Author Topic: Electronic dummy load questions  (Read 25145 times)

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Offline scrat

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Re: Electronic dummy load questions
« Reply #25 on: September 14, 2010, 08:07:55 am »
Just had a look at the 3 datasheets: reading rdsON value, it is clear that a high additional voltage drop must happen, since they have resistances in the order of a few Ohms. Furthermore, you should have driven their gate with a higher voltage, since their ON resistance is specified at Vgs = 10V. If you charge them at only about 4.5V you have to look at their output characteristics to see which current they will draw (es. last fig. of page 2 for FS7SM16A). On the transistor's output characteristic you have to trace the generator-load straight line (a line from Vbus on the horizontal axis and Vbus/Rload on the vertical one). At the point the curve for Vgs = 4.5 or 5 V matches the straight line your transistor will sit when ON. For a switching application this point must stay on the left, where the curve has a high slope, slope depending on the Vgs you impose.
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Offline exodiaTopic starter

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Re: Electronic dummy load questions
« Reply #26 on: September 14, 2010, 03:30:41 pm »
Yes i agree with you partly. I don't think that RDSon alone is responsible for that voltage drop. Probably not driving the gate high enough is the main problem. Maybe low frequency also. If i drive the gate with enough voltage the mosfet should behave like an ideal swith correct? Also if i increase the frequency to a reasonable level the losses of switch on/off will go down?
 

Offline Time

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Re: Electronic dummy load questions
« Reply #27 on: September 14, 2010, 03:48:32 pm »
A poor Rdson is an artifact of not driving the gate hard enough.  Realistically, it will never behave "ideal" in your actual circuit.  Even with sufficient Vgs for the best conduction you will still have some channel resistance.
-Time
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #28 on: September 15, 2010, 10:41:08 am »
Obvioulsly, as a correct value for resistance you should use the specified RdsON, if you drive the gate at the specified voltage (10V, if I remember correctly, for the 3 MOSFETs).
Frequency increase won't solve the voltage drop problem, since average current is the same. It only lowers current ripple if an inductance is put in series to resistors, but in your load circuit there aren't reactive components (L/C).

From your measurements:
FS7SM16A voltage drop across resistors 6,24V  I consumed 1,2A -> Rds = (Vbus - Vr) / I = (9-6.24) / 1.2 = 2.3 Ohm
2N60B voltage drop across resistors 4,3V  I consumed 0,85A       -> Rds = (Vbus - Vr) / I = (9-4.3) / 0.85 = 5.5 Ohm
IRF840A voltage drop across resistors 3,1V  I consumed 0,62A    -> Rds = (Vbus - Vr) / I = (9-3.1) / 0.62 = 9.5 Ohm
If you measured with MOSFETs ON, values are correct and the circuit is the one you posted, there is no doubt.
Othewise, if you were applying a duty-cycle < 100% and measured average or rms current, then things change obviously.

Looking at the data, you can obtain the slope of Id vs Vds curve (MOSFET's equivalent resistance):
For FS7SM16A: at Vgs=4 V you have about 6V/3A = 2Ohm (from first fig. of 3rd page of datasheet, 3A taken because unfortunately there aren't curves for lower currents)
For 2N60B: at Vgs=5 V you have about 5V/1A = 5Ohm (from Fig.1 of datasheet)
For IRF840A: at Vgs=from 4.5V to 5V you have from 6/0.45=13.3 to 6/1.5=4 Ohm (Fig.1 of datashet, unfortunately there is a big variation from 4.5 to 5 V, just where you probably set gate voltage from Arduino)

As you can see, this confirms gate voltage as the cause of such a low current.
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Offline exodiaTopic starter

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Re: Electronic dummy load questions
« Reply #29 on: September 23, 2010, 05:51:14 pm »
Sorry for talking that long to reply

I was using PWM with a duty cycle of 100% when i took these measurements. If i apply less than 100% duty cycle should i use a true RMS multimeter? The true RMS is needed for DC also?

RDSON is variable according to voltage drain and gate? What is the typical RDSON we see in datasheets?

