Help in reduce circuit losses

GH Crash said:

I'm looking at the DMN2040 data sheet and trying to determine if there might be a better FET for the job.

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You are looking for the on-resistance value in the data sheet (example below):
That allows you to calculate the drain-source voltage drop for a given drain-source current.

My feeling are pretty tough. A few negative comments will not even register. What does register, like an earthquake, is any comment that could be taken as derogatory. It is easy for those with knowledge to speak to a less knowledgeable person in a way that the less knowledgeable feels slighted. Also, it should be remembered that all the standard electronics measuring tools may not be available to non-electronics person. Short curt answers or questions often lead misunderstandings. Maybe we all should all watch how we say things.

Post #28 doesn't seem to address measuring loss across the switch. The attachment to the original post shows a switch. FYI; Switch resistance, voltage loss across the switch was determined and accounted for.

If I have said that I will follow, or have followed, the direction given, why is assumed that I haven't?
If I say that I know nothing, then indicate that I know a little about one thing, is that sufficient reason to assume that I’m completely knowledgeable in the subject.
If I say resistance and someone else says voltage drop, isn't it likely that we are talking about the same thing but each using a language, or term, that is familiar to them.

I apologize to everyone for getting off subject. I really do know very little about electronics, and it is sometimes frustrating when thing seem to get buried down in details that I don't understand.


I said that we had 0.2v unaccounted for. By that I meant that the voltage loss in a particular area was higher than expected, or calculated. I also said that we were checking the ATTiny85 coding. That in itself indicates that it appears the unaccounted voltage loss may be the FET not being fully on, duty cycle less than 100%.

Thanks danadak, So that number isn't of real importance in this application?


Again, I apologize.

Thanks danadak, So that number isn't of real importance in this application?

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Used as a simple switch I would think not of concern.


Regards, Dana.

crutschow said:

You are looking for the on-resistance value in the data sheet (example below):
That allows you to calculate the drain-source voltage drop for a given drain-source current.

View attachment 147183

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Thanks. I suspected that was important. I was wondering that since this FET is being used in a PWM application if other factors, like turn-on and turn-off times might also be important.

GH Crash said:

being used in a PWM application if other factors, like turn-on and turn-off times might also be important.

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Yes, but usually of secondary importance, except for high PWM frequencies above 20kHz or so.
The rise and fall times are mostly determined by how fast the gate drive can charge and discharge the large MOSFET gate capacitance (total gate charge).
So for a given gate-drive current, a MOSFET with a lower gate capacitance would be faster.

GH Crash said:

Any suggestions on how to reduce the resistance of the attached circuit.

A supercapacitor (EDLC) is used to power a small motor. There is an ATTiny85 chip in the circuit that is used as a timer and a PWM controller for the motor. The problem that I'm loosing about half a volt between the capacitor's voltage and the voltage across the motor's leads. I need to minimize losses in the circuit as much as possible.

Can you suggest means to reduce voltage loses to the motor?

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I am not reading this thread, but you could check the datasheet of the capacitor and check for series resistance?

Lightium said:

I am not reading this thread, but you could check the datasheet of the capacitor and check for series resistance?

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I can, but realize that the capacitor is not a part of the circuit we are discussing, as different capacitors are used at different times. Also, series resistance is capacitor specific, that is different capacitors (different capacitance and/or different brands) have different resistances.

Hello everyone, I read this post and I've several observations in the analysis of the schematic and the PCB.
The choice of the type of Mosfet used seems inadequate to me, why a channel P rather than a N? If we look in more detail may see that there is a permanent circulation of current via the motor and the construction diode of the mosfet, little by little the super capacitor is unloaded even if the FET is "off". The use of an N-FET eliminates this unnecessary consumption (blocking diode).
Regarding the on/off switch, why not benefit from the Reset pin of the Atiny85, the Pin 1 of the UC maintained at ground via a switch places the UC in standby mode and in fact consumes only very little, a few uA in standbye mode, that eliminate another serial resistor on power traces.
Then C4 utility? We have a 10F super-cap the 100nf seems illusory.
C5 is already in parallel with the 667PF MOSFET caps, but why not if PWM pulses are slow.
The time constant R4/C3 seems huge (without the UC code impossible to know what PB4 does?)
Another concern in the design of the PCB, the traces seems to be of small section for power and ground, why not put a ground plane eliminating the connections layer-1/Layer-2 and allowing to have more important width track section?
Hope this help.

