Boosting the Amper and altering the frequency

[Maxouishere]

You may cretae short circuits (or may have created short circuits) between the "+" and"-" wires in your grid somewhere. Also check your wiring and make sure the "+" and "-" wires do not touch.

Thanks. I think this is the issue. 10.000 of them walk on each frame. And some wires got loose. I will make them tighter. I don't know if it is necessary but I am also planning to get a heatsink for the transistor.

place a series resistor between the collector of the BD135 and the grid, e.g. 100 Ω to protect the transistor from overcurrent.

Would doing this also reduce the current on the grids? and is 100Ω ~20W overkill?

 

[Harald Kapp]

Would doing this also reduce the current on the grids

yes

and is 100Ω ~20W overkill?

about 8 W should suffice: P = V²/R with V = 29 V and R = resistance

 

[bertus]

Hello,

What is the supply voltage for the BD135?
At 29 Volts, the current with the 100 Ohms resistor will be 29 Volts / 100 Ohms = 0.29 Amp.
The power in the resistor will be 0.29 Amp X 29 Volts = 8.41 Watts.

The calculations can be done with this wheel:


Bertus

 

[Maxouishere]

What is the supply voltage for the BD135?

Hello. Judging from the schematic in the post #82, I think It looks like 5V but I'm not sure. It looks like 5V signals comes from pin_8 of Arduino through BC107 to BD135. But I'm no expert

At 29 Volts, the current with the 100 Ohms resistor will be 29 Volts / 100 Ohms = 0.29 Amp.
The power in the resistor will be 0.29 Amp X 29 Volts = 8.41 Watts.

Should I also add the grid's resistance? 100Ω resistor will be connected in series to a grid system which is made by resistance wires, 2nd schematic in post #83. 29V / (100+46)Ω = 0.19A where 46Ω is the R_eq of the grid system. Therefore P = 0.19A * 29V = 5.51W, right?

The calculations can be done with this wheel:

Thank you for this and excel file above. Very useful.

@Harald Kapp If I want to keep the current same or even more after adding that 100 Ω~8W resistor at the collector end of the BD135, should I reduce the 138Ω resistor's value which is at BC107's collector end? Also, If I make the wires around the frame so tight, so that they won't be able to make + and - touch each other, do I still need that 100Ω resistor?

I tried to follow a Youtube tutorial to calculate the values at the ends of transistors but It doesn't seem correct:

 

[Maxouishere]

In datasheet of BD135 it says that max collector current is 1.5A What should I do to achieve at least 1-1.2A current? I assume I need a heatsink at that level of current. I built a new grid system. now each wire (10x) has 180Ω resistance and all connected parallel as below schematic:

@Harald Kapp You told me to add 100Ω resistor. Can I connect 10Ω 8W resistor instead of 100Ω 8W to keep the current above 1A or is there another way?

 

[Harald Kapp]

In case of a short circuit the worst case is that only the current limiting resistor is in the circuit. You will then have 29 V across this resistor R_limit. IF you want 1.2 A through that resistor, you have
R_limit = 29 V / 1.2 A = 24 Ω (rounded).
Power dissipation in the resistor can then be as high as
P = V²/R = (29 V)²/24 Ω = 35 W.

Can I connect 10Ω 8W resistor instead of 100Ω

With 10 Ω the max. current in case of a short circuit is 29 V / 10 Ω = 2.9A which is almost twice as much as the rating of the transistor says.

 

[Maxouishere]

With 10 Ω the max. current in case of a short circuit is 29 V / 10 Ω = 2.9A

Thank you so much for the clarification. I was making a huge mistake by summing R_eq of the grid and 100Ω.

In case of a short circuit the worst case is that only the current limiting resistor is in the circuit. You will then have 29 V across this resistor Rlimit. IF you want 1.2 A through that resistor, you have
Rlimit = 29 V / 1.2 A = 24 Ω (rounded).
Power dissipation in the resistor can then be as high as
P = V²/R = (29 V)²/24 Ω = 35 W.

For the worst case (35W power dissipation), I'd better mount a heatsink (this) on BD135, right?
Heatsink's thermal resistance = 30 K/W

 

[Harald Kapp]

When you use the DB135 as a switch, power dissipation will be low:
Assumin a max. I_CE of 1.25 A (the limit of the transistor), the collector-emitter saturation voltage V_CEsat will be 0.25 V worst case (datasheet).
The power dissipation is then P_CE = I_CE × V_CEsat = 1.5 A × 0.25 V = 0.375 W.
Add the power dissipatin from the base-emitter junction P_BE = V_BE × I_BE = 1 V × 0.03 A = 0.03 W.
The total power dissipation is then 0.4 W worst case (rounded).

The thermal resistance junction-ambient of the transistor is 100 ° /W. A power dissipation of 0.4 W will therefore increase the junction temperature over ambient by delta-T = 100 ° /W × 0.4 W = 40 °. Even when the transistor is operated in a hot environemnt, say + 40 °C, the junction temperature will be at max. 80 °C, way below the allowed 150 °C (max). Therefore no heat sink is necessary.

