the electrolytic capacitor?

d.) Is it true that the output voltage of the buck-boost converter will be in negative? I'm trying to develop this circuit for charging sealed lead acid battery purposes, does it mean that I can apply the output negative voltage of the converter to the negative terminal of the battery? Can charging be made in this manner?

e.) Sorry, i understand that when a boost converter step-up the DC input voltage, the output current simply reduces, is this right? But as for a buck, when the DC input voltage is being step-down, does it mean that my output current of the converter will be increased?

4. ## Re: Buck-Boost Converter - Component selection

Originally Posted by devonsc
a.) Regarding the component selection of the buck-boost converter, can I implement the same method I used for a boost DC/DC converter?
a) Inductor Current Rating for a Buck-Boost Converter at http://powerelectronics.com/mag/powe...urrent_rating/

5. ## Re: Buck-Boost Converter - Application

d.) Is it true that the output voltage of the buck-boost converter will be in negative? I'm trying to develop this circuit for charging sealed lead acid battery purposes, does it mean that I can apply the output negative voltage of the converter to the negative terminal of the battery? Can charging be made in this manner?
Hi, guess I've understand the matters I asked except for the mentioned above. Can I charge the battery by using a negative output voltage from the converter being apply to the negative terminal of the battery?

6. ## Re: Buck-Boost Converter - Application

Originally Posted by devonsc
... Can I charge the battery by using a negative output voltage from the converter being apply to the negative terminal of the battery?
The negative pin of the ouput capacitor is more 'negative' compared to the positive pin of the output capacitor. In order to push the charging current through the battery, you should connect as follows:

positive pin of output capacitor -> positive terminal of battery
negative pin of output capacitor -> negative terminal of battery

7. ## Re: Buck-Boost Converter - Application

devonsc:

The key to understand switching power supply is to understand how an inductor works. Try to visualize how the electric field and magnetic field interact with each other inside an inductor. Once you understand how an inductor works, then move on to study the transformer.

I will take awhile but once you get it, everything else will fall into place.

8. ## Re: Buck-Boost Converter - Application

I've constructed the buck-boost converter (as shown in this post) and tried to simulate but I don't get any result, as in no readings at all at the output.

Then, constructing the circuit and test in practical, I still don't get any readings at the output. please, please, really begging for help....please?

This is what I did in practical, mind to comment? I include a 1kohm resistor as the load, then everything is exactly the same as the circuit posted in this post (as shown above). Please? Seriously, panic...please help? I owe you guys...please?

Hi there, something real embarrassed to ask: real panic, seriously. Is the following pin assignment labels correct? I labeled those pins circled in red, with reference to the figure on its left...

I tried locating the appropriate pins by using the "diode measurement mode" on the multimeter. Then, I connect in such a way where pin 'S' is to the ground whereas pin "D" to the inductor end junction. Is this right? Seriously, please help? Please?

9. ## Buck-Boost Converter - Application

It should be a N-channel power mosfet. The labels are correctly assigned. The Drain (D) is connected to one of the inductor ends, Source (S) is connected to ground/common. Do you mind to tell how did you drive the gate of the power mosfet?

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10. ## Re: Buck-Boost Converter - Application

Glad to hear you're with me, thanks. I guess I will post everything that I've done:

As listed below are the values of the components:
a.) inductor = 50mH
d.) MOSFET - NPN (Logic MOSFET - IRL2703)

To drive the mosfet, I'm using a function generator and provide a 15kHz pulse with a peak-to-peak of 5V to the base (B), then pin 'S' to the ground and pin 'D' to the inductor end. I intend to drive the MOSFET through a PIC microcontroller, but it has yet ready at the moment.

I tried to measure the DC voltage across the inductor and I obtain a constant voltage of the exact same voltage value from my input to the circuit, I repeat applying voltage to the circuit by slowly increasing the input voltage from 2V till 5V. I didn't go any higher. But theres no output. I tried to measure and observe if there is any output from my 'S' pin of the MOSFET but no output. It is a logic MOSFET, I thought I will only need approximately 1V to 2V to drive the MOSFET?

Apart from that, does it mean that I'm leaving the MOSFET in the on state if I were to connect the base (B) of the MOSFET to a fix 5V supply? I guess I didn't burn anything as all components were not hot at all.