EDIT

Ok i did get a RFP70N06. I tried to connect the gate directly to MCU and i did get 10,54 Volt across resistors from 12Volts i supplied and 2.12 Amps. That with 100%PWM, i also tried putting 12 volts directly at the gate of the mosfet just to try to see if the voltage is not enough but i got the same measurements
« Last Edit: September 25, 2010, 05:05:35 pm by exodia »
 

Offline JohnShaft

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Re: Electronic dummy load questions
« Reply #30 on: September 30, 2010, 10:44:33 pm »
FET gates have quite a bit of capacitance. You generally cannot drive a PWM signal directly into the gate of the FET since MCU pins cannot source enough current to switch the FET quickly. You want the FET to spend most of its time fully ON or fully OFF. The time inbetween is going to heat up the FET fast.


ok i have tried some more with PWM now that i got an oscilloscope. I have tried so far with 3 mosfets with not so great success.

The mosfets i used

FS7SM16A voltage drop across resistors 6,24V  I consumed 1,2A
2N60B voltage drop across resistors 4,3V  I consumed 0,85A
IRF840A voltage drop across resistors 3,1V  I consumed 0,62A

Why i got so huge voltage drop across the resistors? And why every mosfet behaves so much different as a switch?
1st reason i think is maybe because i drive the mosfets directly from MCU and not from a totem pole driver
2nd maybe the frequency of arduino PWM is too low only 490 Hz

I would like to hear your opinions
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #31 on: October 01, 2010, 11:44:41 am »
Sorry for talking that long to reply

I was using PWM with a duty cycle of 100% when i took these measurements. If i apply less than 100% duty cycle should i use a true RMS multimeter? The true RMS is needed for DC also?

RDSON is variable according to voltage drain and gate? What is the typical RDSON we see in datasheets?

EDIT

Ok i did get a RFP70N06. I tried to connect the gate directly to MCU and i did get 10,54 Volt across resistors from 12Volts i supplied and 2.12 Amps. That with 100%PWM, i also tried putting 12 volts directly at the gate of the mosfet just to try to see if the voltage is not enough but i got the same measurements

Typical rdsON is specified for a given voltage (measurement conditions are listed on the same row of the table or above it).
From your measurements the MOSFET should have about 1 Ohm of equivalent resistance, which is at least incredible with Vgs=12V.
For PWM a DMM in DC mode should be good enough (they're conversion is made by integration, so they measure an average value), but we can never be sure, a true rms multimeter could be better, since results are so strange.
I think that something is not going right, the measurement or the gate voltage. If you have an oscilloscope by hand (or you can borrow one), try measuring with it.
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Offline exodiaTopic starter

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Re: Electronic dummy load questions
« Reply #32 on: October 04, 2010, 06:15:05 am »
I have an oscilloscope, what should i try and measure with it?
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #33 on: October 04, 2010, 10:25:00 am »
Please post a schematic (even a scan of a sketch on paper can be enough) and measure all voltages: Vgs, then Vdd and Vd (to obtain voltage on the load resistors by subtraction), and measure resistor's value (with the DMM, immediately after switching off the circuit, when they have heaten up, if they do during current pass).

When using the scope, always pay attention at the fact the GND terminal on the scope is bound to earth potential. On my thesis I burned a motor drive board only because I connected the GND of the probe to the power supply, relying on the fact that a scope probe has 10MOhm resistance. Unfortunately, even if the supply was isolated, ground of the circuit was connected by a BNC to another channel of the scope (to view the DACs outputs)! Connections to PC are a subtle way for the current to flow to ground too, since almost all ports shield is connected to earth.
If your Vdd supply is isolated and no other connection to earth exists on your circuit, there shouldn't be any problem.
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Offline exodiaTopic starter

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Re: Electronic dummy load questions
« Reply #34 on: October 04, 2010, 08:09:20 pm »
I have made the following circuit

