This is the expected value (25 V / 2.7 kΩ).
Check your multimeter. Looks like the 200 mA range is defect.
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Boosting the Amper and altering the frequency
- Thread starter Maxouishere
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Same thing happens with another multimeter. What else could be the issue here?
200mA slot: https://streamable.com/9qy15u
10A slot: https://streamable.com/nnxm6h
Your videos make no sense. You use the 200 mA input in both cases, but for measuring 10 A you need to use the 10 A input.
Is the output of teh step-up regulator stable (constant voltage with and without load)? Your wildly jittering measurement suggest the voltage may vary, too.
Is the output of teh step-up regulator stable (constant voltage with and without load)? Your wildly jittering measurement suggest the voltage may vary, too.
Oh, yeah. I thought It was hFe slot. And thought this multimeter uses the same slot for current measurement
I just saw the indicator line when you pointed out. I changed the 2.7kΩ resistor with 145Ω resistor. Now it measures the current.Current: https://streamable.com/qskhs1
Yes, voltage is stable. I changed the arduino code to max out the voltage at pin_8 just to see. Voltage is not flactuating.
Voltage: https://streamable.com/jf68uj
If you mean without connecting battery by saying "without load", it shows 0V.
But now there might be another issue. (I hope not.) 145Ω resistor which I used to replace the 2.7kΩ resistor, gets super hot. I tried to touch it after removing the battery. And I smell something burned but no smoke. Do you think if it is the resistor's smell when it gets super hot? Is this smell normal?
yes.
At 25 V the power dissipation in a 145 Ω resistor is P = V2/R = 625 V²/145 Ω = 4.3 W. You will need at least a 4.5 W resistor to withstand this power. Hot it will become anyway.
Do I need that resistor at collector end of the transistor no matter what? I mean, I will connect it to my frame (i will add the picture below) It has 341Ω resistance from one end to the other. Do I still need to add that 145Ω resistor?
P = 25^2/341 = 1.83W
Can I assume it like the above equation without connecting the 145Ω resistor?
And for the frame, I used 2x wires, one is +, the other is -. Please refer to above picture for resistance value.
No. Not "no matter what".
If you have a load (your resistive grid), the load will limit the current. But in your original setup you showed no load, only the transistor, multimeter and power source. That setup required the resistor to limit current.
no.
Measured cold? When hot the grid's resistance will increase, lowering the current. In this case that's good.
I measured the resistance at around 25 degrees Celcius room temperature, without any load. Therefore wires should be around the same temperature as well.
I measured the each resistive grid frame. Resistance for each frame varies because for 2 of them I used different wires (thinner) to see if it is any good.
I connected them parallel as you pointed out months ago. 25V and 0.8/5 = 0.16 A on each should be alright.
(Is this current calculation OK? I put there 0.8A because it was our initial wanted output current. or now it should be I=V/Req = 25/46.7 = 0.535A so around 0.11A on each frame. I am kind of confused with this.)
I will connect the current setup to the frames, when a replacement for my dead BD135 arrives. Thank you so much for your help so far.
Attachments
So far we had a single grid (load). No you have 5 in parallel which changes matters. Total collector current will now be 25 V / 46.7 Ω = 535 mA. A single BD135 has a min. current gain of 25 so you will need at least a base current of 525 mA / 25 = 21 mA.
Your circuit uses 138 Ω base resistance from the Arduino output to the base of the BD135. This results in a curent of (5 V -0.6 V) / 138 Ω = 32 mA. This is enough to drive the BD135, but too much for the Arduino. An Arduino's digital output can deliver only 20 mA continuously - which would not be enough to drive the BD135 (> 21 mA required, see above).
The solution: you need a driver stage like this:
Instead of a BC107 you can use any general purpose small signal NPN transistor like for example 2N2222. The 470 Ω esistor limits the current from the Arduino to (5 V - 2 × 0.6 V) / 470 Ω = 8 mA which is safe for the Arduino. Q1 amplifies this current sufficiently to drive the BD136 as in your circuit.
BY the way: the problem exists even with a single grid as load as the base current is limited by the two base resistors in series and will be too high anyway.
Your circuit uses 138 Ω base resistance from the Arduino output to the base of the BD135. This results in a curent of (5 V -0.6 V) / 138 Ω = 32 mA. This is enough to drive the BD135, but too much for the Arduino. An Arduino's digital output can deliver only 20 mA continuously - which would not be enough to drive the BD135 (> 21 mA required, see above).
The solution: you need a driver stage like this:
Instead of a BC107 you can use any general purpose small signal NPN transistor like for example 2N2222. The 470 Ω esistor limits the current from the Arduino to (5 V - 2 × 0.6 V) / 470 Ω = 8 mA which is safe for the Arduino. Q1 amplifies this current sufficiently to drive the BD136 as in your circuit.
BY the way: the problem exists even with a single grid as load as the base current is limited by the two base resistors in series and will be too high anyway.
I ordered both of them just in case.
To be sure, I will share the final schematic before applying any load. Thank you so much.
@Harald Kapp
One question about parallel connection.
My frames are connected like the picture below. They are parallel right now, right?
6x "DC male barrel to wire jack"s are wired together. I just insert them into the female barrel slots.
Below picture is a side view illustration. Grid only has 2 wires rolled around the wooden frame. one is + the other is -. They do not touch each other. And at the end of the frame, I nailed the ends of each wire to the wooden frame so that the wires remain stretched.