But, if the connection of my MOSFET is right, how come I obtain no output at all in simulation? Worry here.....very worry

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11. ## Buck-Boost Converter - Application

i = C dv/dt
Rg is required to limit dv/dt. Without the gate resistor (Rg), the rising time of the input signal could be very short.

For example,
i = 450pF 5V/dt = 2.25xE-9 / dt

If the dt is 10ns, then
i = 2.25x10-9 / 10x10-9 = 0.225A

If the dt is 1ns, then
i = 2.25A

I think the 5V of function generator is not 'powerful' enough to drive the mosfet. The driving signal must be able to supply enough charging current to turn on the power mosfet. Also, it's advised that you put a Rg in series with the gate of the power mosfet.

Besides you have to remember that in order to turn on the power mosfet, the Vgs (not Vg) should be > than the gate threshold voltage (Vgs(th)). Function generator usually have two output terminals. How did you connect both output terminals to your circuit or the gate of the power mosfet?

By the way, what was the VDS (pls refer to the datasheet of the power mosfet) when you tested the circuit?

12. ## Re: Buck-Boost Converter - Application

I think the 5V of function generator is not 'powerful' enough to drive the mosfet.
But I'm using a logic MOSFET, thought it will only require approximately 1V to 2V to drive it?

The driving signal should be able to supply enough charging gate current. Also, it's advised that you put a Rg in series with the gate of the power mosfet.
Thanks, mind to talk a little about the need to do this? Is it somekind of safety precaution to avoid spoiling the "internal diode"?

Besides you have to remember that in order to turn on the power mosfet, the Vgs (not Vg) should be > than the gate threshold voltage (Vgs(th)).
I tried measuring by these two pins but zero, whereas measuring voltage across pin D and pin S, I obtain the input voltage value...

Function generator usually have two output terminals. How did you connect both output terminals to your circuit or the gate of the power mosfet?
From the way you tell me, I guess I've done something wrong :( I connect the positive output to the pin B of the MOSFET and the other I ground it. Is this wrong? Sorry...

By the way, what was the VDS (pls refer to the datasheet of the power mosfet) when you tested the circuit?
Do you mean Vds as in, the diode forward voltage? It is 1.3V...

13. ## Buck-Boost Converter - Application

I remember something about signal generator. The one in my lab has output impedence of 600ohm. This output impedance will result in voltage drop and therefore the voltage at the gate might not be 5V as you expected.

14. ## Re: Buck-Boost Converter - Application

I remember something about signal generator. The one in my lab has output impedence of 600ohm. This output impedance will result in voltage drop and therefore the voltage at the gate might not be 5V as you expected.
As for the one in my lab, it has the output of 50ohm impedance, hope I'm right, as this is what I read from the label of the output...

15. ## Re: Buck-Boost Converter - Application

its written 50 ohms but it may not be 50 ohms.

devonsc you are right about the voltage required to turn on the MOSFET. the datasheet says that the threshold gate-source voltage Vgs(th) is 1V so anything above that will surely turn on the MOSFET.

the Vds is the drain to source voltage which depends on the drain to source resistance when the MOSFET is ON. the datasheet says that the Rds(on) is 0.04 ohms for a Vgs of 10V @ Id= 14A and it is 0.60 ohms for a Vgs of 4.5V @ Id= 12A. so as you can see that Rds(on) depends on both the Vgs and Id so Vds will also depend on this. the more the Rds(on) the more the voltage will be dropped across the MOSFET and the more the power will be dissipated in it. so the lower the Rds(on) the better

16. ## Re: Buck-Boost Converter - Application

Thanks...

Do you guys mind to comment about this? I tried to turn on the mosfet by doing the following. With the circuit as shown constructed, I measure the voltage drop across the Drain and Source and I obtain 5V. But I'm not sure if Im doing the right thing as even though I remove 5V to the Gate, I still obtain a reading of 5V if I were to tap my multimeter on the Drain and Source...

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17. ## Re: Buck-Boost Converter - Application

I explained in a reply to your other post, about the DC-DC converter that you swapped the gate and the drain of the MOSFET: the drain is in the middle.

Now a few words about the so-called buck-boost:
the name is confusing, I prefer inverting regulator, to avoid confusion with another type of circuit that really works to produce a constant output, whether the input is above or below this desired output.

