How To Improve Boost DC/DC Converter?

[nicleo]

...and last - yes, you need a driver, or a bunch of hex inverters in parallel (I MEAN LOTS!) even if you use logic level mosfet. here's why.

Assume the gate of one of these big mosfets is about 2nF. We can use I=C*dv/dt to find out current drive needed.

For 50ns risetime for 0-5v step on the gate requires 200mA peak current, 10ns rise needs 1A peak current. This is way too much for the little PIC output inverter, and it is the reason I see many PIC projects have problems. Risetime slower than 50ns means you start to lose efficiency, so just keep it fast and don't worry about mosfet in resistive region.

Now that I think about it, maybe a bunch of hex inverters is OK. Output of PIC drives 3 inverters in parallel, and these guys drive 8 inverters in parallel that switch the gate. 8x at 50mA should be enough for 50nS risetime..

Interesting... I took an example from On Semiconductor as follows:

NTP60N06L
Power MOSFET
60 Amps, 60 Volts, Logic Level
Input Capacitance: 3075pF (maximum input capacitance)
Rise Time: 576ns (typical rise time)

I = C dv/dt = 3075pF x 5V/576ns = 0.027A = 27mA

In my opinion, when we talk about 'Logic Level' power mosfet, the 'Logic Level' means 'more' to the voltage level. For example, standard/conventional power mosfet requires Vgs = 10V (or more), whilst Logic Level power mosfet requires Vgs = 5V. In a circuit that has MCU and standard power mosfet, we need TWO types of power supply, ie. 5V to power the MCU circuit and 10V (or higher) to drive the standard power mosfet. However, in a circuit that has MCU and Logic Level power mosfet, we need only ONE power supply, i.e. 5V. Again, in my opinion, depend on the application (how many mosfets driven by MCU) and part number, Logic Level power mosfet could be interfaced directly to MCU (claimed by On Semiconductor).

 

[electronrancher]

500 nanoseconds risetime is not too good for switching power supplies. mosfet will spend 1 microsecond in heavy resistive region per cycle. even at 10khz this means mos will get very hot and die young. i strongly feel that 20mA of gate drive will make your switcher less efficient than it deserves.

logic-level mos basically means Vth less than 2.5-3v, so for a 5v or 3.3v logic output, it could eventually enhance the mosfet and turn on system power or something. it is not automatically guaranteeing you can use the mosfet for heavy current chopping like it sees in DC-DC.

i don't think you want to use any mosfet right at Vth anyway. 3xVth gives good enhancement, but there is better and better performance all the way up to gate breakdown voltage (or as close as you feel is good engineering judgement over process and temp). For a 5v Vcc, you need a mosfet with less than 2.5v Vth to really move any current efficiently.

take a look at this fairchild logic-level NMOS - this is a really nice power switching device if you want one.
https://www.fairchildsemi.com/pf/FD/FDB7030BL.html

It mentions it is specifically designed for raising efficiency in low-voltage DC-DC. Risetime is quoted as 12-22ns. I think a switch like this is a good choice for a boost up to 30v VOUT.

These switches are really nice for buck converters, as the body diode is already 30ns fast recovery... but should be a cool, quiet switch for boost as well.

 

[devonsc]

Thank you very much for all the detail explanation.

8 bits of resolution means smallest duty cycle step is 0.4%. Should be plenty. What you want to avoid is something like 5% step where in order to get 51% duty your converter must give two 50% pulses and one 55% pulse

Sorry, do you mind teaching me how do I calculate the the duty cycle step? As in how to I obtain 0.4% duty cycle step from 8 bits resolution?

I = C dv/dt

where C=input capacitance, dv=output voltage of my PIC, dt=rise time of the MOSFET.

Would like to reconfirm with you regarding the above calculation. As someone taught me to determine the drive current required to operate the MOSFET by using the following method.

T=1/f
then, I=Q/T

where f=frequency of the PWM, Q=gate charge of the MOSFET

The selected MOSFET for my converter (at this moment) = IRL2703
https://www.chipcatalog.com/IR/Datasheet/IRL2703.htm

By using both approach, I obtain a diffferent answer. For instance, if I were to use I=C*dv/dt. I will obtain an answer = 16mA.

While if I were to use the formula I posted, it will be 0.06mA

Need help, confused, please?