Do I also need to connect the ends of the + and - wire of each frame together, or the illustration in the first picture is enough for parallel connection?
One question about parallel connection.
My frames are connected like the picture below. They are parallel right now, right?
6x "DC male barrel to wire jack"s are wired together. I just insert them into the female barrel slots.
Below picture is a side view illustration. Grid only has 2 wires rolled around the wooden frame. one is + the other is -. They do not touch each other. And at the end of the frame, I nailed the ends of each wire to the wooden frame so that the wires remain stretched.
Do I also need to connect the ends of the + and - wire of each frame together, or the illustration in the first picture is enough for parallel connection?
I'm afraid I can't read your picture well enough. Not a problem of my eyesight nor of the clarity of the image, but what it represents.
It looks like the end of the "+" wire and the end of the "-" wire are not connected. In that case, how would current flow through the wires from the connector's "+" to the connector's "-" pin? In post #8 you stated that the resistance of each grid is between 170 Ω and 340 Ω, depending on wire type. Without the ends of the wires being connected, the resistance would be infinite (well almost).
It looks like the end of the "+" wire and the end of the "-" wire are not connected. In that case, how would current flow through the wires from the connector's "+" to the connector's "-" pin? In post #8 you stated that the resistance of each grid is between 170 Ω and 340 Ω, depending on wire type. Without the ends of the wires being connected, the resistance would be infinite (well almost
).I'm afraid I can't read your picture well enough. Not a problem of my eyesight nor of the clarity of the image, but what it represents.
It looks like the end of the "+" wire and the end of the "-" wire are not connected. In that case, how would current flow through the wires from the connector's "+" to the connector's "-" pin? In post #8 you stated that the resistance of each grid is between 170 Ω and 340 Ω, depending on wire type. Without the ends of the wires being connected, the resistance would be infinite (well almost
I thought the same thing but wanted to be sure.
I don't have the frames with me right now hence I tried to show it on a drawing. I will send a photo when I get a chance.Yes, they are not connected. I will add terminal systems at bottom.
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Then how did you get the resistance values you quoted?
When the ends of the wires are not connected the resistance is quasi-infinite and you don't need a beefy power transistor to drive voltage only to the grids. A small signal transistor like a BC107 alone can do that.
I measured them with multimeter (without any load) I measured start of Frame_1(+) and end of Frame_1(+) and so on for each frame.
Basicly I used only 2 wires on each frame. I stretched one nichrome wire (+) and rolled it around the frame, then I took the second nichrome wire (-) and rolled it around the same frame in a way that + and - do not touch each other. So the resistance that I measured is for only one wire either + or -.
I just finished connecting the end terminals. After connecting, I measured 43.2Ω from start of Frame_1(+) and end of Frame_5(+). Same value for the (-) wires. It is very close to the calculated 46.7Ω equivalent resistance. I will take a photo tomorrow.
I have already connected the end terminals but in a way that I can easily remove and attach together. Are you saying that If I keep the ends not connected, It can still work? In that case, should I replace the BD135 in the solution you draw at post#69 with BC107, or just a replacement on the original setup?
That measurement is useless, as there is no closed circuit for current to flow because you have left the end of the wire unconnected.
I have no idea how this contraption is meant to work. By electrostatic fields or by thermal effects from heating of the wires? Therefore I cannot answer your question with confidence.
A schematic as in post #68 is much more helpful.
I think they are connected as in schematic below: I'm not sure if it helps for validation of the scematic but I also added 2 photos. I am thinking that when they touch + and - It will stimulate them to sting. Please also check the 3rd picture attachment to see how - and + is rolled around frame-1.
The solution: you need a driver stage like this:
I also tried to implement your solution in post #69 in to the original setup and I ended up with below schematic (I hope I merged them correctly
): I have one question regarding to 5V connection of BC107's collector end. I would rather not use a second battery. Can I connect it to my 12V battery or I need a voltage regulator like LM2595 to limit the voltage down to 5V? I read that pin-2-13 on Arduino gives 5V output each. Can I use one of those pins instead?
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Your first image is, beg your pardon, nonsense. As indicated below, all your resistors are short circuited and even if they weren't, they'd be in series, not in parallel, so the total resistance of 44 Ω cannot be achieved.
What you have is acc. to the measured resistances this (resistance values rounded):
Since + (point C) and - (point A) are not connected, no current will flow between A and C and you have no need for the driver stage.
As to the driver stage. Compare your drawing
with mine
. The latter clearly shows that the resistor R2 is connected to +5 V, not to the output of the LM2596 step-up module.
What you have is acc. to the measured resistances this (resistance values rounded):
Since + (point C) and - (point A) are not connected, no current will flow between A and C and you have no need for the driver stage.
As to the driver stage. Compare your drawing
with mine
Attachments
I guessed
)) drawing was short-circut but since I measured the R_eq almost close, I thought I couldn't draw it right.I learnt the concept wrong. I thought + and - shouldn't touch each other on the grid. I can't visualize how to connect them properly. I will work on the frames.
I thought I could reduce the voltage from battery (12V) to R2 with LM2595 (to 5V). I am guessing you don't mean the 5V slot on Arduino so I need another battery(5V)?
Also can you recommend me a software to draw circuits to simulate them?
I meant the 5 V from the arduino.
You can also use the 12 V from the battery, but then you need to increase the value for R2 (post #77) by a factor of 12/5 = 2.4 to keep the current the same.
There are many available. I used circuitlab for the drawings here. I also use LTSPICE from time to time.
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