Electronic dummy load questions - Page 1
Forum section: electronics / projects

author: exodia
author_rank: Contributor
timestamp: August 13, 2010, 10:06:40 am
content: Hello thereI'm into building an electronic dummy loader. I want to describe my idea and thoughts and tell me what you think about.I'm planning on making a power loader circuit to test the durability of Photo voltaic batteries according to different loads.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 reasonsAlso i'm thinking of using a max4173 IC for sensing current draw from resistor network. Any other alternatives?My main question is if i can use PWM with 2-3 logic level gate mosfets to get whatever wattage i want instead of 25-50-75-100-125...* and if yes what kind of mosfets and what precautions for interferences from PWM should i take account for? The voltage will be about 12-14 and the amps about 10 so i think an IRLZ44 will do the trick?The circuit will have the battery then the charge controller which will have only the battery installed and not the PV panel and should make output voltage about 12 , then my circuit and finally the resistor network.one last thing i want to know if i can use separate voltage for the mcu and the various ic's instead of using voltage from the battery. My concern is that if i use separate voltage for mcu i will have two different grounds one from battery to resistor network and one from the supply of mcu. I'm wondering if i tie the grounds together because they are on different voltage and maybe the 12 volt rail due to pwming it will be noisy will interfere with my MCUenough questions for 1st postThanks in advance guys

author: Simon
author_rank: Unknown
timestamp: August 13, 2010, 12:48:51 pm
content: why not have a look at daves recent blog about a dummy load he made ?, I think you have a bit to learn before going head long into it.

author: exodia
author_rank: Contributor
timestamp: August 13, 2010, 01:44:31 pm
content: Yea i saw Dave's approachi 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?

author: TechGuy
author_rank: Regular Contributor
timestamp: August 13, 2010, 06:48:42 pm
content: Quote from: exodia on August 13, 2010, 01:44:31 pmWell using a PWM to control the load might cause some issues with measurements, since your load won't be constant. I believe that most common adjustable dummy loads use a bank power BJT transistors operating in linear mode to adjust the power load. Really all you need is a set of power BJTs attached to a heatsink and a potentimeter to adjust the base current to the transistors to control the load. You can use your power resistors in series with the power BJT's avoid adjusting the base current into saturation. The Power resistors would limit the max current the device would handle. Just make sure that the power BJT's you buy have a high operating voltage incase you wish to re-use it to test power supplies with higher output voltages.

author: toli
author_rank: Frequent Contributor
timestamp: August 14, 2010, 06:05:53 am
content: QuoteThe way I see it, 7 resistors of 10ohm in parallel are 1.43Ohm

author: exodia
author_rank: Contributor
timestamp: August 16, 2010, 06:24:27 am
content: @toli yes you are correct, thanks for correcting my excel calculator@TechGuy i'll do some more researchEdit Dave's approach at the last mins of video is to connect the mcu to a voltage follower though PWM and a smoothing filter. That way with 50%duty cycle the op amp will get say 2,5 volts or a couple of 5/0 pulses???. The mosfet will conduct fully at 2,5 volts? BTW does pwming a mosfet that is connected to bank of parallel resistors affects the Rtotal value according to pwm?

author: scrat
author_rank: Frequent Contributor
timestamp: August 16, 2010, 10:38:12 am
content: 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).A good dummy load for measurement should be linear, not PWM, but it will be huge if power is quite high. PWM seems to me a good solution (especially if you plan to use a converter on the real system).So, filter inductance * PWM period becomes the real matter. Higher values will make current more constant.The issues with low inductance or low frequency values will be on the sensing side (current will be a heavy triangular wave added to a DC, but this can be filtered out), and on the measurement side, since the triangular ripple causes added losses (it has null mean value, but not null RMS value). The total load for the battery will be more than expected (you must take into account battery internal resistance and inductance ESR)...This can or can not be a trouble, depending on the application. If you plan to convert battery voltage with a switching converter, a good thing could be to reproduce the same conditions in terms of ripple on the load, so the measure of duration will be very accurate.Maybe I went a little off topic, or said thing you already know...

author: scrat
author_rank: Frequent Contributor
timestamp: August 16, 2010, 11:00:54 am
content: BTW theoretically PWMing a power supply and extracting from it a certain average current is the same as using a resistor which draws the same current. Pratically, the troubles are with series resistance...

