# Brushed DC motor as generator, characterising the motor

I've read through the several topics here concerning using a DC motor as a generator but I've failed to answer my question so, apologies, I'm going to ask again.

I need to use a small DC motor as a wind-powered generator in association with an energy recovery chip (TI BQ25505). This chip will only successfully recover energy if the input voltage to the chip is ~400 mV or greater. I have tried using several DC motors as generators but the only one I have found which produces a terminal voltage greater than 400 mV when the breeze is just sufficient to overcome the back-EMF of the generator took 4 weeks to arrive from China and was provided without a specification.

Hence I would like to characterise the generator in order to purchase an equivalent locally. Here is what I know so far:

• Winding resistance 75 Ohms.
• Operating speed at "minimum breeze" 12 to 15 Hz.
• Voltage produced across 1 kOhm resistor at "minimum breeze" 650 mV DC.

Is there a way that I can convert this knowledge into (or perform more measurements in order to arrive at) the parameters which DC motors seem to be specified in, i.e. terminal voltage, output power, max RPM and max torque?

• Was there any description of the motor where you purchased it? If you have electrical instruments, there are ways to find out something about the motor, but only the manufacturer can provide complete information. You need to be able to measure speed, voltage and current.
– user80875
Apr 9, 2018 at 17:29
• The only electrical characteristics are buried in the Chinese. I can measure speed, voltage and current, though the current only in units of mA (I measured ~1 mA at "minimum breeze").
– Rob
Apr 9, 2018 at 18:08
• Google Translate suggests that the only electrical details at the link are "DC output voltage: 0.01-5.5 V, output current: 0.01-100 mA, rated wheel speed: 100-6000 rounds".
– Rob
Apr 9, 2018 at 18:19

"100-6000 rounds" probably means 100-6000 revolutions per minute. Used as a generator, the motor will probably produce about 5.5/6000 = 0.0009 volts per RPM. I assume speed in Hz is equivalent to revolutions per second. The voltage produced should be 0.0009 X 60 = 0.055 V or 55mv per RPS. At 12 revolutions per second, you could expect about 55 x 12 = 660 mV.

If the motor/generator produces 5.5 V and 100 mA at 6000 RPM that would be a maximum power of 100 X 5.5 = 550 milliwatts.

You can calculate torque from Power(W) = T(Nm) X RPM X 2Pi/60

• Very useful, thanks. So we have max power 0.550 mW @ 6000 RPM which gives torque 0.875 mNm. A question though: one of the other DC motors I tried is listed as 8.1 W, 12 V DC, 12 mNm, 4450 RPM. Your comment above suggests it might generate 12/4450 = 2 mV per RPM. Under "minimum breeze" I know that it actually generated less than 200 mV. Now I didn't measure what "minimum breeze" was for this case but it was several hundred RPM, certainly rather more than 100 RPM. What factors am I missing?
– Rob
Apr 9, 2018 at 20:31
• At low RPM, losses are a significant factor. What was the load? Losses can also be a factor in interpreting the published information. I may have time to look at the details later.
– user80875
Apr 11, 2018 at 14:45
• Load is a bit difficult to estimate in this case as it was connected to VIN of the BQ25505 which attempts to clamp VIN at 80% of the open circuit voltage (which the BQ25505 tests every 16 seconds). The open circuit voltage I seem to recall being around 400 mV. But don't worry yourself about it, this is all getting a bit vague. I've ordered another motor based on your advice which I hope will do the trick; I'll report back on how that goes.
– Rob
Apr 11, 2018 at 17:53
• So, I've now tried out a 1W, 6V, 2500 RPM 2607T006SR motor. At "minimum breeze", which was about 25 Hz for this motor, it produced just over 200 mV across a 1 kOhm load. Given the spec of 6V @ 2500 RPM, I was expecting more like a few Volts. Is there anything in the motor spec (at the link above) which would indicate why I'm getting such a low terminal voltage?
– Rob
Apr 16, 2018 at 10:56
• The internal resistance of the motor and the load resistance will form a voltage divider. That will reduce the terminal voltage somewhat. Also, some of the torque of the wind turbine will be used to overcome friction. I will try to look at the problem again tonight.
– user80875
Apr 16, 2018 at 12:58