I have two servo motors that I rigged up to use as a telescope remote focuser. The idea is to turn one servo by hand and use the power generated to turn the other, which is geared to a telescope focuser knob. I noticed that when the two servos are electrically connected, it is noticeably harder to turn a servo compared to turning it by itself. I tried changing the polarity of the connection hoping it would help, but it is still harder to turn the servo when they are connected. Does anyone know why this is?

  • $\begingroup$ Could you provide a diagram of how you have connected the two servos? $\endgroup$
    – Shahbaz
    Commented Nov 10, 2015 at 18:42
  • $\begingroup$ I'm not sure how to make a diagram on here, and I am terrible with Microsoft Paint. It is a very simple setup. I have two motors. The are connected negative to negative, positive to positive. (Polarity doesn't matter and can be changed) I manually turn one motor, which generates a current. This current then turns the other motor. My question is, why is it harder to turn the manually turned motor when the two motors are electrically connected, as opposed to when they are not? I think Chuck gave a very thorough answer. $\endgroup$ Commented Nov 11, 2015 at 0:13
  • $\begingroup$ Microsoft Paint is a terrible tool. Try to take a look at inkscape, even with basic functionality you can make really good looking drawings. $\endgroup$
    – Shahbaz
    Commented Nov 11, 2015 at 0:26

2 Answers 2


The other servo, being an electrical device, has some finite resistance.

When the servos are not electrically connected, the output of the manually turned servo is an open circuit, meaning that there is infinite resistance and no current flows. Because no current is flowing, the electrical output power of manually turned servo is

$P = IV$

where $I = 0$, so

$P = (0)(V) = 0$.

When you connect another servo, there is now a finite resistance connected to the manually turned servo's leads, so now current may flow. Now the current is given by

$V = IR$

$I = V/R$

so the electrical output power of the servo is

$P = IV$

$P = V^2/R$

There are two "types" of voltage in a motor when you operate it normally; the applied voltage and a back-voltage, commonly called "back EMF". This back EMF is a byproduct of motor motion.

The requirements for motor action are:

  1. Current-carrying conductor
  2. Applied magnetic field
  3. Applied voltage differential on the conductor

The output of motor action is relative motion between the current-carrying conductor and the magnetic field.

The requirements for generator action are:

  1. Current-carrying conductor
  2. Applied magnetic field
  3. Relative motion between the conductor and field

The output of generator action is a voltage differential in the conductor.

So you can see that if you meet motor action then you get generator action and both happen simultaneously.

When you back-drive a motor, you can get an output voltage at the motor terminals, if you have an applied magnetic field. In some instances, electric vehicles for instance, the motor is created such that the magnetic field is in the form of an electromagnet, called field windings. In this version, it is possible to vary the strength of the magnetic field by varying the current in the field windings; this varies the output current of the motor during back-driven operations and produces a variable strength braking effect.

In your case, with servo motors, the magnetic field typically comes in the form of permanent magnets in the motor housing. This means that the only way to vary output current (assuming a fixed output resistance) is to vary speed, as this varies the voltage you generate as dictated by the back-EMF constant;

$ V_{\mbox{back EMF}} = K_{\omega} \omega \phi$

where $K_{\omega}$ is the back-EMF constant, $\omega$ is the speed at which you are driving the motor, and $\phi$ is the strength of the magnetic field.

Looking again at the electrical output power of the servo:

$P = V^2/R$

Output power of the servo goes up with voltage squared. Since voltage is proportional to the driving speed of the motor, you can see that output power goes up based on speed squared; if you double the speed at which you drive the servo you quadruple the output power of the servo. This is why the servo gets "very hard" to drive.

There will always be some resistance when you try to drive a servo or motor, even with an open circuit, because they are real devices and thus have mass and inertia. You apply a torque to accelerate the motor:

$ \tau = I\alpha$

where $\tau$ is the torque you apply, $I$ is the effective motor inertia, and $\alpha$ is the rate at which you are accelerating the motor shaft. Output power varies with torque and speed:

$P_\mbox{mechanical} = \tau \omega$

so it requires more power to accelerate an already-moving object than it does to accelerate a stationary object.