The voltage reading i get are

PWM    VDD    VG    Vsupply    current    Ohm
0      12         0        12            0          2,7
10    11,08    0,5      12           0,345    2,8
20    10,15    0,96    11,99      0,697    2,7
30    9,23     1,45    11,98      1,051    2,8
40    7,99     1,96    11,96      1,505    2,7
50    6,95     2,45    11,94      1,872    2,7
60    5,93     2,94    11,91      2,252    2,7
70    4,91     3,39    11,9       2,648    2,7
80    3,8      3,9      11,87      3,063    2,6
90    2,83    4,39    11,85      3,383    2,7
100    1,48    4,78   11,82     3,902    2,8
VDD=Drain pin of mosfet
VG=Gate

 

Offline sonicj

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Re: Electronic dummy load questions
« Reply #35 on: October 05, 2010, 04:36:45 am »
20 page discussion on the matter: rcgroups.com - How to build a constant-current discharger

it gets pretty interesting near the end.

i have the older version of this: cba III. it uses a single irl2910 mosfet. the cbIII uses 2 irl2910's as some of the heavier load users were cookin 'em.
works fine, it would be nice if it monitored individual cell voltage.

some of the cheap chinese multi-chargers have a discharge function with serial output for logging. ie: imax b6
logview <==open source battery logging software
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #36 on: October 05, 2010, 07:59:07 am »
Well, what I was asking for was really a single measurement taken with the oscilloscope, at a high duty-cycle, even at 100%, to see the real waveforms. I'll take a look at the measurements you provided, which are useful too, to try understanding what happens there.
Please remind me what are the specifications you need to comply, and the core of the issue... it was quite a long time since you started the thread, now I start to be confused :)

EDIT: which is the MOSFET you used for measurement? Is it the IRFN140 as in the schematic? Sorry for the silly question, but you changed many times the device, so to make sure I ask..
« Last Edit: October 05, 2010, 08:06:03 am by scrat »
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Offline exodiaTopic starter

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Re: Electronic dummy load questions
« Reply #37 on: October 05, 2010, 05:12:10 pm »
Here you go

The channel 1 is at the gate of the mosfet and the second at the drain. The 1st channel is at -4,4V offset so u can see both signals

The mosfet i used is the RFP70N06

The specifications are that i can burn about 128watts. The load will always be the resistor network i'll construct. The source will always be a battery 12 Volt.
@sonicj

Thanks for the links i'll study them

« Last Edit: October 05, 2010, 08:39:08 pm by exodia »
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #38 on: October 06, 2010, 10:17:36 am »
The specifications are that i can burn about 128watts. The load will always be the resistor network i'll construct. The source will always be a battery 12 Volt.

128W... are you sure? You measured a resistance of at least 2.6 Ohm.
The maximum power you can dissipate from a 12V source through a 2.6Ohm resistor network is about 50W (P =  V^2/R = 12^2/2.6 = 55.38 W) !!
If these are the specifications, you'll need a step-up conversion.
« Last Edit: October 06, 2010, 12:40:35 pm by scrat »
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Offline exodiaTopic starter

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Re: Electronic dummy load questions
« Reply #39 on: October 06, 2010, 04:42:19 pm »
This is just a protoype on a breadboard. The production will have <0,3 ohm resistance but i will limit it at about 128 watt (0,5 watt steps for 255 PWM)
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #40 on: October 06, 2010, 05:15:48 pm »
OK, that's reassuring  :)
From the data you posted, there is not much matching with the datasheet curves.
In http://www.datasheetcatalog.org/datasheet/fairchild/RFP70N06.pdf, fig.7 shows that for a Vds of about 1V the current should be about 10A at Vgs= 4.5V. Your current is about 1/3 of that, which is quite strange.
These are typical performance, so it could still be inside the specs of the MOSFET.
The question I have to ask you is why you chose such a big MOSFET, since you have much lower currents and voltages to deal with. Driving the gate of that MOSFET at a so low voltage (could be slightly above its threshold, which is from 2 to 4 V) is not a good choice, since all written specs data are for Vgs=12V or 10V. The MOSFET will remain into saturation (the flat part of the curves in Fig.7) or at least at the corner between triode and saturation.
If you will drive the gate at 10 or 12 V, it will have (at max) 14mOhm resistance, so with a 0.3 Ohm load you're going to reach 12/(0.3+0.014) = 38.2 A peak on the production device, while now 12/(2.6+0.014) = 4.59 A.
So, I think you just have to drive the gate at a higher voltage. A transistor or a specific gate driver will do it. Also remember that the strenght of your current driving the gate affects the turn-on and turn-off time. If left with no current feedback (and no inductance) your discharger will not be that accurate, but if I remember right this wasn't the purpose of the project.
In the figure below, if you will use the 12V battery for 12V aux you'll have to take into account the small power loss caused by the gate charging. Otherwise, you could put there a tiny step-up to obtain the 12V aux from MCU 5V Vdd.


