Boost Chopper or Buck converter?

In summary: I've designed and simulated a design for controlling a brushed DC series motor in OrCAD and it does what i was aiming for it to do. Which is to step down voltage to an average of 50% with duty cycle at 50% etc. However from the link below it seems that i have designed a Boost Chopper, which is really confusing me. Can someone please just explain to me what the hell i have done?
  • #1
emaN resU
6
0
Ok, first post so be nice guys!

I've designed and simulated a design for controlling a brushed DC series motor in OrCAD and it does what i was aiming for it to do. Which is to step down voltage to an average of 50% with duty cycle at 50% etc.

I thought I had it all figured out but recently I've got really confused about what I am doing. :P

Ok so I thought I had designed a Buck converter see figure below:

BuckConverter-1.jpg


However from the link below it seems that i have designed a Boost Chopper, which is really confusing me. Can someone please just explain to me what the hell i have done?

http://www.microsemi.com/catalog/pa...0&P1_Voltage1=600&P2_PMType=PMIGBT&P3_CONF=47

emaN resU
 
Engineering news on Phys.org
  • #2
Welcome to the PF. That's definitely not a Buck topology circuit. A Buck converter (to decrease voltage) uses a series switch element, not a switch to ground.

What are you trying to do? Just PWM the supply voltage for the motor winding?
 
  • #3
berkeman said:
Welcome to the PF.

Why thank you!

berkeman said:
That's definitely not a Buck topology circuit. A Buck converter (to decrease voltage) uses a series switch element, not a switch to ground.

Hmmmm, this is not good. When you say a series switch element are you talking about a high side driver or something altogetehr different. I thought aslong as it had a FWD in parallel to the motor the motor voltage would be dropped down.

berkeman said:
What are you trying to do? Just PWM the supply voltage for the motor winding?

Precisely, it would be for an electric car. I've seen many diagrams like this, so I figured it would be ok. In my simulation the voltage does drop down from half of source though at 50% duty cycle. So it does step down, so is that not a buck converter? If it's not, what is it? :D

Sorry for my ignorance, iv'e never done anything to do with power electronics before!

Any links, data sheets that i could read through would be good guys, i obviously need to learn what's going on!

Cheers

emaN resU
 
  • #5
berkeman said:
I googled "dc-dc converter" +tutorial, and got lots of good hits. Here's one of the first ones, from Maxim:

http://www.maxim-ic.com/appnotes.cfm/an_pk/2031/

.

Yeh, I've read that. Google is a great tool! I wouldn't be on a forum pleading for help if I could google it, trust me.

I thought i had designed a buck-boost circuit, but as its a 1 quadrant drive and will never go into boost (regen) mode so it's just a buck converter.

I'm so confused.

I'll just fiddle around with some equations, see if they match my results.

Cheers for the advice.
 
  • #6
You haven't said what results you are getting. Have you built it? What do you see on a CRO?

If you are looking for a name for it, it is just an amplifier.
It looks like you would just get the motor current being switched at the mark-space ratio of the input.
Nothing wrong with that, though. It is a perfectly OK way to do it.

The diode across the motor shorts out any inductive kickback from the motor. This is normally done with inductive loads to protect the switching transistor which would otherwise get a positive going spike when the transistor turned off.

What frequency are you driving it at?
 
  • #7
vk6kro said:
You haven't said what results you are getting. Have you built it? What do you see on a CRO?

hey dude, I've only simulated using OrCAD, which I get the results I was hoping for, so when I change the gate drive voltage of the IGBT to be on for half or a quarter duty cycle i get 100V and 50V respectively of average voltage. So, surely a step down converter.
vk6kro said:
If you are looking for a name for it, it is just an amplifier.

I showed the circuit to a guy who actually builds and designs motor controllers for a living and he said aslong as the switch is in series with the inductor and that the diode is in parallel to the motor its a buck converter.
vk6kro said:
It looks like you would just get the motor current being switched at the mark-space ratio of the input.
Nothing wrong with that, though. It is a perfectly OK way to do it.

I have no idea what mark-space ratio means, i'll go look that up :P

EDIT: HAHA ooooooh its just Duty Cycle, my bad.

vk6kro said:
The diode across the motor shorts out any inductive kickback from the motor. This is normally done with inductive loads to protect the switching transistor which would otherwise get a positive going spike when the transistor turned off.

Yer it's a freewheeling diode, in my simulation I remove the FWD and there is a HUGE voltage spike of negative 1.6KV which then goes back up to positive 1.6KV and then settles back to the expected level, but when I put the FWD back in I get a nice dip in voltage, no spikes.

vk6kro said:
What frequency are you driving it at?

16.4KHz, just above audible range(for most adults), but not too high so I don't get ridiculous losses.
 
Last edited:
  • #8
If you switch the transistor so that the motor gets power for only half the time, it will have half the average voltage it would have if it was conducting all the time.
Likewise, give it power for a quarter of the time and it will get a quarter of the average power.

So, whether you call that a chopper or a buck converter doesn't really make much difference except that the motor will really have a square wave across it, not a nice smooth reduced DC voltage which you would hope for from a buck circuit.