Finally, the total power you apply to the servo is given by:

$P = P_{\mbox{mechanical}} + P_{\mbox{backEMF}}$

$P = \tau \omega + V^2/R$

$P = I\alpha \omega + (K_{\omega} \omega \phi)^2/R$

$P = (I \alpha)\omega + \frac{(K_\omega \phi)^2}{R}\omega^2 $

Where again, if the back-driven servo is not electrically connected to anything, then $R$ becomes infinitely large and the second term in the above equation becomes zero.

  • $\begingroup$ So, if you take an electric motor without connecting the leads to any power supply or impedance, you cannot get any back-EMF or current? You simply reduce to a shaft under torque that accelerates (along with some friction of course)? $\endgroup$ Commented Nov 10, 2015 at 14:54
  • $\begingroup$ @BrianLynch - That's correct; you will still generate a back EMF, a voltage, but if there's no load (infinite resistance) then there can't be any current flow. Similar to a battery - there's a voltage, but no current flows without a connected load, and since there's no current there's no output power. $\endgroup$
    – Chuck
    Commented Nov 10, 2015 at 15:10
  • $\begingroup$ Oops, of course there is voltage but no current. I should know this, just so used to using motors as motors. $\endgroup$ Commented Nov 10, 2015 at 15:12
  • $\begingroup$ Can you even apply the motor equation $V = RI + k\omega$ here? Or is it that resistance $R$ that is infinite? I know that equation is simplified too, so maybe that's where my confusion comes from. $\endgroup$ Commented Nov 10, 2015 at 15:15
  • $\begingroup$ @BrianLynch - You can't apply that equation because there is no $I$, there is no applied current. $RI$ in that equation represents the applied voltage, which doesn't exist if you're only back-driving it. $\endgroup$
    – Chuck
    Commented Nov 10, 2015 at 15:44

Do I understand correctly that you do not use any external power supply?

You said it yourself, you use one servo to generate power to the other. If it is not connected it is easy to turn since it is not generatrin power (not 100% exact but let's run with this in this case).

If you connect the servo, your mechanical energy will be transformed to electrical energy transfered to the other servo and transformed back to mechanical energy. So you still have to generate the energy required by the mchanical system connected to the second servo. If the servos are the same you gain nothing, you might as well turn the mechanical system directly, if they are different you might have to exert less torque on the first motor, then the second one is producing, but the velocity of the second one will be then slower. (similarly to mechanical gears)

  • $\begingroup$ I'm a little bit rusty on back-driving motors to generate electricity, but something seems off about this explanation. If you turn a motor then it will generate current and you'll get some resistance from the back EMF. If you connect the leads to a load then that current will be used to drive the load. In order to get resistance back on the motor you'd have to send current into it, which a load would not be doing. $\endgroup$ Commented Nov 10, 2015 at 12:04
  • $\begingroup$ which part is off? this: "you might have to exert less torque on the first motor but the velocity of the second one will be then slower. (similarly to mechanical gears)" $\endgroup$
    – 50k4
    Commented Nov 10, 2015 at 12:09
  • $\begingroup$ made some edits.... $\endgroup$
    – 50k4
    Commented Nov 10, 2015 at 12:12
  • $\begingroup$ Sorry, I don't mean your answer is badly worded. I just don't understand where the extra resistance is coming from. The servos are connected electrically not mechanically, so any extra resistance should be from extra current being developed as the motor is turned like a generator. So where does that extra current come from? Or is it something like when you connect two motors you end up with essentially one larger motor with a higher motor constant and higher back-driving stiffness? $\endgroup$ Commented Nov 10, 2015 at 12:35
  • $\begingroup$ There is no extra current. It is just the difference between the motor constants. on one side the motor constant will link the rotational velocity to electrical potential and on the other side the other motor constant will link the potention to output rotational velocity. And then the torques can be calculated based on energy or based on motor constants...It should be the same result. If the servos are the same in an ideal case (not considering losses) the motion should be 1:1 if motor constants differ the two motors will act as two mechancal gears wih different radii would $\endgroup$
    – 50k4
    Commented Nov 10, 2015 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.