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Nevermind (for now) what Wikipedia or anyone else says. Stop and actually think about it. One problem with the question is that is says "buck-boost" but doesn't otherwise define that. A buck-boo...
Answer
#3: Post edited
by
Olin Lathrop
·
2022-03-13T16:14:10Z (about 2 years ago)
- Nevermind (for now) what Wikipedia or anyone else says. Stop and actually <i>think</i> about it.
- One problem with the question is that is says "buck-boost" but doesn't otherwise define that. A buck-boost converter generally means that the output can be both above or below the input. There are different topologies to achieve that.
- Since the question as posed is too vague, I'll answer for a straight buck converter. That's a well defined topology:
- <img src="https://electrical.codidact.com/uploads/pjeAhndD3rMu86Gyn3gWLy2j">
- In continuous mode with ideal components, the output voltage is the input voltage times the duty cycle. Another way to think about this is that L1 and C1 form a low pass filter. The top end of L1 is switched between Vin and ground, with the only choice being the relative time spent at each. Vout is the averaged result.
- You say Vin is 24 V and Vout 20 V. Vout is therefore (20 V)/(24 V) = 83% of Vin. The duty cycle therefore needs to be 83%.
- <blockquote> Isn't buck-boost just as well-defined a topolgy as the straight buck or boost converters?</blockquote>
- No. As I said, "buck-boost" generally refers to a DC to DC switching power supply which can have its output voltage above or below its input voltage.
There are a number of different ways to achieve that. You could use two switches with a single inductor to make a "DC transformer". You could use capacitive coupling. There are versions that use coupled inductors. Depending on how you define the term, you could even say that using a transformer qualifies at buck-boost, although personally I think that's a stretch.- Show the circuit for your buck-boost, and we can get into more details of how the duty cycle controls the input to output voltage ratio.
- Nevermind (for now) what Wikipedia or anyone else says. Stop and actually <i>think</i> about it.
- One problem with the question is that is says "buck-boost" but doesn't otherwise define that. A buck-boost converter generally means that the output can be both above or below the input. There are different topologies to achieve that.
- Since the question as posed is too vague, I'll answer for a straight buck converter. That's a well defined topology:
- <img src="https://electrical.codidact.com/uploads/pjeAhndD3rMu86Gyn3gWLy2j">
- In continuous mode with ideal components, the output voltage is the input voltage times the duty cycle. Another way to think about this is that L1 and C1 form a low pass filter. The top end of L1 is switched between Vin and ground, with the only choice being the relative time spent at each. Vout is the averaged result.
- You say Vin is 24 V and Vout 20 V. Vout is therefore (20 V)/(24 V) = 83% of Vin. The duty cycle therefore needs to be 83%.
- <blockquote> Isn't buck-boost just as well-defined a topolgy as the straight buck or boost converters?</blockquote>
- No. As I said, "buck-boost" generally refers to a DC to DC switching power supply which can have its output voltage above or below its input voltage.
- There are a number of different ways to achieve that. You could use two switches with a single inductor to make a "DC transformer". You could use capacitive coupling. There are versions that use coupled inductors. Depending on how you define the term, you could even say that using a transformer qualifies as buck-boost, although personally I think that's a stretch.
- Show the circuit for your buck-boost, and we can get into more details of how the duty cycle controls the input to output voltage ratio.
#2: Post edited
by
Olin Lathrop
·
2022-03-13T16:12:10Z (about 2 years ago)
- Nevermind (for now) what Wikipedia or anyone else says. Stop and actually <i>think</i> about it.
- One problem with the question is that is says "buck-boost" but doesn't otherwise define that. A buck-boost converter generally means that the output can be both above or below the input. There are different topologies to achieve that.
- Since the question as posed is too vague, I'll answer for a straight buck converter. That's a well defined topology:
- <img src="https://electrical.codidact.com/uploads/pjeAhndD3rMu86Gyn3gWLy2j">
- In continuous mode with ideal components, the output voltage is the input voltage times the duty cycle. Another way to think about this is that L1 and C1 form a low pass filter. The top end of L1 is switched between Vin and ground, with the only choice being the relative time spent at each. Vout is the averaged result.
You say Vin is 24 V and Vout 20 V. Vout is therefore (20 V)/(24 V) = 83% of Vin. The duty cycle therefore needs to be 83%.
- Nevermind (for now) what Wikipedia or anyone else says. Stop and actually <i>think</i> about it.
- One problem with the question is that is says "buck-boost" but doesn't otherwise define that. A buck-boost converter generally means that the output can be both above or below the input. There are different topologies to achieve that.
- Since the question as posed is too vague, I'll answer for a straight buck converter. That's a well defined topology:
- <img src="https://electrical.codidact.com/uploads/pjeAhndD3rMu86Gyn3gWLy2j">
- In continuous mode with ideal components, the output voltage is the input voltage times the duty cycle. Another way to think about this is that L1 and C1 form a low pass filter. The top end of L1 is switched between Vin and ground, with the only choice being the relative time spent at each. Vout is the averaged result.
- You say Vin is 24 V and Vout 20 V. Vout is therefore (20 V)/(24 V) = 83% of Vin. The duty cycle therefore needs to be 83%.
- <blockquote> Isn't buck-boost just as well-defined a topolgy as the straight buck or boost converters?</blockquote>
- No. As I said, "buck-boost" generally refers to a DC to DC switching power supply which can have its output voltage above or below its input voltage.
- There are a number of different ways to achieve that. You could use two switches with a single inductor to make a "DC transformer". You could use capacitive coupling. There are versions that use coupled inductors. Depending on how you define the term, you could even say that using a transformer qualifies at buck-boost, although personally I think that's a stretch.
- Show the circuit for your buck-boost, and we can get into more details of how the duty cycle controls the input to output voltage ratio.
#1: Initial revision
by
Olin Lathrop
·
2022-03-13T12:40:36Z (about 2 years ago)
Nevermind (for now) what Wikipedia or anyone else says. Stop and actually <i>think</i> about it. One problem with the question is that is says "buck-boost" but doesn't otherwise define that. A buck-boost converter generally means that the output can be both above or below the input. There are different topologies to achieve that. Since the question as posed is too vague, I'll answer for a straight buck converter. That's a well defined topology: <img src="https://electrical.codidact.com/uploads/pjeAhndD3rMu86Gyn3gWLy2j"> In continuous mode with ideal components, the output voltage is the input voltage times the duty cycle. Another way to think about this is that L1 and C1 form a low pass filter. The top end of L1 is switched between Vin and ground, with the only choice being the relative time spent at each. Vout is the averaged result. You say Vin is 24 V and Vout 20 V. Vout is therefore (20 V)/(24 V) = 83% of Vin. The duty cycle therefore needs to be 83%.