The leads to the motor will also radiate radio and TV interference at multiples of that 16.4 KHz chopping rate, so you may like to make them short or to shield them.
 
  • #9
vk6kro said:
If you switch the transistor so that the motor gets power for only half the time, it will have half the average voltage it would have if it was conducting all the time.
Likewise, give it power for a quarter of the time and it will get a quarter of the average power.

So, whether you call that a chopper or a buck converter doesn't really make much difference except that the motor will really have a square wave across it, not a nice smooth reduced DC voltage which you would hope for from a buck circuit.

Ok, reading what you have just said a few times, i realized i had misunderstood what you were saying. Whether I call it a chopper or a buck circuit won't make much difference, I agree, sort of (unless its a buck chopper, i think :P).

However, the difference between a buck and boost circuit has a difference, the formulas are different, the output graphs are different, their arrangement is different they are different. So my question at the start was, is it a chopper boost circuit or a buck converter? Is a chopper boost circuit the same as a boost converter? I'm unsure whether its the semantics I'm getting confused on.

vk6kro said:
The leads to the motor will also radiate radio and TV interference at multiples of that 16.4 KHz chopping rate, so you may like to make them short or to shield them.

Yer that's a good point, it's also an idea to twist the cables that run in parallel. But that's at a later stage, IF I put the design into a car. Which doesn't look too hopeful. :P

All I want to do at the moment is to prove this works in CCM. To do that I need to know what type if converter it is, so I can use the appropriate formula.

From the simulation results, current over the inductor(motor coil) ramps up as you would expect however does not reach equilibrium for the simulation time I've entered, which took my computer ages to do, so its in CCM at this point as current never falls to 0, but I just need to prove that with formulas.

It's nice of you to reply by the way vk6kro and berkeman.

PeaCe

emaN resU
 
  • #10
OK.

I think you have about half of a buck converter. You have the chopper bit, but the full thing does some integrating of the chopped waveform to give a steady filtered output.

I doubt if you can model a DC motor purely on its stationary inductance.
When it starts rotating there is a pretty powerful back EMF in there as well.

Interesting project though.
 
  • #11
vk6kro said:
OK.

I think you have about half of a buck converter. You have the chopper bit, but the full thing does some integrating of the chopped waveform to give a steady filtered output.

Hey man,

It'll be fiiiiiiine! Did you read my comment in post #5? The guy I asked who designs motor controllers confirmed my suspicions.

vk6kro said:
I doubt if you can model a DC motor purely on its stationary inductance.
When it starts rotating there is a pretty powerful back EMF in there as well.

Your point about modelling it on stationary inductance is so true, I did attempt to use the information on the website below, but some of the properties of the circuit I had no idea how to implement in OrCAD. I also tried to use a DC motor that's already in the OrCAD library and that failed epicly, so I was left with the static model.

http://www.ecircuitcenter.com/Circuits/smps_buck/smps_buck.htm

vk6kro said:
Interesting project though.

I really want to build it and give it a test! But I'm a bummy student, with no money :( haha

Need to wait until I get a job!

Cheers for the help guys

emaN resU
 

1. What is a Boost Chopper or Buck converter?

A Boost Chopper or Buck converter is a type of DC-DC power converter that is used to step up or step down the input voltage to a desired output voltage level. It consists of switches, inductors, capacitors, and diodes that work together to convert the input voltage into the desired output voltage.

2. What is the difference between a Boost Chopper and a Buck converter?

The main difference between a Boost Chopper and a Buck converter is the direction of voltage conversion. A Boost Chopper converts a lower input voltage to a higher output voltage, while a Buck converter converts a higher input voltage to a lower output voltage. Additionally, the components used in each circuit may differ in terms of number and configuration.

3. What are the applications of Boost Chopper and Buck converters?

Boost Choppers and Buck converters are commonly used in electronic devices that require a stable and regulated power supply, such as computers, mobile phones, and LED lighting systems. They are also used in renewable energy systems to convert the DC output of solar panels or wind turbines to a usable AC voltage for household appliances.

4. How do Boost Choppers and Buck converters work?

Both Boost Choppers and Buck converters work based on the principle of inductance. When a switch is turned on, the inductor stores energy in the form of a magnetic field. When the switch is turned off, the inductor releases the stored energy, which is then converted to the desired output voltage by the use of capacitors and diodes.

5. What are the advantages of using Boost Choppers and Buck converters?

Boost Choppers and Buck converters have several advantages, including high efficiency, compact size, and wide input voltage range. They also provide a stable output voltage even when the input voltage fluctuates, making them ideal for use in various electronic devices and systems.

Similar threads

  • Electrical Engineering
Replies
14
Views
799
  • Electrical Engineering
Replies
32
Views
2K
  • Electrical Engineering
Replies
10
Views
1K
  • Electrical Engineering
Replies
13
Views
2K
  • Electrical Engineering
Replies
17
Views
2K
Replies
1
Views
1K
Replies
2
Views
3K
Replies
2
Views
1K
  • Electrical Engineering
Replies
2
Views
13K
  • Electrical Engineering
Replies
1
Views
3K
Back
Top