Added after 2 hours 22 minutes:

Hi, I would like to ask about the selection of inductor in the boost DC/DC converter. I found the following from a website but I don't quite understand what it means. As attached is a portion of it:

Do you guys mind to help me out here? Please? Thanks a lot in advance...

(sorry as i do not know how to make it appear on the page without downloading)

 

[electronrancher]

I was calculating the duty cycle step by dividing 100% by the number of LSB. For 8 bits/256 LSB, smallest duty cycle step is 100%/256-> .39%

As for mosfet current - we are both right! For any single turn-on, use I=Cdv/dt to find out how much driver current is needed for x risetime.

But this is a pulse of current, and nothing else happens until the next switch. Therefore average current when you look at many cycles is I=fQ like you said. I usually calculate power needed for switching (switching loss) as P=(f/2)CV^2 where f is frequency, C is gate capacitance, and V is gate drive voltage (for example 5v)

And as for your inductor calculation, you can estimate your average output current, say 1A. Now keep the ripple smaller or equal to 1/2 of this value to stay in CCM.
V=L(dI/dt) which transforms into equations 4.2 and 4.3 in your JPG - where V=VIN, dI = ripple current, and dt = Duty*tswitch at your estimated Duty cycle.

 

[devonsc]

Thanks a lot,.

I would like to post what I have at this moment. I’m just using a basic boost DC/DC converter. As shown is the specification list: (Guess it is too simple for you guys, hope you guys will not laugh at my basic DC/DC converter and I took so much time to get it done… )

Specification list: (Hope I didn’t miss out anything)

Solar Panel.
Vout = 7.5V
Iout = 33mA

MOSFET = IRL2703
Rds (on) = 0.06 ohm
V (threshold gate-voltage) = 1V to 2V
Input capacitance = 450pF
Rise time = 140ns
Drive current = 16mA (I’m driving the MOSFET from a PIC microcontroller, where the output from the microcontroller is approximately 25mA)

Diode = BAT48
Forward voltage = 0.5V
Peak reverse voltage = 40V

Inductor.
Through calculations, I obtained a value of approximately 34.67mH. Thus, I intend to get 50mH inductor.

Capacitor.
Through calculations, I obtained the value of approximately 355microFarad. Thus I intend to use 500microFarad if any, else I will get the 1000microFarad.

PWM signal.
Frequency = 15.625 kHz
Resolution = 8-bits
T = 64microseconds

I’m still reading and trying to understand how I get feedback from the circuit to my PIC microcontroller to track the maximum power. In the meantime, if you guys don’t mind, I would like to ask about => Can I do this:

First, I record the output voltage and the output current of the solar panel. With these values, I calculate the duty ratio needed base on the output voltage of the solar panel and form a look-up table for my PIC microcontroller. So, by tapping the output voltage from the solar panel, I would be able determine the appropriate duty ratio for step-up purposes. Meaning, I will change the duty ratio base on the output voltage of the solar panel. Is this considered as increasing the efficiency of the boost DC/DC converter? Will this work?

But, there are reading materials which mention that I should tap the power as a feedback to control the PWM signal. Is this necessary? I thought the output voltage of the solar panel varies with a wide range and good enough for a feedback to determine the duty ratio? Sorry if all I mentioned are nonsense.

 

[electronrancher]

simple? no, this will make a nice pwm. forgive the side dicussions on gate charging, etc - this thread is really about your pwm.

your component choice looks ok, the switch looks good. really the big question I have is the panel current, and the overall current in the system.. it seems low.

33mA, and 15mA ripple current is going to be tough to get good efficiency. doesn't a pic need 5mA or so at 20MHZ?

first, what is your expected output current? 20mA? Is that enough to trickle charge a lead acid? i never tried it, but it seems small. can you parallel several panels to get it up in the 100's of mA range? Since pic supply current is fixed, the higher the output current, the better your efficiency will be.

 

[E-design]

As for mosfet current - we are both right! For any single turn-on, use I=Cdv/dt to find out how much driver current is needed for x risetime.

Remember that for a given Vds a fixed amount of charge has to be transferred to the input capacitance to turn the device on. For this reason switching times can be reduced using a constant current source rather than a constant voltage source.

