# Queries related to buck converters

• Wrichik Basu
Gold Member
2022 Award
In a circuit that I am building, I have a rectified DC of about 22V, and I want to regulate this down to say, around 8V to 10V.

Previously, I thought of using a voltage regulator, LM7808, as discussed here, but I strongly believe that a potential drop of 22V - 10V = 12V would produce a lot of heat in it.

Being new in this field, I recently found that there are buck converters that overcome problems that one faces while using a voltage regulator. I intend to use this one.

A few queries:

1. Did I choose a proper buck converter (given that I have checked that the current ratings are compatible with my circuit)?

2. Say I adjust the potentiometer such that it will provide 10V output when input is 22V. Now, my input drops to 15V. Will the output drop as well, or will it remain constant at 10V?

3. Will the buck converter give 0V output when the input is below the defined output (10V here)?

Gold Member
You can learn more about the buck regulator circuit from the IC datasheet here. However, the other components in the circuit (primarily the inductor) also matter and are not described in the link you provided.
As you said, if the current requirements are compatible, then:
1) Probably, but it can be complicated. Your voltage ranges are ok.
2) The circuit will adjust for input voltage changes (as best it can) to keep the output voltage where you set it.
3) No, it will put out as much voltage as it can, almost the same voltage as the input.

Mentor
Being new in this field, I recently found that there are buck converters that overcome problems that one faces while using a voltage regulator. I intend to use this one.

A few queries:

1. Did I choose a proper buck converter (given that I have checked that the current ratings are compatible with my circuit)?

2. Say I adjust the potentiometer such that it will provide 10V output when input is 22V. Now, my input drops to 15V. Will the output drop as well, or will it remain constant at 10V?

3. Will the buck converter give 0V output when the input is below the defined output (10V here)?
For those Simple Switcher designs, National has design calculators at their website that let you chose the external components based on your input voltage range, desired output voltage, output current, etc. If you don't check your components with such a calculator (or do the calculations yourself after you have more experience), you can run into issues with "line regulation" where the IC goes into duty cycle limiting, or cycle skipping, or other modes that can affect your output "load" regulation.

Take a look at the calculators at their website for the Simple Switchers, and play around with the design values to be sure that the input voltage range you have will work the way you want it to.

Gold Member
2022 Award
For those Simple Switcher designs, National has design calculators at their website that let you chose the external components based on your input voltage range, desired output voltage, output current, etc. If you don't check your components with such a calculator (or do the calculations yourself after you have more experience), you can run into issues with "line regulation" where the IC goes into duty cycle limiting, or cycle skipping, or other modes that can affect your output "load" regulation.

Take a look at the calculators at their website for the Simple Switchers, and play around with the design values to be sure that the input voltage range you have will work the way you want it to.
Can you provide a link to the site?

Gold Member
2022 Award
You can learn more about the buck regulator circuit from the IC datasheet here. However, the other components in the circuit (primarily the inductor) also matter and are not described in the link you provided.
I've seen that datasheet. Since the IC is embedded in the module, and some capacitors and diodes are visible, do you think I have to add them externally once again?

Last edited:
That buck regulator will produce a stable regulated output so long as the input voltage is at least 2.75V above the set output voltage. You should adjust the voltage before connecting it to your circuit.

For example; If you set Vout to 10V then the input must stay above 12.75V to keep regulating. If the input voltage falls below 12.75V the regulated output voltage will begin to fall. It does not fall instantly to zero, but does it's best to keep working with a lower output voltage.

The complete assembled circuit board shown costs about US$2 from China. You only need to adjust the output voltage, then connect the input and output wires to use it. The text with that Calcutta Electronics advert was confused in translation so is very confusing to read. If you have an input voltage that can fall below the output voltage plus +2.75V you should consider a buck/boost regulator topology, it costs slightly more, say about US$5.

Also, your rectified DC of 22V is probably peak voltage without load.
The voltage ripple when drawing 2 amps must not fall below 12.75V.
That gives dV = 22V – 12.75V = 9.25V ripple voltage.

