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Offline exodia

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Electronic dummy load questions
« on: August 13, 2010, 08:06:40 PM »
Hello there

I'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 reasons

Also 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 MCU

enough questions for 1st post

Thanks in advance guys
 

Offline Simon

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Re: Electronic dummy load questions
« Reply #1 on: August 13, 2010, 10:48:51 PM »
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.
The ĀµCurrent has landed in Europe and now also selling Probe Master probes: http://www.sparkylabs.co.uk/shop/index.php?cPath=30
 

Offline exodia

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Re: Electronic dummy load questions
« Reply #2 on: August 13, 2010, 11:44:31 PM »
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?

 

Offline TechGuy

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Re: Electronic dummy load questions
« Reply #3 on: August 14, 2010, 04:48:42 AM »
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?

Well 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.



 

Offline toli

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Re: Electronic dummy load questions
« Reply #4 on: August 14, 2010, 04:05:53 PM »
Quote
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

The way I see it, 7 resistors of 10ohm in parallel are 1.43Ohm :)
 

Offline exodia

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Re: Electronic dummy load questions
« Reply #5 on: August 16, 2010, 04:24:27 PM »
@toli yes you are correct, thanks for correcting my excel calculator :)

@TechGuy i'll do some more research

Edit 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?
« Last Edit: August 16, 2010, 06:24:20 PM by exodia »
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #6 on: August 16, 2010, 08:38:12 PM »
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...

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Offline scrat

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Re: Electronic dummy load questions
« Reply #7 on: August 16, 2010, 09:00:54 PM »
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...
One machine can do the work of fifty ordinary men. No machine can do the work of one extraordinary man. - Elbert Hubbard
 

Offline exodia

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Re: Electronic dummy load questions
« Reply #8 on: August 16, 2010, 10:08:22 PM »
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?

Quote
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

What 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?
« Last Edit: August 16, 2010, 10:28:27 PM by exodia »
 

Offline TechGuy

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Re: Electronic dummy load questions
« Reply #9 on: August 17, 2010, 07:00:05 AM »
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).

Not 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.



 

Offline scrat

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Re: Electronic dummy load questions
« Reply #10 on: August 18, 2010, 06:26:26 PM »
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.

@TechGuy
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. So
equivalent (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)?
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Offline exodia

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Re: Electronic dummy load questions
« Reply #11 on: August 18, 2010, 08:14:09 PM »
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.
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #12 on: August 18, 2010, 10:15:39 PM »
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).
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Offline TechGuy

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Re: Electronic dummy load questions
« Reply #13 on: August 19, 2010, 12:37:02 AM »
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).

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. 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.

Quote
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).

An 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.


Quote
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 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.



 

Offline scrat

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Re: Electronic dummy load questions
« Reply #14 on: August 19, 2010, 03:03:00 AM »
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.

In 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 ...)

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).

However, the doubt remains if a switching load (chopped resistance) could be an adequate testing device. I'm not so experienced to say this ;)
« Last Edit: August 19, 2010, 03:49:50 AM by scrat »
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Offline TechGuy

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Re: Electronic dummy load questions
« Reply #15 on: August 19, 2010, 12:12:15 PM »
Quote
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).

In 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.shtm

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


Quote
Otherwise, switching converters won't have any Continuous Conduction Mode, and will only work in DCM (Discontinuous ...)

The 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.

 
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #16 on: August 19, 2010, 11:14:45 PM »
@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.
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Offline exodia

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Re: Electronic dummy load questions
« Reply #17 on: August 20, 2010, 02:03:44 AM »
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 proteus

Thanks
 

Offline TechGuy

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Re: Electronic dummy load questions
« Reply #18 on: August 20, 2010, 04:16:43 AM »
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).


I 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.
 

Offline exodia

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Re: Electronic dummy load questions
« Reply #19 on: September 13, 2010, 12:38:45 AM »
Hello again fellow members

I'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 curves :( or any hints on how to work it on active mode



On 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 mode

Can 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's

Thanks in advance
 

Offline scrat

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Re: Electronic dummy load questions
« Reply #20 on: September 13, 2010, 08:54:45 PM »
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...
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Offline exodia

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Re: Electronic dummy load questions
« Reply #21 on: September 14, 2010, 03:01:07 AM »
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
 

alm

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Re: Electronic dummy load questions
« Reply #22 on: September 14, 2010, 08:04:24 AM »
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?
I 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.
 

Offline Time

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Re: Electronic dummy load questions
« Reply #23 on: September 14, 2010, 08:11:57 AM »
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.
-Time
 

Offline exodia

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Re: Electronic dummy load questions
« Reply #24 on: September 14, 2010, 05:28:56 PM »
@Time

The software it's proteus, i'll try that 10K pull down resistor

@ alm

The 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
 


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