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Offline scrat

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Re: Electronic dummy load questions
« Reply #41 on: October 06, 2010, 10:36:03 pm »

The key issue is that you can't operate a boost circuit beyond a 50% duty cycle.

Why do you say that? It is common to build CCM boosts with D greater then 0.5. DCM is impractical at higher power due to high pk currents. CCM  is the preferred mode other then low power.

I have a CCM 300W APFC boost on my bench right now that does what you say it cant. D varies from near zero to near 1. The converter transitions from DCM to CCM at about 30% full load. I have a couple of isolated bucks (FLYBACKS) that also operate in CCM d> 0.5
[...]
Is your concern stability?
Maybe I misunderstood what you said.

I think that the concern is about stability, of course boost and buck-boost can operate at duty cycle over 0.5. If I remember right, the major issue about boost dynamics is the fact that, to increase current, you have to increment the duty-cycle, but during transient this causes a momentary fall in output current, since you retain for a longer period the output voltage bounded to ground.
BTW, isolated bucks are called forward, while a flyback is an isolated buck-boost, which uses the magnetizing inductance of the transformer.
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Offline scrat

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Re: Electronic dummy load questions
« Reply #42 on: October 07, 2010, 08:56:25 am »
We discussed this previously in this thread. A boost can be operated in CCM, but with a more complex controller, due to the RHPZ issue you mentioned (I had never seen before this acronym, since I studied this things on Italian texts).
Boost in CCM is the only topology which maintains a continuous input current, but the disadvantages are that a huge inductor must be used, to avoid saturation and to limit current ripple.
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Offline scrat

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Re: Electronic dummy load questions
« Reply #43 on: October 07, 2010, 10:18:30 am »
@exodia
To avoid using gate drivers, a solution, pointed out by someone on another thread (https://www.eevblog.com/forum/index.php?topic=1176.0), is to use low-gate voltage MOSFETs, which have low RdsOn at Vgs = 4-5V, like the IRL540 (77mOhm at Vgs=5V).

EDIT: for example, the IRLB8721 (https://ec.irf.com/v6/en/US/adirect/ir?cmd=catSearchFrame&domSendTo=byID&domProductQueryName=IRLB8721) goes up to 44 A (> 12V/0.3Ohm) even at 100°C, and has Rdson=16mOhm at Vgs=4.5V.
« Last Edit: October 07, 2010, 10:25:34 am by scrat »
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Offline exodiaTopic starter

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Re: Electronic dummy load questions
« Reply #44 on: October 09, 2010, 03:51:54 pm »
OK, that's reassuring  :)
From the data you posted, there is not much matching with the datasheet curves.
In http://www.datasheetcatalog.org/datasheet/fairchild/RFP70N06.pdf, fig.7 shows that for a Vds of about 1V the current should be about 10A at Vgs= 4.5V. Your current is about 1/3 of that, which is quite strange.
These are typical performance, so it could still be inside the specs of the MOSFET.
The question I have to ask you is why you chose such a big MOSFET, since you have much lower currents and voltages to deal with. Driving the gate of that MOSFET at a so low voltage (could be slightly above its threshold, which is from 2 to 4 V) is not a good choice, since all written specs data are for Vgs=12V or 10V. The MOSFET will remain into saturation (the flat part of the curves in Fig.7) or at least at the corner between triode and saturation.
If you will drive the gate at 10 or 12 V, it will have (at max) 14mOhm resistance, so with a 0.3 Ohm load you're going to reach 12/(0.3+0.014) = 38.2 A peak on the production device, while now 12/(2.6+0.014) = 4.59 A.
So, I think you just have to drive the gate at a higher voltage. A transistor or a specific gate driver will do it. Also remember that the strenght of your current driving the gate affects the turn-on and turn-off time. If left with no current feedback (and no inductance) your discharger will not be that accurate, but if I remember right this wasn't the purpose of the project.
In the figure below, if you will use the 12V battery for 12V aux you'll have to take into account the small power loss caused by the gate charging. Otherwise, you could put there a tiny step-up to obtain the 12V aux from MCU 5V Vdd.