Hi Vidalv,

First off, I appreciate your comments very much.

The drawing is incorrect. The MOSFET used in the drawing should have been an N-channel to accurately match the real circuit. My fault for not catching that before posting the drawing.

The switch was placed in the main line to completely isolate the capacitor when not in use.
Agree, C4 is extraneous. (Original circuit testing was done with a bench top power supply.) The C4 traces still exist on the circuit board, but the capacitor is not installed.

"C5 is already in parallel with the 667PF MOSFET caps, but why not if PWM pulses are slow." I don't understand. Could you elaborate?

Someone else pointed out the issue with R4/C3. The original 22k R4 has been replaced with a 100 ohm resistor. Do you agree with that change? What values would you suggest for R4 and C3?

Strictly out of curiosity, how did you determine seemed too small? (I ask solely to increase my knowledge.) On the actual circuit board, the traces from the capacitor to the motor connections are 0.5mm in width, all others are 0.25mm. Does this sound adequate?

The idea of a ground plate was considered but decided against as one of the goals of the project was to design as small and light a circuit as possible. It was felt that a copper ground plate would be significantly heavier than separate power traces.

Your complete circuit analysis has been most helpful. Please continue to contribute to this forum thread.

GH Crash said:

It was felt that a copper ground plate would be significantly heavier than separate power traces.

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Highly doubt that the weight of a copper ground plane would be a significant percentage of the total system weight.

Your welcome @GR Crash,

Can you post the new schematic please, just to have the same release to talk about.

Your main concern is the loss of energy, you know the Ohm law, the greater is the resistivity, the more you get a loss due to the Joule effect. To reduce at the maximum this, you need to use larger traces possible, on the actual schematic you can imagine this :

Ok, I agree this is just an iteration, just with 1A circulating in the circuit you loose 0.1V, with 10A this is 1V. I agree with @crutschow, enlarging traces and adding a ground plate do not change dramatically the total weight but reasonably reduce loss.

If you look on the Mosfet datasheet, by construction, this is physical, parasitic capacitor are present in the device (like intrinsic diode) :

More you add a capacitor in parallel of this, more you increase this total capacity.

R4/C3 is a filter, Time constant is : T=R*C, please have a look here : https://www.digikey.fr/fr/resources/conversion-calculators/conversion-calculator-time-constant
With 100Ω and 10uF = 1mS but with 22KΩ and 10uF this change to = 220mS !
This dependant from what you need exactly!

For this point : "On the actual circuit board, the traces from the capacitor to the motor connections are 0.5mm in width, all others are 0.25mm. Does this sound adequate?"
Why your traces are even small ? For the signal command why not, but for power put the maximum possible to reduce power loss.
Regards.
Vincent.

crutschow said:

Highly doubt that the weight of a copper ground plane would be a significant percentage of the total system weight.

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You'd be "surprised", copper on a PCB is 25mg/cm2. That will cut some fraction of a second from battery life.

"copper on a PCB is 25mg/cm2" : This PCB is around 0.16cm2 that dramatically increase the total weight of the project ...
No, seriously what are we talking about?
What is the total weight of the entire project?
The main subject is to reduce the power loss, not to reduce the weight I think ... right?

vidalv said:

"copper on a PCB is 25mg/cm2" : This PCB is around 0.16cm2 that dramatically increase the total weight of the project ...
No, seriously what are we talking about?
What is the total weight of the entire project?
The main subject is to reduce the power loss, not to reduce the weight I think ... right?

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I'm not going to do the math for the OP, he clearly said he didn't want that kind of help, he could do it. It's just surprising that the simple math is being ignored. Power-to-weight ratio is the key. Giving up milliVolts of is squared in the power function for milligrams of copper that decrease ohmic losses. Easy call when power losses are the concern - for now.

Sorry for my absence. Life got in the way of play.

The first thing. We found a flaw in the PWM coding. In effect, it was only going to 80-85% duty cycle. That has been corrected, and now the voltage loss is now much closer to what is to be expected from the circuit.

Now it is time to see what can be done to reduce system losses. It is also where comments about circuit improvements are welcome.

Attached is the circuit diagram for the circuit as it presently exists. The big change is that a FET was added to replace the original high resistance mechanical one. The black rectangle at the top of the drawing is a connector where a stand-alone programming board is plugged in for setting the desired time and speed. (The programmer is only attached to the CMC board to adjust settings, then it is removed.) The capacitor is not part of the circuit.

