Let's Make Robots!


Guten tag LMR,

Calculon is going to build this ciruit, but he wonders if his fellow LMR can comment on it before he starts. He's tested each component, but hasn't put them all together yet. Calculon's multimeter just blew a fuse, so he can't really measure the output at the moment. He has a wicked cool motor controller and he doesn't want to accidentally convert it to smoke. Here's the circuit:


M is a serial dual h-bridge powering 2x 2v 9amp gearmotors. The controller has an onboard regulator to power the ic. Not shown: 2 serial lines between M and IC.

Also, LED symbols are actually these guys and are drawing between 90 and 200 Ma each. Each runs to ground, but calculon only shows the digital inputs in his schematic.

Calculon wants to use 2 7805s in order to cut down on the heat.

1: The numbers: 300 mah to the sensors. Voltage drop is 7v. = 2.1 Watts burned off. At 50C/Watt heat dissapation, that means the 7805 should be 100 degrees above ambient?

2: Can one 7805 safely handle double that amount? Is 2 a bad idea?

3: Also, will the voltage drop be 7, or will it be less than that, due to the motor controller? 

4:Should the ICs ground line Y connect back to M, or should it run down to 0v?

Comment viewing options

Select your preferred way to display the comments and click "Save settings" to activate your changes.
caclulon was planning on using using a giant heatsink with 2 7805s and a cooling fan. probably overkill.

The Qik motor driver board can supply a 5V output from its onboard regulator, but judging from the user guide I wouldn't recommend drawing more than 70mA from it, and certainly not 400mA.

The power calculation was correct, the problem was with the thermal resistance calcs. The thermal resistance from the junction (semiconductor core parts) to still air can be up to 65°C/W in a TO-220 package like most 7805s have, but that only applies if you have no heatsink and no circulation in the air.
The thermal resistance from the junction to the outer casing is only 5°C/W, so you can slap a decent heatsink on and keep the total thermal resistance quite low. In that case you add the thermal resistance of the junction-to-case and the heatsink's thermal resistance together, and use that to determine ΔT, which is the increase in junction temperature relative to the ambient temperature.

If you stick the 7805 in a tight case by itself you'll have to use 65°C/W, which will net you P = 7V*400mA = 2.8W,
ΔT = 2.8W*65°C/W = 182°C, which means unless you're outdoors in Antarctica the thermal protection module in the 7805 will take over and cut your current output. Definitely going to need a heatsink on this one, unless you go with the original idea of splitting the regulation between 2 7805's. Personally I'd go with the heatsink, probably cheaper and less hassle space-wise, but the double 7805 option will work too, so long as the ambient temperature is below ~34°C.

Calculon likes these sensors a lot. They are quite handy when his XMOS doesn't have an adc and he doesn't want to bring in another ic.

thanks for the input on this disertation and elaboration on regulation moderation.

Those sensors should only draw 100mA maximum each. With 2 of them running from a 7805 @ 12V

the power is 0.2A x 7V = 1.4W

My Speed Writter uses two of these sensors, I found the current draw was only about 150mA but I suspect if I used an oscilloscope then the current might peak at 200mA for both of them. Use a suitable heatsink on your 7805's and they will be fine running directly from your 12V supply.


x is whatever the answer is  to question #3. If there was a foward voltage from the motor controller, Calculon thought the voltage drop on tthe 7805 could be less than 7v.

Can you elaborate on what Calculon has wrong in his equation?

Isn't power dissapation P = Voltage Drop * current ? 


1) Shouldn't the sensors only draw 100mA max each, for 200mA max per 7805? Since each regulator can handle up to 1.5A, and the efficiency is essentially unrelated to the current draw below the limit, one regulator will do the job perfectly.
What exactly does the 'X' line represent? BTW thermal resistance doesn't work like that =)

2) Yep, 1.5A max output, 35V max input.

3) Depends on whether you are drawing the full 12V from supply, or less than that from one of the motor driver board outputs.

4) Your IC should be connected to the ground from the motor driver and the other ground. Keep just one ground throughout your circuit and you'll make life a lot easier for yourself.

Warren G.