The minimum storage capacitance needed on the full wave rectified DC will be C = Q / V, where Q = I * t.
Half a power cycle is about t=10ms, I=2A, V=9.
So; Cmin = I * t / V = 2 * 10e-3 / 9 = 2200 uF.

Wrichik Basu
Gold Member
2022 Award
The minimum storage capacitance needed on the full wave rectified DC will be C = Q / V, where Q = I * t.
Half a power cycle is about t=10ms, I=2A, V=9.
So; Cmin = I * t / V = 2 * 10e-3 / 9 = 2200 uF.
I have to add this capacitor after the rectifier bridge, right (to get a peak voltage without ripples)?

I have to add this capacitor after the rectifier bridge, right (to get a peak voltage without ripples)?
There can never be zero ripple without infinite capacitance or zero current.

You could use 2200uF. It would depend on the transformer and diode current rating, but I would use two 2200uF in parallel or 4700uF. This is a bit of guessing because you have not given us the maximum regulated output current, the mains frequency, or the transformer unloaded AC voltage.
More capacitance results in less ripple. Less ripple makes a higher average rectified voltage.

A linear regulator with 2 amp output requires 2 amp input. But for a 100% efficient DC-DC converter, if the output current was 2 amp at 10 volts, the input current from 20 volts would be only 1 amp. 2A * 10V = 20 watt = 1A * 20V. So there is less rectified current taken from the storage capacitor by the DC-DC converter and so less storage capacitance is essential.

Gold Member
2022 Award
There can never be zero ripple without infinite capacitance or zero current.

You could use 2200uF. It would depend on the transformer and diode current rating, but I would use two 2200uF in parallel or 4700uF. This is a bit of guessing because you have not given us the maximum regulated output current, the mains frequency, or the transformer unloaded AC voltage.
More capacitance results in less ripple. Less ripple makes a higher average rectified voltage.

A linear regulator with 2 amp output requires 2 amp input. But for a 100% efficient DC-DC converter, if the output current was 2 amp at 10 volts, the input current from 20 volts would be only 1 amp. 2A * 10V = 20 watt = 1A * 20V. So there is less rectified current taken from the storage capacitor by the DC-DC converter and so less storage capacitance is essential.
In India, we get 50Hz 220V RMS mains. As per my current plans, the transformer is a 12-0-12 230V to 12V one. The current in my circuit is not supposed to reach 1A, otherwise there will be problems. If I take an idealistic case and use all the digital and analog I/O pins, each drawing about 30mA, the current would be around 600mA. So, it will not cross 1A under any circumstance.

After stepping down the voltage, I will use a full wave rectifier bridge (with 1N5408 diodes), and then use a capacitor to get peak voltage. This peak voltage will be 22V maximum and 15V minimum. I was hoping to use a 6800μF capacitor, the largest I have at hand, to get peak voltage.

A part of this peak voltage will be regulated down to 10V or so and supplied to an Arduino. Previously I wanted to regulate using a voltage regulator, but on learning how they misbehave on heating up, I changed to a buck converter.

The other part of the peak voltage will first pass through a voltage divider, and then to the A0 pin on arduino as input.

Last edited:
Mentor
In India, we get 50Hz 220V RMS mains. As per my current plans, the transformer is a 12-0-12 230V to 12V one. The current in my circuit is not supposed to reach 1A, otherwise there will be problems. If I take an idealistic case and use all the digital and analog I/O pins, each drawing about 30mA, the current would be around 600mA. So, it will not cross 1A under any circumstance.

After stepping down the voltage, I will use a full wave rectifier bridge (with 1N5408 diodes), and then use a capacitor to get peak voltage. This peak voltage will be 22V maximum and 15V minimum.
If you're using a bridge rectifier that implies that you won't be using the center tap of your transformer secondary and that you'll be getting the full 12 + 12 = 24 volts rms of the transformer. That translates into rectified peak of about 40 V (less two diode drops).

If you use the center tap and full-wave rectify with two diodes then you'll get about 17 V (less one diode drop).