author: exodia
author_rank: Contributor
timestamp: August 16, 2010, 12:08:22 pm
content: I'm thinking of using a MOSFET driver for better compatibility and to be sure i can drive the mosfets at higher frequency than 20Khz to avoid noise. I don't think i have to use an op amp like Dave's approach. My plan is to prompt the user the wattage he wants, then i can take a reading of the current from a max4173^2 and divide it with the Rtotal in parallel which i know. That will give me the power. The i will make adjustments in PWM and fortunately i can come really close to the wattage requested. That is at least my theory, i don't know how close to implementation is that.At home i'll try connecting a 7,5 ohm 30watt bank to 2 AA batteries and measure the current with an ohm meter (only think is will the ohm meter be able to measure current from pwming the mosfet, don't know if RMS capable meter has to do with it)BTW i get the following quote from another forum but i don't know if it stands real, can anyone confirm?QuoteWhat i have understood from PWM is that it chops the time source is fully on or fully OFF and if the frequency is high enough the load gets the average voltage. Haven't heard anything about resistance choping.My only thinking is because of Ohm's law is that you can somehow simulate the lower voltage getting from PWM by doubling the resistor?

author: TechGuy
author_rank: Regular Contributor
timestamp: August 16, 2010, 09:00:05 pm
content: Quote from: scrat on August 16, 2010, 10:38:12 amNot that big of deal when operating in linear mode. It's fairly easy to set up a feedback loop to regulate current to offset temperature changes. This can be done using a current sensor and a comparator that compares the current sensor voltage with a pot output voltage. The current sensor could be a low ohm resistor that monitors the voltage drop across the resistor.Mosfet are voltage controlled devices, BJT are current controlled devices. It will be easier to design an adjustable current sink using a current controlled devices than voltage controlled devices. Mosfets are usually very sensitive to minor voltage noise when operating in linear mode. Your control circuit will need to be powered using a very low noise power supply to prevent significant current fluxuations. Even a 50 mv noise at the gate voltage can cause significant impendance changes when using an MOSFET in linear mode.The issue with PWM is that it will be difficult to smooth the current demand on the source side. A LC filter will smooth out the load side, but the source side will remain fairly choppy, because the switching transistor will transition from infinite resistance to near zero resistance. The inductor will also make it difficult to adjust the load with a fixed frequency because the inductor will have a fixed value. You can only push so much current per cycle through a fixed inductor with a fixed frequency. A low value inductor will permit a lot of current to pass, but it will also make the source input current a lot more choppy because it will have a low impedience. A high value inductor will have a higher impediance, but will limit the power you can push through.In my opinion, A linear mode sink is really the only way to design a simple adjustable load dummy.Choosing to control the load using a PWM or MOSFET in linear mode is just going to make the project a whole lot more complex.

author: scrat
author_rank: Frequent Contributor
timestamp: August 18, 2010, 08:26:26 am
content: I must admit I didn't think much in deep on what I wrote..About MOSFETs instead of BJTs... I was thinking of dissipating all the power needed on the transistors, but now I realize it will be quite a silly approach, since it will have no advantage over the linear one (the power rating of transistors will have to be the same as for linear control). If transistors are used to drive resistors, the issue with thermal drift doesn't exist (negative feedback from emitter resistance), but this surely limits power.@TechGuyI 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. Soequivalent (average) resistance = source voltage / average current = source voltage / (source voltage / load resistance * duty cycle) = load resistance / duty cycle.The same is for dissipated power: why will you use current and resistance measurement to obtain power? Why don't you use voltage and current, which are both easily measurable (and cut off resistance thermal dependence)?

author: exodia
author_rank: Contributor
timestamp: August 18, 2010, 10:14:09 am
content: As a matter of fact i was thinking exactly why use resistor and amps to make the total power when i read about the thermal coefficience of the resistor. Guys why should i use an inductor. I'll connect the PWM output to a pair of bjt's or driver that will drive the MOSFET gate.

author: scrat
author_rank: Frequent Contributor
timestamp: August 18, 2010, 12:15:39 pm
content: 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).

author: TechGuy
author_rank: Regular Contributor
timestamp: August 18, 2010, 02:37:02 pm
content: Quote from: scrat on August 18, 2010, 12:15:39 pmBut 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. When the switching transistor switchs off, the inductor supplies the current to the dummy load, by draining the magnetic field stored in the inductor's core. Sure you have a nice smooth load output, but that isn't important You can't really use a PWM if you want a smooth constant current drain on a power supply.QuoteAn unsaturated inductor has impediance (ie resistance) that limits current flow. The higher the inductance the higher the impedance. The only way you can adjust the impedance of a fixed inductor is to have a DC winding so that you can apply current to hold the inductor at a point closer to saturation causing the inductor's impedance to fall. But this gets tricky. Usually it done with a second core with a DC winding in close proximity to the inductor. The magnetic field alters the inductance of the inductor. this is how magnetic amplifiers work.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.QuoteBut not on the source (power supply side) side. 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.When testing a power supply you want a stable adjustable current demand from the load dummy. How smooth the output is, is irrelavent.Try for yourself, build a simply Buck regulator using a 555 or a waveform function generator as your PWM controller. Then take a look at the waveform from the transistor high side (assuming you use either a NPN BJT or an N-Channel Mosfet). See what the output looks like. Really you need a current probe so you can see the current flow as the switching transistor cycles on or off. You will also need to apply some output load on the buck regulator, enough so that it drops the voltage across the power supplies output caps.