Essentially, your buck-boost (we'll stick to this name) works just like the boost, with some differences:

1. the entire energy required for one switching cycle is stored in the inductor. In the case of the boost, during the energy transfer phase, part of the energy comes directly from the input power, since the inductor is effectively in series with the input voltage: E=Vin*Iout*Tsw, so the inductor only needs to store:
Eind=(Vout-Vin)*Iout*Tsw. For the buck-boost you need to store:
Eind=Vout*Iout*Tsw
This means that the inductor for the buck-boost can be larger physically. Other than that, it is selected the same way as for the boost, based on the higher energy it must handle.

2. the output is negative with respect to ground. Yes, the capacitor has the + connected to the ground.

3. the transistor is now on the "high side", it has either the source or the drain connected to the input power, so you will need a "high-side driver".

18. ## Re: Buck-Boost Converter - Application

Oh no! Thanks, VVV :) By the way, I tried to obtain the correct configuration of the pin, by searching for the internal diode in the MOSFET. Should I do this? This is how I obtain the configuration of the pin I posted (circled in red). And I thought the Gate is actually "together" with the casing, thus, when I check the connectivity of the casing and the middle pin, it is a short. This is another reason why and how I obtain the pin configurations. Just to voice out how I do it, mind correcting me? Thanks....

1. the entire energy required for one switching cycle is stored in the inductor. In the case of the boost, during the energy transfer phase, part of the energy comes directly from the input power, since the inductor is effectively in series with the input voltage: E=Vin*Iout*Tsw, so the inductor only needs to store:
Eind=(Vout-Vin)*Iout*Tsw. For the buck-boost you need to store:
Eind=Vout*Iout*Tsw
I'm lost, sorry, blur here. I calculated and I obtained a very small inductance for the buck-boost inductor. Based on the formula given in the site:
http://schmidt-walter.fbe.fh-darmsta..._e/smps_e.html

This means that the inductor for the buck-boost can be larger physically. Other than that, it is selected the same way as for the boost, based on the higher energy it must handle.

3. the transistor is now on the "high side", it has either the source or the drain connected to the input power, so you will need a "high-side driver".
I'm using the Drain being connected to the input power of the buck-boost circuit and Source to the Inductor side/junction. I guess I shall stick to the following circuit, any comments? Is my MOSFET (IRL2703, a NPN) suitable? By the way, do you mind to briefly explain a little about "high-side driver"? :( Sorry...

19. ## Re: Buck-Boost Converter - Application

Looking for the diode is tricky. Because the G-S capacitance can be charged (by you, trying with the ohmmeter) and so you will see a "short" between drain and source. So probably the first thing to do is to find the gate, since it is supposed to be isolated from everything. Then find the diode. If it appears shorted, try reapplying voltage to gate with respect to the other pins (with different polarity) until the "short" disappears, then find the diode. But how do you know if it's an N-channel or P-channel? And I am against handling the MOSFET too much.
The best way is to look at the datahseet. For MOSFETS, it looks like all manufacturers came to an agreement so the pinout is the same.

For TO-220 the drain is the middle pin and it is connected to the tab, too.
With the tab at the top and the plastic facing you, the gate is to the left, the source to the right. This applies to all To-220 MOSFET's that I am aware, either N-channel or P-channel.

Yes, the inductance is smaller, but sometimes, to store the energy, you MAY need a larger core (so it could be physically larger, but this is NOT a rule).

A high-side driver is meant to control a transistor that does not have the source connected to ground. Like in your case.
Transformers can be used, but these have limitations on the maximum duty-cycle.
Specialized high-side driversexist, that use either floating power supplies or some kind of boostrapping techique, where a capacitor is periodically charged from a ground-referenced power supply and then acts as a floating power supply. All this business of floating power supply is needed because the source "moves" with respect to ground, but you need to supply the MOSFET with a voltage between its gate and source, so you power must follow the source.
An example of high-side driver is the IR20153S from International Rectifier: http://www.irf.com/product-info/data...a/ir20153s.pdf
Notice the diode from Vcc to Vb (bootstrap); this diode charges up the capacitor when the source of the MOSFET is at ground potential (when the diode in a buck conducts). When the MOSFET is turned on and the source is at a high voltage, the capacitor maintains almost constant a voltage with respect to the source (one pin is connected to the source). so now the diode is off, since the top of the cap is at a voltage V=Vs+Vcap, where Vs is the voltage of the source of the MOSFET and it can be 150V and Vcap is approx Vcc. So the diode has to handle at least 150V.