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Is it possible to use a Mosfet in switching mode as an electronic load?
Electronic test load designP-Channel Depletion Mode MOSFET for Negative Pulsesmeaning of MOSFET “linear region” in the context of switching lossesActive load using MOSFETHow to increase mosfet switching speed, and decrease switching losses?SOA analysis for MOSFET-based Electronic LoadRipple when switching on MOSFETs with PWMLoad-line of MOSFET when analyzing triode modeMosfet for low load currentHigh Side Switching Inductive Load with MOSFET
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$begingroup$
I'm trying to use a mosfet as an electronic load to measure the IV curve of a solar cell, but if I use it in linear region it dissipates too much power. I was thinking that I could use it in switching mode like sending a PWM value from a microcontroller, and set the output over drain pin, signal which should be a pwm as well as its input, but with the real value like modulate and demodulate the signal using a low pass filter.
I have tried that and it works, the problem is that it doesn't produce the real IV curve, it shows a linear function. Could you advice me how can I use this mosfet in switching mode to produce that electronic load?. In the picture there's a scheme that I have used to achieve it.
In the circuit Current sensor is a ACS712 for 20Amps and the voltage circuit is just a voltage divider, both filtered signals go to a ESP32 Microcontroller.
Thanks, I'm a little frustrated and I would like to know the reason why this method is not as effective as a electronic load in linear region.
mosfet pwm electronic-load
New contributor
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add a comment
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$begingroup$
I'm trying to use a mosfet as an electronic load to measure the IV curve of a solar cell, but if I use it in linear region it dissipates too much power. I was thinking that I could use it in switching mode like sending a PWM value from a microcontroller, and set the output over drain pin, signal which should be a pwm as well as its input, but with the real value like modulate and demodulate the signal using a low pass filter.
I have tried that and it works, the problem is that it doesn't produce the real IV curve, it shows a linear function. Could you advice me how can I use this mosfet in switching mode to produce that electronic load?. In the picture there's a scheme that I have used to achieve it.
In the circuit Current sensor is a ACS712 for 20Amps and the voltage circuit is just a voltage divider, both filtered signals go to a ESP32 Microcontroller.
Thanks, I'm a little frustrated and I would like to know the reason why this method is not as effective as a electronic load in linear region.
mosfet pwm electronic-load
New contributor
$endgroup$
$begingroup$
The solar module cannot traverse its VI curve without delivering power somewhere. Anything that reduces the dissipated power to near zero will also make it impossible to measure the real VI curve. You can PWM the transistor, but let it deliver power to a resistor. Put a capacitor in parallel with the solar module and an inductor in series with the transistor.
$endgroup$
– mkeith
4 hours ago
add a comment
|
$begingroup$
I'm trying to use a mosfet as an electronic load to measure the IV curve of a solar cell, but if I use it in linear region it dissipates too much power. I was thinking that I could use it in switching mode like sending a PWM value from a microcontroller, and set the output over drain pin, signal which should be a pwm as well as its input, but with the real value like modulate and demodulate the signal using a low pass filter.
I have tried that and it works, the problem is that it doesn't produce the real IV curve, it shows a linear function. Could you advice me how can I use this mosfet in switching mode to produce that electronic load?. In the picture there's a scheme that I have used to achieve it.
In the circuit Current sensor is a ACS712 for 20Amps and the voltage circuit is just a voltage divider, both filtered signals go to a ESP32 Microcontroller.
Thanks, I'm a little frustrated and I would like to know the reason why this method is not as effective as a electronic load in linear region.
mosfet pwm electronic-load
New contributor
$endgroup$
I'm trying to use a mosfet as an electronic load to measure the IV curve of a solar cell, but if I use it in linear region it dissipates too much power. I was thinking that I could use it in switching mode like sending a PWM value from a microcontroller, and set the output over drain pin, signal which should be a pwm as well as its input, but with the real value like modulate and demodulate the signal using a low pass filter.