author: scrat
author_rank: Frequent Contributor
timestamp: August 18, 2010, 05:03:00 pm
content: Quote from: TechGuy on August 18, 2010, 02:37:02 pmOf 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)Quote from: TechGuy on August 18, 2010, 02:37:02 pmIn switching control you should be controlling current properly (to reach the desired mean value), but even if you don't control current, there is a steady-state value in which average current will settle. Otherwise, switching converters won't have any Continuous Conduction Mode, and will only work in DCM (Discontinuous ...)Quote from: TechGuy on August 18, 2010, 02:37:02 pmThat'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).However, the doubt remains if a switching load (chopped resistance) could be an adequate testing device. I'm not so experienced to say this

author: TechGuy
author_rank: Regular Contributor
timestamp: August 19, 2010, 02:12:15 am
content: QuoteIn a Boost Circuit, the inductor current oscillates with the switching transistor. Current demand to the Power supply follows the Inductor current:http://www.powerdesignersusa.com/InfoWeb/design_center/articles/DC-DC/converter.shtmThe key issue is that you can't operate a boost circuit beyond a 50% duty cycle.QuoteThe only switching converters that come can operate at or very near continunous Conduction Mode are Full H-Bridge Transformers SMPS (very near continuous) and Multiphase Buck/Boost/etc SMPS (where several converters switch on and off during different periods of the switching cycle) (can be countinous if the phases over lap). But over lapping phases will not have steady current demand on the source side, unless you get the timing perfect, the inductors and switching transistors are perfectly matched.

author: scrat
author_rank: Frequent Contributor
timestamp: August 19, 2010, 01:14:45 pm
content: @TechGuy:I'm very pleased to learn from people who know in depth the matter...Thanks.Although I studied quite in depth power electronics, my practical experience is very limited.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).It is clear that, in a boost, input current is inductor current, so it has a triangular ripple due to switching (low rising slope and higher negative slope).However, as far as I know and to my reasoning, a step-up can work in CCM and so over 50% duty, but a more sophisticated controller must be used, while in DCM or border-line (variable frequency) current control is quite simple (higher duty, higher average current). The reason for that is in the dynamical behaviour (phase inversion at start of buty-cycle changing), but this could not be a problem, or can be compensated by an adequate controller.Does it make sense to you?@exodia:I apologize if I'm going a little off topic, hope it is interesting also for you.

author: exodia
author_rank: Contributor
timestamp: August 19, 2010, 04:03:44 pm
content: I quite lost u guys. I'll study abit more the answers and i'll try some simulations on proteus. I'll return for help that maybe i need with proteusThanks

author: TechGuy
author_rank: Regular Contributor
timestamp: August 19, 2010, 06:16:43 pm
content: Quote from: scrat on August 19, 2010, 01:14:45 pmI have a 5KW Programmable electronic load that consists TO-3 BJT transistors. Its a rack mount unit designed for testing Battery banks, and other DC sources and weighs about 40 pounds.

author: exodia
author_rank: Contributor
timestamp: September 12, 2010, 02:38:45 pm
content: Hello again fellow membersI'm thinking of smoothing out the PWM i sent to the switch with an RC filter and feed that to a transistor or mosfet. If i understand correctly i must work the "switch" at it's active region as many have suggested. However they are some stuff that don't understand from the datasheets and how to work a transistor at it's active mode and here i want your help.I remember from university that in the VCE/IC curves we had the base current curves and then we selected the base current to be at the middle of the plot and according to the collect voltage and current we wanted and that was the best active mode region to work. Unfortunately i don't know how to read the datasheets and decide how to bias the BJT for active or switch mode. For example i pick a 2N6547 I can't find that VCE/IC curvesor any hints on how to work it on active modeOn the mosfet side if i understand correct then i should search for a VGS/VDS or ID plot just like Dave suggested in the episode?For example i select the HUF75339G3 but how to make it work on active modeCan someone help me understand once and for all how to read BJT and MOSFET datasheets and how to select the various working modes on BJT's and MOSFET'sThanks in advance