The heat sink you propose has a thermal resistance of 30 °/W. Add the thermal resistance junction-case of the BD135 you get a total of 40 °/W. The temperature rise from the dissipated power is then only delta-T = 40 °/W × 0.4 W = 16 °. This heatsink is absolutely sufficient in this switching application.

 

[Maxouishere]

When you use the DB135 as a switch, power dissipation will be low:
Assumin a max. I_CE of 1.25 A (the limit of the transistor), the collector-emitter saturation voltage V_CEsat will be 0.25 V worst case (datasheet).
The power dissipation is then P_CE = I_CE × V_CEsat = 1.5 A × 0.25 V = 0.375 W.
Add the power dissipatin from the base-emitter junction P_BE = V_BE × I_BE = 1 V × 0.03 A = 0.03 W.
The total power dissipation is then 0.4 W worst case (rounded).

The thermal resistance junction-ambient of the transistor is 100 ° /W. A power dissipation of 0.4 W will therefore increase the junction temperature over ambient by delta-T = 100 ° /W × 0.4 W = 40 °. Even when the transistor is operated in a hot environemnt, say + 40 °C, the junction temperature will be at max. 80 °C, way below the allowed 150 °C (max). Therefore no heat sink is necessary.

The heat sink you propose has a thermal resistance of 30 °/W. Add the thermal resistance junction-case of the BD135 you get a total of 40 °/W. The temperature rise from the dissipated power is then only delta-T = 40 °/W × 0.4 W = 16 °. This heatsink is absolutely sufficient in this switching application.

Thanks for the calculations. I learn alot from your posts. I bought a 27Ω resistor but couldn't test the device with it yet because of colony issues. I will test and post the results this week. I have one question for the device count. We have built this device for only 1 hive whereas I have 10. Can I connect this to others by using other digital output ports of Arduino simultaneously? What I mean is each output pin will be connected to BC107 + BD135 + XL6009. Is this possible or I need a new Arduino for each hive and if possible, Does 12V 4.5ah battery enough for this kind of setup?

 

[Maxouishere]

@Harald Kapp I have enough transistors and voltage regulators to try with 10 output pins at the same time but I don't want to harm the Arduino. Is this safe?

 

[Harald Kapp]

There are several limits to observe, read here.

 

[Maxouishere]

There are several limits to observe, read here.

• The absolute maximum for any single IO pin is 40 mA (this is the maximum. You should never actually pull a full 40 mA from a pin. Basically, it's the threshold at which Atmel can no longer guarantee the chip won't be damaged. You should always ensure you're safely below this current limit.)
• The total current from all the IO pins together is 200 mA max

From 5V: 5V/138Ω = 36.23 mA
From pin_8 =5V/470Ω = 10.64 mA (not continuous)

If I draw the current from the battery, instead of 5V pin of Arduino (I remember you said I can by increasing 138Ω resistor's value by a factor of 2.4), does that mean I can connect 10x safely? Total current draw from the Arduino then would be 106.38 mA.

 

[Harald Kapp]

From pin_8 =5V/470Ω = 10.64 mA

It is even less. You forgot to inlcude the V_BE of the transistors: I = (5 V - 2 × 0.6 V) / 470 Ω = 6 mA.

If I draw the current from the battery, instead of 5V pin of Arduino

That should work.

 

[Maxouishere]

@Harald Kapp Hello. It is already connected minus voltage on the battery. Do I still need the connection which is highlighted with red colour? (Schematic on post #86)

 

[Harald Kapp]

Of course you need this connection. How else would the Arduino be connected to minus?

 

[Maxouishere]

Of course you need this connection. How else would the Arduino be connected to minus?

Right. What was I thinking

Let's say I will use 5x output pins of Arduino. Do I need 5x voltage regulator (XL6009) or can I divide +out of one regulator into 5 for each pin just like image below? (from post#86)

 

[Harald Kapp]

That depends on the max. curren the regulator can deliver and the sum of the currents the grids will draw. This sum must be less than the max. output current of the regulator. If it exceeds the regulator's max. current, you'll need equivalently more regulators.

 

[Maxouishere]

That depends on the max. curren the regulator can deliver and the sum of the currents the grids will draw. This sum must be less than the max. output current of the regulator. If it exceeds the regulator's max. current, you'll need equivalently more regulators.

Thanks. I was thinking to reduce the max current. I will keep it below maximum output current of the regulator (4A)
It is good to know that with one I can do that. This will save good amount of space and wiring.

 

[Maxouishere]

@Harald Kapp Hello. I was reading about diodes and I wonder If I need them on the build with multiple outputs as below. Or is it not necessery in my case?:

I couldn't make it work with only one but this does not mean device is not working. I better test it with at least on 5 unit at the same time of day, simultaneously, to see if the problem is device or not.

 

[Harald Kapp]

I see no diodes in your schematic. What are you talking about?