I will try to respond to your last week's posts shortly

Hello, the wiring of the switch seems too much complicated (for me) and the use of another MOS-Fet is not required, why would like to add 33 milliohms (minimum) to the loss circuit? Generally the power switch cut the positive line, not the ground, but why not!

Vidalv,

(Post #56) The previous mechanical switch had a contact resistance of 70 milliohms and was only rated for 300 milliamps. The rest of the circuit is rated for 2 amps or greater. A Fairchild FDMS8025 FET has a Resistance of 3.5 milliohms. There is no particular reason for the switch in the negative leg, it just as easily could have been in the positive leg.

I do not understand your ”wiring of the switch seems too much complicated” statement. How you would have done it? (The C6 capacitor is noise suppression when a bench top power supply is being used for testing.)

(Post #53) I’m not sure where you came up with the surface area of 0.16cm2 for the circuit. This circuit, circuit board and all components measures just over1/2 inch by 3/4 inch and weighs the same as a $1 US dollar bill. The circuit will be used where even at that weight it will increase the weight of the aircraft by 20-25%. By my calculations, a copper ground plate would add about 6-7% to the weight of the circuit.

I concur with making the +/- current carrying traces larger. The power traces on the current printed circuit board have been enlarged. Although I feel, but have not proved, that the losses due to the original 0.5mm wide traces was going to be less than the wiring from capacitor to board and board to motor, just due to their lengths. (Yes, larger wire would reduce losses, but would also increase the weight.)

Knowing what time constant I need is where my ignorance of electronics shows. (R4 has been renamed R6 on the latest circuit drawing.) Can you provide some guidance?

Hello,

I am surprised that the power switch is only 0.3A, for what potential?

"FDMS8025 FET has a Resistance of 3.5 milliohms" Yes for 4.5V but at 21A! Adding solder connections is probably more.

By convention, the negative pole of a circuit is also named by the term "Ground" is often used to refer to a common reference point for voltage in a circuit. In many circuits, the negative terminal is often connected to ground (0 volts) to provide a reference point to measure the voltage. This ensures the safety and stability of electrical systems.
Why the negative is the ground:
- The ground serves as a return path for the current and a reference point for voltages in a circuit.
- In many DC circuits, especially in battery supplied systems, the negative terminal is generally at a potential lower than the positive terminal. The landing of the negative terminal helps to establish a common reference point for all the components of the circuit.
Well, this is up to you, it's just a convention!

I understood for C6, my concept should be to cut the positive using only a switch able to support 5V and 2A (or more).

I make a mistake on my surface calculation, no real dimension on the size of the PCB sorry, BTW, take a new PCB with full copper on both sides and weigh it, do it for your current PCB and you will have a good idea of weight difference. It's up to you to see if x% of weight increase is less than x% flight autonomy. Same for the increase in traces. (Is the flight weight in pico-gram )

For the time constant, it is determined by R6 (22K) and C5 (100nF) which gives, more or less 2.2 ms, sorry to send you back the question, but why do you put this on PB4 of the UC? The key is probably in your code, but without it, impossible to understand the reason.
My supposition, If PB4 is an input, and PB1 an output, makes it possible to consider this R/C for power on reset (POR), to be sure of the UC state at startup. In this case, are 2.2 ms sufficient or not? The Attiny turns below 100us, it should be enough!
You agree?

V,

The previous mechanical power switch was used because of years of experience with current flows greater than it’s 5v, 0.3a rating. No power switch of suitable amp rating (~3a) could be found with a suitable footprint, 4x9mm.

The FDMS8025 was just an example of a FET with low resistance. It is used in many of the lipo battery protection circuits.

The PB4 (pin 3 of the ATtiny85) is a feedback leg that is used to constantly adjust the PWM duty cycle so that the voltage across the motor remains constant even as the capacitor voltage drops. So, I think that it is basically correct to assume PB4 as an input that affects PB1’s output. Beyond that, I do not know enough to voice an opinion, especially as to what time constant would be appropriate.

Anyone out there have an opinion they want to share?

George

Thanks George for answers, BTW could you post the code (if not confidential)? Or send by Private Message? Or just the part to understand retro-action from filter and PWM, I'm not a fluent coder but just to have an idea.
Vincent.