author: scrat
author_rank: Frequent Contributor
timestamp: September 13, 2010, 10:54:45 am
content: I think that when using a transistor (either BJT o MOSFET, they both are tranistors) for power applications in linear region the most strict boundary would be maximum power.In fact, switching is used so frequently because there we can use transistors in their most efficient regions (as a closed or open switch).Take a look at the Ic/Ids vs Vce/Vds graph. It shows you that for a fixed input Ib/Vgs current follows a certain curve when Vce/Vds varies. The simplest approximation of a transistor is a controlled current generator. Output (Ic/Ids) current increases at input (Ib/Vgs) increase. This is almost true in a portion of the graph far from the two axes. Near the vertical axis it seems like a closed switch (so has low voltage drop), while near the horizontal axis (where it goes if input is low, just look at the curves which pass near Vce/Vds axis) it seems to an open switch.On that graph you can trace a voltage limit (a vertical line which passes at Vbreakdown), a current limit (a line that passes at Imax) and a power limit (an hyperbola which is on the high-right corner). You must remain in between. While with switching control you "jump" from near the horizontal axis (OFF, no current for any voltage) to near the vertical axis (low voltage for any current value), in linear mode you just make a (PI?) regulator to choose a value for input (Ib/Vgs) so that output current (or voltage) is the one you need. By taking typical values for amplification gain (hfe/gm) you can guess where your bias point should be to achied a certain output, but gains are usually varying very much from one device to another within the same model. So you need a regulator which makes output to stay close to the reference.Hope it can help...

author: exodia
author_rank: Contributor
timestamp: September 13, 2010, 05:01:07 pm
content: 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 usedFS7SM16A voltage drop across resistors 6,24V I consumed 1,2A2N60B voltage drop across resistors 4,3V I consumed 0,85AIRF840A voltage drop across resistors 3,1V I consumed 0,62AWhy 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 driver2nd maybe the frequency of arduino PWM is too low only 490 HzI would like to hear your opinions

author: Unknown
author_rank: Unknown
timestamp: September 13, 2010, 10:04:24 pm
content: Quote from: exodia on September 13, 2010, 05:01:07 pmI 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.

author: Time
author_rank: Frequent Contributor
timestamp: September 13, 2010, 10:11:57 pm
content: I am not sure I understand whats going on with the voltage indicator in that schematic (the circle thing that says volts). It looks like its just shorted out... I don't recognize that software package so maybe its just me... or maybe not.Also, maybe you want a pull down resistor on the gate of your transistor.

author: exodia
author_rank: Contributor
timestamp: September 14, 2010, 07:28:56 am
content: @TimeThe software it's proteus, i'll try that 10K pull down resistor@ almThe mosfets don't have a voltage drop like transistors VCEsat?Why i got 3 different results from the mosfets? All mosfets are capable of at least 2A and the load resistor was constantly at 5?. Theoretically if the mosfet doesn't drop any voltage across VDS i should get 1,8=9/5 correct?Maybe they don't open hard enough, i should try the totem pole driver 1st

author: scrat
author_rank: Frequent Contributor
timestamp: September 14, 2010, 08:07:55 am
content: 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.

author: exodia
author_rank: Contributor
timestamp: September 14, 2010, 03:30:41 pm
content: 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?

author: Time
author_rank: Frequent Contributor
timestamp: September 14, 2010, 03:48:32 pm
content: 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.

author: scrat
author_rank: Frequent Contributor
timestamp: September 15, 2010, 10:41:08 am
content: 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 Ohm2N60B voltage drop across resistors 4,3V I consumed 0,85A -> Rds = (Vbus - Vr) / I = (9-4.3) / 0.85 = 5.5 OhmIRF840A voltage drop across resistors 3,1V I consumed 0,62A -> Rds = (Vbus - Vr) / I = (9-3.1) / 0.62 = 9.5 OhmIf 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.

author: exodia
author_rank: Contributor
timestamp: September 23, 2010, 05:51:14 pm
content: Sorry for talking that long to replyI 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?EDITOk 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

author: JohnShaft
author_rank: Newbie
timestamp: September 30, 2010, 10:44:33 pm
content: 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.Quote from: exodia on September 13, 2010, 05:01:07 pm

author: scrat
author_rank: Frequent Contributor
timestamp: October 01, 2010, 11:44:41 am
content: Quote from: exodia on September 23, 2010, 05:51:14 pmTypical 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.