 

[devonsc]

33mA, and 15mA ripple current is going to be tough to get good efficiency. doesn't a pic need 5mA or so at 20MHZ?

Sorry, I don't quite get what you mean here. Mind explaining? Through simulation, I obtained an Output Current = approximately 33mA. With reference to certain reading material, through calculation, I obtained the Average Inductor Current as 30mA and again from these reading materials, I was adviced to assume the Ripple Inductor Current to be 0.2 of the Average Inductor Current. Thus, my Ripple Inductor Current is 0.6mA. Is this okay? Hopefully what I'm doing is right...

About my solar panel, according to the solar panel's datasheet, the short circuit current is approximately 160mA. As mentioned in the specification list above is the simulation result by using a 7.5V battery to source my DC/DC converter circuit and a 1000ohm resistor as my load. With these set-up, I obtain the following:

a.) Vout of battery = 7.5V
b.) Iout of battery = 33mA
c.) Vout of the boost DC/DC converter circuit = 15V
d.) Iout of the boost DC/DC converter circuit = 15mA

Do you mind giving some advice on how can I simulate close to practical if my simulation is all wrong? What resistance value should I use to represent my battery?

But by testing the solar panel, meaning, by plotting the I-V characteristic of the solar panel by using various resistor as load, I obtained the maximum power output as the following:

a.) Maximum output power = 222.56mW
b.) Voltage at maximum power output = 4.87V
c.) Current at maximum power output = 45.7mA

Are these readings good enough for me to run my DC/DC converter? Am I right to say that I'm more concern about the current readings as compare to the others if I were to use a solar panel to run my converter for charging purposes?

Help needed...it seems that I've been doing everything wrongly. Worry...please?

 

[andy1]

How about trying the synchronous DC/DC? Try good cap with low ESR at input and output.

 

[electronrancher]

hmmm. synchronous boost is not a trivial circuit. think about it, andy.

OK Devonsc - there is no need to worry. Your calculations look fine. I was just thinking you were working in a much higher current range (1-2A). You should be able to get good results with your setup.

I think it's time to build now. Why don't you check back in when you have something on PCB.

 

[devonsc]

hmmm. synchronous boost is not a trivial circuit. think about it, andy.

Would like to understand more: Do you mind explaining briefly on what do you mean by trivial circuit? Hope you don't mind....

By the way, I'm building it now, I've bought all my components and waiting for my ordered MOSFET to reach me by next week (from UK). In the meantime, I'm writing my PWM code, using a PIC microcontroller.

 

[devonsc]

Hi to genius-es,

Would like to ask for you guys' opinion on what are the matters that I should or could consider in my basic DC/DC converter circuit to impress my lecturer Sorry if I am not allow to ask such questions

Added after 22 seconds:

Hi to genius-es,

Would like to ask for you guys' opinion on what are the matters that I should or could consider in my basic DC/DC converter circuit to impress my lecturer Sorry if I am not allow to ask such questions As it seems to be very "plain".

 

[devonsc]

Hi there!

I'm confused on certain matters regarding the selection of components required in my DC/DC converter. Regarding the selection of inductor needed, with reference to certain application notes found on national.com's website, one of the paramaters that I must first know would be the output current of the circuit. However, without contructing the circuitry, will I be able to know the output current of the DC/DC converter? Can I make such assumption? =>

Assuming I intend to boost a level of input voltage of 7.5V to a level of 15V. Meaning, a duty ratio of 50%. Then, with reference to the input voltage, I assume that my current drops 50% due to the step-up ratio of 50%. Can I make such assumptions? Advice needed.

Added after 1 minutes:

By the way, this is where I refer from:

**broken link removed**

 

[nicleo]

I'm confused on certain matters regarding the selection of components required in my DC/DC converter. Regarding the selection of inductor needed, with reference to certain application notes found on national.com's website, one of the paramaters that I must first know would be the output current of the circuit. However, without contructing the circuitry, will I be able to know the output current of the DC/DC converter? Can I make such assumption? =>

Assuming I intend to boost a level of input voltage of 7.5V to a level of 15V. Meaning, a duty ratio of 50%. Then, with reference to the input voltage, I assume that my current drops 50% due to the step-up ratio of 50%. Can I make such assumptions? Advice needed.