#### Attachments

1.4 KB · Views: 378
Baluncore
Gold Member
2022 Award
If you're using a bridge rectifier that implies that you won't be using the center tap of your transformer secondary and that you'll be getting the full 12 + 12 = 24 volts rms of the transformer. That translates into rectified peak of about 40 V (less two diode drops).
I have arranged for a transformer that doesn't have a centre tap. So, I suppose I will get a 12v rms now.

Transformer is specified under a load current, which we do not know.
So the 22V you measure unloaded is unrealistic.

12V RMS rectified will give 12 * √2 = 17.0V peak.
The 1N5408 will only conduct for a short time up to Vpeak, so allow 1.1V drop across each conducting diode. Subtract 2.2V for the two conducting bridge diodes making 14.8V.

Dropout voltage of 8V DC-DC converter is 10.75V
That leaves only 4 volts for ripple.
50 Hz = 20msec per cycle = 10ms per half cycle.
Current from reservoir capacitor is a maximum of 1 amp.
C = 1 amp * 10 ms / 4 volt = 2500 uF minimum.
So 4700uF or 6800uF will be OK.

1. A centre tapped 12-0-12 transformer needs only two diodes to full-wave rectify. So less voltage is lost than with a four diode bridge. 2.2V lost becomes only 1.1V lost. Ripple of 4V can be bigger = 5V and so Cmin = 1A * 10 ms / 5V = 2000uF minimum. So there is an advantage in using a centre tapped secondary.

2. Why do you divide rectified reservoir DC and supply it to Arduino pin A0. Is that AtoD input?

Gold Member
2022 Award
1. A centre tapped 12-0-12 transformer needs only two diodes to full-wave rectify. So less voltage is lost than with a four diode bridge. 2.2V lost becomes only 1.1V lost. Ripple of 4V can be bigger = 5V and so Cmin = 1A * 10 ms / 5V = 2000uF minimum. So there is an advantage in using a centre tapped secondary.
Will keep that in mind.
2. Why do you divide rectified reservoir DC and supply it to Arduino pin A0. Is that AtoD input?
Ideally, in my circuit, after a voltage divider, I will put it into the Arduino and measure the voltage (1023 bits correspond to 5V, then using unitary method...). Then I will control an SSR using the Arduino based on the input DC voltage.

Gold Member
Dearly Missed

#### Attachments

21.3 KB · Views: 395
Wrichik Basu
Simulation of full wave diode rectifiers driving switching DC-DC converters.
Notes;
Design here is for a supply to a switching regulator with 10 volt output, 1 amp load. That is 10 watt. The reservoir cap has been reduced to 1000uF to show increase ripple and variation in regulator input current.

The regulator input is modeled as a voltage controlled current, I = 10 watt / Vout. That model will not start because an initial infinite current is needed. A simulation work-around is to either set a limit such as I = Min( 10amp, 10watt / Vout ), or to start the supply at 90° phase so the secondary AC peak voltage is present initially.

Secondary no load voltage is set to Vpeak = 22V. The winding resistance reduces that to 12V RMS under full load. The same weight of copper is used in the transformer secondaries. Consider the same two secondary windings, connected in parallel or series. The resistance of the windings is set to 0.5 ohm each for the two diode simulation. In parallel that makes 0.25 ohm for the bridge simulation.

Conclusions:
The diode pair has lower peak diode current than the bridge because it takes more voltage and so more time to turn on the two series diodes in the bridge. As expected, the output voltage of the pair is also higher than the bridge by one series diode drop. Increasing the reservoir capacitance increases the peak diode current, reduces the conduction angle and makes for a worse PF.

The diode pair beats the bridge in energy efficiency. The only advantage of using a diode bridge appears to be that you have less secondary terminals on the transformer which weighs the same.

If the load (arduino) has an on board linear regulator, set the switching regulator output to the minimum acceptable voltage for the load. That will reduce heating of the linear load regulator and reduce the supply current from the reservoir to the switching regulator.

See attached graphics image. LTspice files available on request.

#### Attachments

• Full-Wave Rectifiers 34.png
36.5 KB · Views: 271
Last edited:
Wrichik Basu