author: exodia
author_rank: Contributor
timestamp: October 04, 2010, 06:15:05 am
content: I have an oscilloscope, what should i try and measure with it?

author: scrat
author_rank: Frequent Contributor
timestamp: October 04, 2010, 10:25:00 am
content: 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.

author: exodia
author_rank: Contributor
timestamp: October 04, 2010, 08:09:20 pm
content: I have made the following circuitThe voltage reading i get arePWM VDD VG Vsupply current Ohm0 12 0 12 0 2,710 11,08 0,5 12 0,345 2,820 10,15 0,96 11,99 0,697 2,730 9,23 1,45 11,98 1,051 2,840 7,99 1,96 11,96 1,505 2,750 6,95 2,45 11,94 1,872 2,760 5,93 2,94 11,91 2,252 2,770 4,91 3,39 11,9 2,648 2,780 3,8 3,9 11,87 3,063 2,690 2,83 4,39 11,85 3,383 2,7100 1,48 4,78 11,82 3,902 2,8VDD=Drain pin of mosfetVG=Gate

author: sonicj
author_rank: Frequent Contributor
timestamp: October 05, 2010, 04:36:45 am
content: 20 page discussion on the matter:rcgroups.com - How to build a constant-current dischargerit 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 b6logview<==open source battery logging software

author: scrat
author_rank: Frequent Contributor
timestamp: October 05, 2010, 07:59:07 am
content: 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 confusedEDIT: 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..

author: exodia
author_rank: Contributor
timestamp: October 05, 2010, 05:12:10 pm
content: Here you goThe 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 signalsThe mosfet i used is the RFP70N06The 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.@sonicjThanks for the links i'll study them

author: scrat
author_rank: Frequent Contributor
timestamp: October 06, 2010, 10:17:36 am
content: Quote from: exodia on October 05, 2010, 05:12:10 pm128W... 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.

author: exodia
author_rank: Contributor
timestamp: October 06, 2010, 04:42:19 pm
content: 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)

author: scrat
author_rank: Frequent Contributor
timestamp: October 06, 2010, 05:15:48 pm
content: OK, that's reassuringFrom the data you posted, there is not much matching with the datasheet curves.Inhttp://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.

author: scrat
author_rank: Frequent Contributor
timestamp: October 06, 2010, 10:36:03 pm
content: Quote from: AcHmed99 on October 06, 2010, 09:53:37 pmI 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.

author: scrat
author_rank: Frequent Contributor
timestamp: October 07, 2010, 08:56:25 am
content: 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.

author: scrat
author_rank: Frequent Contributor
timestamp: October 07, 2010, 10:18:30 am
content: @exodiaTo 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.

author: exodia
author_rank: Contributor
timestamp: October 09, 2010, 03:51:54 pm
content: Quote from: scrat on October 06, 2010, 05:15:48 pmI 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 transistorThe 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?04,148,3570312,5909817,999822,3516826,8213231,511236,3578140,0885546,12164Edit yes that a nice mosfet but much less power dissipation than the one i have selcted. It's only 65W and i need an 128I 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

author: scrat
author_rank: Frequent Contributor
timestamp: October 11, 2010, 08:29:38 am
content: Quote from: exodia on October 09, 2010, 03:51:54 pmYes, 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.Quote from: exodia on October 09, 2010, 03:51:54 pmThat'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).Quote from: exodia on October 09, 2010, 03:51:54 pmThis "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.Quote from: exodia on October 09, 2010, 03:51:54 pmYour 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!Quote from: exodia on October 09, 2010, 03:51:54 pmThat 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

author: exodia
author_rank: Contributor
timestamp: October 11, 2010, 06:27:56 pm
content: 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 paragraphProblem 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 andthe 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 50microseconds). 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) = 120Watts. Therefore the average switching loss is 120W x 0.04 = 4.8 Watts.T=1/f=>T=0,00005 secThat is a full period including a Ton and a Toff cycle, correct?Then 1 ms (duration of transition) /25 msec of the full periodIs the above calculations correct??i got them from the following pagehttp://homepages.which.net/~paul.hills/SpeedControl/MosfetBody.htmlNow 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?

author: scrat
author_rank: Frequent Contributor
timestamp: October 11, 2010, 08:14:52 pm
content: Quote from: exodia on October 11, 2010, 06:27:56 pmThis 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 t2Since 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.Quote from: exodia on October 11, 2010, 06:27:56 pmFor 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.

author: michel75
author_rank: Newbie
timestamp: October 14, 2010, 05:34:09 am
content: 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.