Yea i saw Dave's approach
i want to use MCU cause i'll do some other stuff with it like logging, also i already have bought that 7 power resistors and i want to use them. Any feedback on my approach?
So far i've made a network of 7*25watt 10 Ohm resistors in parallel. Thats 175Watt max and 0,7Ohm Rtotal in theory but i plan to reduce it to 150 max or maybe less for safety reasons
Yes, you can definitely PWM a resistor to create an intermediate resistance.
Example: If you PWM a 20 Ohm resistor on 50% duty cycle, you can get 40 Ohms
If you use a bank of parallel transistors, I'd rather place MOSFETs instead of BJTs, because of the BJT's positive thermal coefficient (the more they heat, the more they conduct, so if one is more conductive, it will burn almost surely, followed by the others).
Inductor smooths current waveform, since otherwise it will be a square PWM wave.
A 0 - V PWM current wave will not dissipate on the resistance the same power as a DC current equal to its average (this is clear if you consider its rms value).
I don't agree on the fact that inductance limits current, since you are not forced to make your PWM "converter" to work in discontinuous mode (eg with current going to zero at each switching period). So I think current can be as high as desired, provided that the inductor used is rated for that (I'm thinking about a boost topology).
Chopping a resistor could be a good idea. To explain the sentence from the other forum, simply think at current and voltage, instead of resistance: at the battery side, voltage is fixed, while average current depends on the time the load is connected.
But on the output side not the input! When the PWM switching transistor shuts off, it also shuts off the current demand from the power supply. ... Sure you have a nice smooth load output, but that isn't important. ...Of course a linear load will be better in terms of precision, dynamics and noise, but it will be very large above certain power values (over a few kW could also be impossible to make it linear)
You must switch off current flow to the inductor to prevent it from saturating. You can't operate an inductor in continuous DC mode because it will quickly saturate and it will lose all impediance, leaving just the DC winding resistors. In DC mode, and Inductor operates as a power resistor.
When you turn off the switching transistor the current load disappears on the source side. A PWM driving into a resistor bank will create average power load. Current and voltage fluxuate with the switching transistor.That's true if you consider a buck topology, but for a boost it isn't. In a buck, the transistor is in series with the input, so of course current from the source is discontinuous. Otherwise, in a boost, input is in series with the inductor, so current can be maintained above zero for all the time (provided that an appropriate switching period is choosen).
That's true if you consider a buck topology, but for a boost it isn't. In a buck, the transistor is in series with the input, so of course current from the source is discontinuous. Otherwise, in a boost, input is in series with the inductor, so current can be maintained above zero for all the time (provided that an appropriate switching period is choosen).
Otherwise, switching converters won't have any Continuous Conduction Mode, and will only work in DCM (Discontinuous ...)
Now you have convinced me that a dummy load in linear control is the right way for exodia's issue, but I'm pretty sure that big loads (such as for large solar converters testing) are made with switching mode converters (there are few components who can withstand very large continuous power in linear mode).
FS7SM16A voltage drop across resistors 6,24V I consumed 1,2AI haven't really followed this thread, but these voltage drops are consistent with the 5 ohm (three times 15 ohm in parallel) resistance. 5 ohm * 0.62 A = 3.1 V. If you want a lower drop, lower the resistance.
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?
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
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
| 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 |
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.
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.
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 (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.
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.
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?This "voltage drop" is really due to both Ohm's law and the partialization you make through PWM.
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 128Your 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 BJTThat 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.
Solution: At 20kHz, there is a MOSFET switching occurrence every 25 microseconds (a switch on every 50 microseconds, and a switch off every 50This 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.
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.
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.