I have tried that and it works, the problem is that it doesn't produce the real IV curve, it shows a linear function. Could you advice me how can I use this mosfet in switching mode to produce that electronic load?. In the picture there's a scheme that I have used to achieve it.
In the circuit Current sensor is a ACS712 for 20Amps and the voltage circuit is just a voltage divider, both filtered signals go to a ESP32 Microcontroller.
Thanks, I'm a little frustrated and I would like to know the reason why this method is not as effective as a electronic load in linear region.
mosfet pwm electronic-load
mosfet pwm electronic-load
New contributor
New contributor
New contributor
asked 8 hours ago
RattenfengarRattenfengar
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$begingroup$
The solar module cannot traverse its VI curve without delivering power somewhere. Anything that reduces the dissipated power to near zero will also make it impossible to measure the real VI curve. You can PWM the transistor, but let it deliver power to a resistor. Put a capacitor in parallel with the solar module and an inductor in series with the transistor.
$endgroup$
– mkeith
4 hours ago
add a comment
|
$begingroup$
The solar module cannot traverse its VI curve without delivering power somewhere. Anything that reduces the dissipated power to near zero will also make it impossible to measure the real VI curve. You can PWM the transistor, but let it deliver power to a resistor. Put a capacitor in parallel with the solar module and an inductor in series with the transistor.
$endgroup$
– mkeith
4 hours ago
$begingroup$
The solar module cannot traverse its VI curve without delivering power somewhere. Anything that reduces the dissipated power to near zero will also make it impossible to measure the real VI curve. You can PWM the transistor, but let it deliver power to a resistor. Put a capacitor in parallel with the solar module and an inductor in series with the transistor.
$endgroup$
– mkeith
4 hours ago
$begingroup$
The solar module cannot traverse its VI curve without delivering power somewhere. Anything that reduces the dissipated power to near zero will also make it impossible to measure the real VI curve. You can PWM the transistor, but let it deliver power to a resistor. Put a capacitor in parallel with the solar module and an inductor in series with the transistor.
$endgroup$
– mkeith
4 hours ago
add a comment
|
2 Answers
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$begingroup$
the problem is that it doesn't produce the real IV curve, it shows a
linear function.
Your PWM is shorting out the solar panel when on and open-circuiting it when off, so you are just getting an average of the open circuit voltage multiplied by the PWM ratio.
To get a proper IV curve you have to draw power (ie. Volts x Amps) from the panel, and that power will have to be dissipated somewhere.
You could put several FETs in parallel so that each one stays within its rating, with large heatsinks and forced air cooling. To reduce maximum FET dissipation you could put high power resistors in series that drop about half the voltage at the panel's expected maximum power point.
Or you could just switch in high power resistors of progressively lower values in parallel. If each resistor value is half the previous one then you can input a binary code and have a 'digital resistor', like this:-
simulate this circuit – Schematic created using CircuitLab
The FETs switching the higher value resistors can be rated lower because they are switching less current. For the higher power resistors you could parallel several lower wattage resistors eg. 5 x 10Ω 25W in parallel would make a 2Ω 125W resistor. Forced air cooling can also be used to run the resistors up to (or over!) their power ratings (ratings in my diagram are based on a panel with open circuit voltage of 22V and MPP of 15Vx8A).
This is the scheme I use for discharge testing of high capacity Lipo batteries, except I just have a row of toggle switches which I operate manually to get the desired load current.
$endgroup$
add a comment
|
$begingroup$
Here is kind of a sketch of an idea that could work. I didn't do any calculations to determine what are the best inductor or capacitor values. You might have to change C1 and L1. But if you are interested you can enter it in a simulator and play with it a bit. In addition to M1 and R1, you could possibly add additional stages with smaller resistor values to get higher currents. The power rating for R1 should be greater than the peak power that the cell can deliver.
simulate this circuit – Schematic created using CircuitLab
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2 Answers
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2 Answers
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$begingroup$
the problem is that it doesn't produce the real IV curve, it shows a
linear function.