I purchased such a mosfet because i want to be able to drive about 128 watts, the other reason is that i went to the local shop and they had that one. You mean i should have bought a mosfet with about the same watt but smaller I and V??? If yes the reason is that i can drive it easier without a driver?
In a previous post i have tried to hardwire the gate of the mosfet at 12 volt without applying any kind of PWM to it, the result was the same with 5 volt at gate and pwm 100% that results are with PWM frequency of 490 Hz.

I tried with 32Khz <0,2 ohm load and the result are different. 1st observation is that the voltage drop at the power supply if much less about 2 volts compared to 4-5 with 490 frequency, the other observation is that mosfet is getting pretty fast very hot without putting too much stress to it. If a run through it about 5 A at 10 Volts i can't touch it in 15 secs. If i run 5A at 490 Hz it's ok no problem. Maybe at 32Khz i can't charge the gate too fast and it's partially conducting, i should try the driver with the transistor

The rdson of the mosfet are always the same or they change value according to vgs,vds??

The final design will have a current sense amplifier for current feedback, i will also read the voltage with micro and then i will have a feedback.

Something that confuses me is that when i increase the current i see a voltage drop. That is from V=I*R, correct? To calculate the total watts this think burn i have to multiply current with Volts. So from the previous measurements, correct?
0
4,14
8,35703
12,59098
17,9998
22,35168
26,82132
31,5112
36,35781
40,08855
46,12164


Edit yes that a nice mosfet but much less power dissipation than the one i have selcted. It's only 65W and i need an 128

I have a small mosfet TO92 BS170, i could use that as a driver, with a mosfet i will not needing any resistors at drain like BJT
« Last Edit: October 11, 2010, 06:04:16 am by exodia »
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #45 on: October 11, 2010, 08:29:38 am »
The rdson of the mosfet are always the same or they change value according to vgs,vds??
Yes, of course, otherwise it won't be a voltage controlled "switch". Rdson is a parameter which makes sense only for the "ON" state (in other words, when the MOSFET has a low Vds voltage). Rdson strongly depends on Vgs, you can see it from the Id vs Vds curve: the straight line passing through the zero changes its slope for the various values of Vgs.

The final design will have a current sense amplifier for current feedback, i will also read the voltage with micro and then i will have a feedback.
That's a good thing, which also prevents damage to the devices and greatly increases precision.
I suggest putting an inductance in series to the load (but then never forget an antiparallel diode!!), to make the load current nearly constant (otherwise it will be a square waveform).

Something that confuses me is that when i increase the current i see a voltage drop. That is from V=I*R, correct? To calculate the total watts this think burn i have to multiply current with Volts. So from the previous measurements, correct?
0
4,14
8,35703
12,59098
17,9998
22,35168
26,82132
31,5112
36,35781
40,08855
46,12164
This "voltage drop" is really due to both Ohm's law and the partialization you make through PWM.
You only measured the average values there, so even if your MOSFET would be an ideal switch (Rdson = 0), you will have seen a voltage drop, since you have put into conduction it only for a certain duty-cycle, not for the entire period.
If you close the switch for 30% of the time and have a supply of 10V, you will see across the ideal switch an average voltage of 7V ( 10 * (1-0.30) ). That's basic switching logic, I hope you already know how to manage with it and you only got confused here.