When we design a power source (e.g. DC/DC converter), we usually know what will be the maximum load current in our application. Once we know the maximum load current, we can choose the correct rating for inductor or switches used in the DC/DC converter. Should not be a problem, I think.

Let's say the DC/DC converter is an IDEAL converter, meaning efficiency is 100%. So, if the maximum input power is 75W (i.e. 7.5V and 10A), the maximum output power that you can get is also 75W. If the output voltage of the DC/DC converter is 15V, then the maximum (continuous) current that you can draw from it will be 75/15 = 5A.

 

[VVV]

For low power DC-DC converters the quiescent currents of the parts involved are usually what hurts the efficiency. Also, the switching losses become a problem, which is why you want a low frequency operation or other techniques, which, in fact, produce an equivalent low frequency.
Personally I would build the boost using a real power supply controller, which offers other features, such as current limit.
Having said that, I will go back to your question.

The issue with the switching losses can be improved by switching the MOSFET faster. Many people misunderstand the PIC's "high current capability". That is just a measure of the internal channel resistance of the transistors. Hence 5V/25mA=200ohm. That is it. So basically you are driving the MOSFET with a gate resistance of about 200ohms (PIC running at 5V). With these in mind we’ll do an analysis of the losses in your converter.

The switching time of the MOSFET is best calculated from the gate charge graph, a picture that looks like the one shown in the first half of my drawing.

The area below the horizontal "plateau" in the middle is called the Ogd. The MOSFET switches on when that charge has been transferred to its gate. But the charge is transferred through the gate resistor and, as you can see, the gate voltage is not 0V, but Vth. Therefore, the current you actually supply to the gate is (Vmax-Vth)/R. So it's not 25mA, as you might think, but much less. Since your PIC is runnig at 5V, Vmax is 5V. Vth is about 3.5V for your MOSFET. That value has to be adjusted for your actual MOSFET current. Vth will be closer to 2.5V at your operating current, and switching times about equal.
So the gate current is:
Ig=(5-2.5)/200=12.5mA
Now, Qgd is from the table 9.3nC max. So your maximum switching time will be:
tsw=Qgd/Ig=0.75us

This is a fairly long time, but assuming a switching frequency of 15.625kHz (T=64us), your switching losses will be (assuming CCM operation):
Psw=Vout*Iout*tsw/T=2.6 mW.
This is about 1.1% of your total output power.

There will be losses associated with driving the gate, given by the total gate charge, Qg:
Pgate=Qg*Vmax/T=1.17mW (Vmax is 5V, since PIC is running at 5V)

Duty cycle at min input voltage and max output power is:
D=1-Vin/(Vout+Vdiode)=0.675

The MOSFET RMS current will be approximately (under CCM conditions):
Irms=Iout*sqrt(D)=12.3mA
MOSFET conduction losses, approximately (under CCM conditions):
Pcond=Rdson*Irms^2=0.009mW very low, which is normal for this low power converter.
Losses associated with Coss (250pF at 15V):
Pc=0.5*Coss*Vout^2/T=0.44mW

Diode conduction losses (under CCM conditions, max output power/ min input voltage):
Pdiode=Vdiode*Iout*(1-D)=1.95mW .It will get worse at higher input voltages, because D decreases. However, I assumed that at maximum output power your input voltage will always be at its minimum, since you are running from a solar panel. Vdiode is about 0.4V.

There will be losses associated with the inductor. I will assume a resistance of 3 ohms for the inductor, since for CCM you need a large inductance, which means higher resistance, too.
Pind=Rind*Iout^2=0.67mW
For the PIC, the quiescent current could be about 2mA @5V, that is Ppic=10mW. But if the PIC is running from a linear regulator you have to consider those losses, too.
So, now we have: Ploss=Psw+Pcond+Pgate+Pc+Pind+Pdiode+Ppic=18.04mW

Efficiency is Eta=Pout/(Pout+Ploss)=92.6% (assuming no other loss for PIC regulator)

As you can see, you can only reduce the losses associated with the MOSFET, by using, for example, the NTR4503 from ON Semi or the TN0201K from Vishay (Siliconix), which have much lower gate charges and Coss. Or you can use a gate driver, like the 2 transistors, which allow you a much lower resistor, say 33 ohm. But the big player remains the PIC quiescent current. To reduce diode losses you would need a synchronous rectifier, but that one will also have gate drive losses, so the gains may not be too great.