Your PWM is shorting out the solar panel when on and open-circuiting it when off, so you are just getting an average of the open circuit voltage multiplied by the PWM ratio.
To get a proper IV curve you have to draw power (ie. Volts x Amps) from the panel, and that power will have to be dissipated somewhere.
You could put several FETs in parallel so that each one stays within its rating, with large heatsinks and forced air cooling. To reduce maximum FET dissipation you could put high power resistors in series that drop about half the voltage at the panel's expected maximum power point.
Or you could just switch in high power resistors of progressively lower values in parallel. If each resistor value is half the previous one then you can input a binary code and have a 'digital resistor', like this:-
simulate this circuit – Schematic created using CircuitLab
The FETs switching the higher value resistors can be rated lower because they are switching less current. For the higher power resistors you could parallel several lower wattage resistors eg. 5 x 10Ω 25W in parallel would make a 2Ω 125W resistor. Forced air cooling can also be used to run the resistors up to (or over!) their power ratings (ratings in my diagram are based on a panel with open circuit voltage of 22V and MPP of 15Vx8A).
This is the scheme I use for discharge testing of high capacity Lipo batteries, except I just have a row of toggle switches which I operate manually to get the desired load current.
$endgroup$
add a comment
|
$begingroup$
the problem is that it doesn't produce the real IV curve, it shows a
linear function.
Your PWM is shorting out the solar panel when on and open-circuiting it when off, so you are just getting an average of the open circuit voltage multiplied by the PWM ratio.
To get a proper IV curve you have to draw power (ie. Volts x Amps) from the panel, and that power will have to be dissipated somewhere.
You could put several FETs in parallel so that each one stays within its rating, with large heatsinks and forced air cooling. To reduce maximum FET dissipation you could put high power resistors in series that drop about half the voltage at the panel's expected maximum power point.
Or you could just switch in high power resistors of progressively lower values in parallel. If each resistor value is half the previous one then you can input a binary code and have a 'digital resistor', like this:-
simulate this circuit – Schematic created using CircuitLab
The FETs switching the higher value resistors can be rated lower because they are switching less current. For the higher power resistors you could parallel several lower wattage resistors eg. 5 x 10Ω 25W in parallel would make a 2Ω 125W resistor. Forced air cooling can also be used to run the resistors up to (or over!) their power ratings (ratings in my diagram are based on a panel with open circuit voltage of 22V and MPP of 15Vx8A).
This is the scheme I use for discharge testing of high capacity Lipo batteries, except I just have a row of toggle switches which I operate manually to get the desired load current.
$endgroup$
add a comment
|
$begingroup$
the problem is that it doesn't produce the real IV curve, it shows a
linear function.
Your PWM is shorting out the solar panel when on and open-circuiting it when off, so you are just getting an average of the open circuit voltage multiplied by the PWM ratio.
To get a proper IV curve you have to draw power (ie. Volts x Amps) from the panel, and that power will have to be dissipated somewhere.
You could put several FETs in parallel so that each one stays within its rating, with large heatsinks and forced air cooling. To reduce maximum FET dissipation you could put high power resistors in series that drop about half the voltage at the panel's expected maximum power point.
Or you could just switch in high power resistors of progressively lower values in parallel. If each resistor value is half the previous one then you can input a binary code and have a 'digital resistor', like this:-
simulate this circuit – Schematic created using CircuitLab
The FETs switching the higher value resistors can be rated lower because they are switching less current. For the higher power resistors you could parallel several lower wattage resistors eg. 5 x 10Ω 25W in parallel would make a 2Ω 125W resistor. Forced air cooling can also be used to run the resistors up to (or over!) their power ratings (ratings in my diagram are based on a panel with open circuit voltage of 22V and MPP of 15Vx8A).