Dissipated power on the MOSFET isn't calculated as averageV*averageI. Instead, resistors dissipate the most part of the power, and this is your goal, otherwise you would have put there the MOSFET alone. The MOSFET only dissipates during conduction (rdson*Id) and at transitions (from ON to OFF and viceversa) if you have an inductance on your load.
On a resistive load the MOSFET will only dissipate P = rdson * Id * duty-cycle. If you don't drive it with a proper Vgs rdson will be much higher than the one declared.

Edit yes that a nice mosfet but much less power dissipation than the one i have selcted. It's only 65W and i need an 128
Your MOSFET doesn't have to dissipate all of the power, for a resistive, it dissipates rdson*Id. Remember that you have the resistors there to dissipate, the MOSFET only has to control for how much time they have to dissipate and for how much time not!

I have a small mosfet TO92 BS170, i could use that as a driver, with a mosfet i will not needing any resistors at drain like BJT
That could be a good solution, but I'm not sure if it will be faster than a BJT for that application. You can try it.

I hope this can help you, but I kindly suggest to take a look at a simple article or tutorial or whatsoever on switching mode power supplies, to achieve at least a basic understanding of a buck (=chopper) converter. This should have been done before anything on this project, and would have allowed you to save much time, IMHO. This doesn't mean that I won't be pleased to answer further :)
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Offline exodiaTopic starter

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Re: Electronic dummy load questions
« Reply #46 on: October 11, 2010, 06:27:56 pm »
Thanks for taking the time to answer my question, i have studied some basic DC to DC converter topologies maybe not too much thou.

Now some power dissipation question, i have found from the Internet some examples on calculating mosfet dissipation but i would like to help me understand, i quote a paragraph

Problem A MOSFET is switched at 20kHz, and takes 1 microsecond to switch between states (on to off and off to on). The supply voltage is 12v and
the current 40 Amps. Calculate the average switching power loss, assuming the voltage and current are at half values during the switching period.

Solution: At 20kHz, there is a MOSFET switching occurrence every 25 microseconds (a switch on every 50 microseconds, and a switch off every 50
microseconds). Therefore, the ratio of switching time to total time is 1/25 = 0.04. The power dissipation when switching is (12v / 2) x (40A / 2) = 120
Watts. Therefore the average switching loss is 120W x 0.04 = 4.8 Watts.

T=1/f=>T=0,00005 sec

That is a full period including a Ton and a Toff cycle, correct?Then 1 ms (duration of transition) /25 msec of the full period

Is the above calculations correct??

i got them from the following page
http://homepages.which.net/~paul.hills/SpeedControl/MosfetBody.html

Now something about coils. I have to select the the inductance but what about the current that is passing through them, are there like resistors that have power ratings?
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #47 on: October 11, 2010, 08:14:52 pm »
Solution: At 20kHz, there is a MOSFET switching occurrence every 25 microseconds (a switch on every 50 microseconds, and a switch off every 50
microseconds). Therefore, the ratio of switching time to total time is 1/25 = 0.04. The power dissipation when switching is (12v / 2) x (40A / 2) = 120
Watts. Therefore the average switching loss is 120W x 0.04 = 4.8 Watts.
This problem seems to me very confusing. For example, it must be specified if the MOSFET is driving a resistive or an inductive load, but for any of the two cases it is quite silly to say that voltage and current will remain constant at half value during switching. We know that switching is an event of finite duration, so the situation described can't happen: otherwise voltage would pass from zero to half to full supply value instantly, and then viceversa.
First of all, switching events are located at 50us one from each other just because the duty-cycle is 50% (if the "half voltage" is related to the output voltage of the chopped supply).