As for your approach, you will have to use continuous current mode (CCM), because that will give you better regulation, although not good regulation. With discontinuous current your output voltage would depend too heavily on the output current. The best thing is to sense the output voltage and change the duty cycle.
In either case, your regulator has no MOSFET protection.

There can be a change in efficiency as the input voltage changes, but it will always be less than 100%. Physics...

 

[devonsc]

Hi, initially I did my calculations by considering the worst case. I did my calculations using an output current of 15mA of my circuit Is this way too low?

But I am supplying my DC/DC converter circuitry through a solar panel, according to the datasheet of the panel, it has a load current of 150mA with 7.5V load voltage. However, after testing the panel (I'm not sure if I've dont the right thing) I only obtain a value of approximately 50mA load current. Does it mean that if I were to boost it with a duty ratio of 50%, I will only obtain an output current of 25mA from my converter circuit?

This is how I test the panel, mind to correct me if I am wrong? Thanks in advance...

I test the panel by first getting the short circuit current reading and open votage reading. Then, I apply a different load to the solar panel to obtain a set of readings with different loads (different values of resistor). With these set of value, I plot an I-V characteristic graph and select the maximum power point through the biggest rectangle drawn. Do you want me to post the set of readings here?

 

[VVV]

Efficiency deals with powers. So if you boost the voltage to twice the input you get half the current; that is, the power is almost constant (minus the losses).

If the rating of the solar panel are 150mA at 7.5V and you get something way different, then maybe the panel is defective. Or, you may have obstructed one of the cells. No cell should be obstructed, otherwise its internal resistance appears to increase and the entire panel's performance is affected.

Testing it with different loads is OK, but I would not short it, I woiuld just stay within the manufacturer's ratings. Then extend the characteristic by extrapolation.

 

[nicleo]

But I am supplying my DC/DC converter circuitry through a solar panel, according to the datasheet of the panel, it has a load current of 150mA with 7.5V load voltage. However, after testing the panel (I'm not sure if I've dont the right thing) I only obtain a value of approximately 50mA load current.

Are you sure that the solar panel is exposed to the correct intensity of light required to produce rated voltage and current? In my opinion, you should connect a 50ohm resistor in series with output terminals of the solar panel. Remember, 0.15A x 0.15A x 50ohm = 1.125W, so the resistor that you choose must at least be able to withstand 1.125W. If the load current that you measure is 50mA, then I think the voltage across the output terminals of the solar panel will not be at 7.5V anymore. Also, do remember that the ambient temperature will affect the load current too. Some papers mention that if the ambient temperature increases, the rated load current will drop. Pls update your finding. Thanks.

 

[devonsc]

Hi nicleo,

I'm using 5W resistors for testing (just in case). Will this affect my readings? Hope no. At a load of 47ohm, I obtain 90+mA but the voltage level is very low, approximately 4V. If I were to boost this voltage level to charge a 12V battery, am I right to say that the output current will end up be approximately 25mA to 30mA?

By the way, I have a dumb question (sorry, don't get angry). By using lesser components in a circuitry, isn't it consuming less power, which eventually reduces power dissipation and increases efficiency? If this is true, why do we have very complicated circuits which eventually increases the efficiency of a circuit? Am I right to say that complicated cicuits, which uses a lot more components that increases the efficiency of a circuit actually required more power to be supplied to it.

Hope my thoughts are not nonsense

 

[VVV]

As I mentioned in a previous post, I did not think the approach was the best.
You may want to consider the MCP1630 to couple to your PIC to get a real charger going.

As for the simplicity of the circuits, it is true that more compliccated circuits can draw more power, although this is not really the case with modern IC's. But they offer a whole lot of other benefits, not achivable with simple circuits: protections, improved algorithms, adequate drive, etc. With all that, it's worth sacrificing a little the efficiency.

Many times, however, the efficiency is actually improved. Take your circuit, for example: why do you need a complicated switching regulator? Well, you need a voltage higher than the input.
Why not use two solar panels in series, then? Well, you would still have to regulate the voltage or current.
Why not use a linear regulator to do that? You could, but the efficiency would be lower.

So, more complex circuits usually happen for a reason: because the simple circuit approach has other shortcomings. You will always want to use the simplest circuit which satisfies the requirements.