This is the scheme I use for discharge testing of high capacity Lipo batteries, except I just have a row of toggle switches which I operate manually to get the desired load current.
$endgroup$
the problem is that it doesn't produce the real IV curve, it shows a
linear function.
Your PWM is shorting out the solar panel when on and open-circuiting it when off, so you are just getting an average of the open circuit voltage multiplied by the PWM ratio.
To get a proper IV curve you have to draw power (ie. Volts x Amps) from the panel, and that power will have to be dissipated somewhere.
You could put several FETs in parallel so that each one stays within its rating, with large heatsinks and forced air cooling. To reduce maximum FET dissipation you could put high power resistors in series that drop about half the voltage at the panel's expected maximum power point.
Or you could just switch in high power resistors of progressively lower values in parallel. If each resistor value is half the previous one then you can input a binary code and have a 'digital resistor', like this:-
simulate this circuit – Schematic created using CircuitLab
The FETs switching the higher value resistors can be rated lower because they are switching less current. For the higher power resistors you could parallel several lower wattage resistors eg. 5 x 10Ω 25W in parallel would make a 2Ω 125W resistor. Forced air cooling can also be used to run the resistors up to (or over!) their power ratings (ratings in my diagram are based on a panel with open circuit voltage of 22V and MPP of 15Vx8A).
This is the scheme I use for discharge testing of high capacity Lipo batteries, except I just have a row of toggle switches which I operate manually to get the desired load current.
answered 6 hours ago
Bruce AbbottBruce Abbott
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29.8k1 gold badge24 silver badges42 bronze badges
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$begingroup$
Here is kind of a sketch of an idea that could work. I didn't do any calculations to determine what are the best inductor or capacitor values. You might have to change C1 and L1. But if you are interested you can enter it in a simulator and play with it a bit. In addition to M1 and R1, you could possibly add additional stages with smaller resistor values to get higher currents. The power rating for R1 should be greater than the peak power that the cell can deliver.
simulate this circuit – Schematic created using CircuitLab
$endgroup$
add a comment
|
$begingroup$
Here is kind of a sketch of an idea that could work. I didn't do any calculations to determine what are the best inductor or capacitor values. You might have to change C1 and L1. But if you are interested you can enter it in a simulator and play with it a bit. In addition to M1 and R1, you could possibly add additional stages with smaller resistor values to get higher currents. The power rating for R1 should be greater than the peak power that the cell can deliver.
simulate this circuit – Schematic created using CircuitLab
$endgroup$
add a comment
|
$begingroup$
Here is kind of a sketch of an idea that could work. I didn't do any calculations to determine what are the best inductor or capacitor values. You might have to change C1 and L1. But if you are interested you can enter it in a simulator and play with it a bit. In addition to M1 and R1, you could possibly add additional stages with smaller resistor values to get higher currents. The power rating for R1 should be greater than the peak power that the cell can deliver.
simulate this circuit – Schematic created using CircuitLab
$endgroup$
Here is kind of a sketch of an idea that could work. I didn't do any calculations to determine what are the best inductor or capacitor values. You might have to change C1 and L1. But if you are interested you can enter it in a simulator and play with it a bit. In addition to M1 and R1, you could possibly add additional stages with smaller resistor values to get higher currents. The power rating for R1 should be greater than the peak power that the cell can deliver.
simulate this circuit – Schematic created using CircuitLab
answered 4 hours ago
mkeithmkeith
12.1k1 gold badge11 silver badges35 bronze badges
12.1k1 gold badge11 silver badges35 bronze badges
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Rattenfengar is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
The solar module cannot traverse its VI curve without delivering power somewhere. Anything that reduces the dissipated power to near zero will also make it impossible to measure the real VI curve. You can PWM the transistor, but let it deliver power to a resistor. Put a capacitor in parallel with the solar module and an inductor in series with the transistor.
$endgroup$
– mkeith
4 hours ago