Then, typical situation in switched mode power supplies is when a strongly inductive load (L/R >> 1/fsw) is switched by a MOSFET. I'll slightly change the problem, just to make it more useful.
A MOSFET switches an inductive load (with antiparallel inductance). Vsupply is 12V, average current is 20A, switching frequency is 20kHz and transition time is 1us, rdson is 1Ohm. Calculate losses.
     ^Vsupply
     |
     o------
     |      |
    C      ---
L   C      /_\
    C       |
     |      |
    <       |
R    >     |
    <       |
     |      |
     o------
     |
  __|
    |_  Q
     |
    _|_
    ///
The equivalent circuit which usally is analyzed for this case is an ideal current generator (constant current source) with a diode in antiparallel and in series with the MOSFET, all between Vsupply and ground. This is because we suppose inductance to force its current to remain almost constant (at the average value) during switching.
     ^Vsupply
     |
     o------
     |      |
 IL  |     ---
 |  (--)   /_\
 \/  |      |
     |      |
     o------
     |
  __|
    |_  Q
     |
    _|_
    ///
When the MOSFET is OFF, all current will flow through the diode, clamping Vd at Vsupply.
During switching (let's analyze the OFF to ON transient, so rising gate for an N-MOSFET), Vds will start at Vsupply and remain constant at Vsupply until the diode stops conducting (which means until MOSFET current linearly reaches inductor current value). Then Vds will linearly go to nearly zero (rdson*IL), while current on the MOSFET remains constant. This is one of the simplest approximation of the switching event.
Vgs^
     | Vdrive
     |                        /                  
     |              _____/                  
     |             /
     |           /
     |     Vth/
     | ____/
     |
Vds^
     |
     | Vsupply__    
     |               \
     |                 \
     |                   \
     | 0                   \____
     |
Id  ^
     |
     | IL           _________________
     |             /
     |            /
     | 0_____/
     |
     ^
PWR|
     |Vsupply*IL
     |               ^                         t1: Vgs reaches Vth, current starts to flow through MOSFET, but part of IL still passes through
     |             /    \                            the diode, so Vd is clamped at Vsupply
     |            /       \                     t2: the MOSFET conducts all the current, and rapidly reduces its equivalent resistance, since
     | 0_____/          \__________         gate reaches its final value (high level from driver or MCU output) and rdson remains constant.
                t1           t2
Since from t1 to t2 there will be 1us, energy dissipated by the MOSFET at each switching is the area of the PWR triangle (Vds(t)*Id(t)), which is Esw = (t2-t1) * Vsupply*IL. For each period there will be two transitions, so average power due to switching is PWRav = 2*Esw / Tsw = (t2-t1) * Vsupply*IL * fsw = 1u * 12 * 20 * 20k = 4.8W. This is because we negliged rdson*IL  voltage (<<Vsupply), which is usually acceptable (otherwise, MOSFET is used as a linear device instead of switching).

The rising time of current extends from the time when Vgs reaches threshold (Vth) to the time Vgs enters Milller zone (flat "plateau" zone on Vgs). There Vds starts to go down.

I found also this link: http://www.irf.com/technical-info/appnotes/mosfet.pdf.

Now something about coils. I have to select the the inductance but what about the current that is passing through them, are there like resistors that have power ratings?
For choosing the inductor there are many factors to take into account: thermal resistive dissipation is one of those, but there are some losses related to magnetic effects (hysteresis and eddy currents). A strong effect is saturation, which puts a very heavy constraint to maximum current*inductance = flux. You always have to include ripple losses and peak values. To calculate ripple, just consider that inductance current rises with a slope equal to VL/L (VL = voltage across inductor), and assume input and output voltages are constant (you will add some capacitors there). So IL maximum excursion during a period will be (Vin-Vout)/L* max_dutycycle.
Usually, magnetic core producers explain how and give some tools to calculate inductor characteristics given your constraints, but this is a difficult matter.
« Last Edit: October 11, 2010, 10:23:10 pm by scrat »
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Offline michel75

  • Newbie
  • Posts: 1
Re: Electronic dummy load questions
« Reply #48 on: October 14, 2010, 05:34:09 am »
The above Information is very helpful for my project.The channel 1 is at the gate of the mosfet and the second at the drain....These are typical performance, Usually one of the regulated outputs is monitored and fed back. Some fraction of the output capability is the necessary load. After that, it's all a guess about which output and what load will keep your switcher running.


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