I've the schematic that is in the picture below:


simulate this circuit – Schematic created using CircuitLab

And I did the following analysis:

  • the opamp acts as a voltage follower, so when \$V_{in}\$ is between \$0\$ volts and \$10\$ volts the output of the opamp can only be as high as the voltage I apply to the positive \$V_+\$ of the opamp;
  • $$R_1=\beta_{min/T2}\cdot\frac{V_b-V_++(V_{sat}-V_{be/T1})}{I_{motor max}}$$ Where \$\beta_{min}\$ is the minimum of the current gain of transistor \$T_2\$, \$V_b\$ is the external voltage source voltage, \$V_+\$ is the positive voltage at the opamp rail, \$V_{sat}\$ is the saturation voltage of \$T_1\$, \$V_{be/T1}\$ is the base-emitter voltage of \$T_1\$ and \$I_{motor max}\$ is the maximum motor current.
  • $$R_2=\frac{V_+-V_{be/T1}-V_{be/T2}}{I_{motor max}}$$

Question: Are my formulas to find \$R_1\$ and \$R_2\$ correct?

  • \$\begingroup\$ You mean, after you fix the confusion between Vm+ and Vm-? Also, you need to specify Vmotormax, the voltage across the motor at maximum current and minimum rpm. \$\endgroup\$ – WhatRoughBeast Jun 11 '18 at 17:45
  • \$\begingroup\$ What is "a motor motor controll"? \$\endgroup\$ – Transistor Jun 11 '18 at 17:47
  • \$\begingroup\$ Your circuit, as far as I can tell, is flawed. Try reversing the op-amp input connections. \$\endgroup\$ – Andy aka Jun 11 '18 at 17:49
  • \$\begingroup\$ @WhatRoughBeast Vm+ is the positive pole of the motor and Vm- is the negative pole. \$\endgroup\$ – asd Jun 11 '18 at 17:53
  • \$\begingroup\$ @Andyaka I'm sorry I corrected it. \$\endgroup\$ – asd Jun 11 '18 at 17:54

The opamp, T1 and T2 are a voltage follower and, due to negative feedback and the massive open loop gain of the opamp, the voltage across R2 equals Vin. No need to delve into algebra to resolve this. This means that the current in the motor is Vin / R2.

No need to even consider what the Vbe of each transistor is. The opamp output drives enough voltage to overcome the Vbe volt drops and maintain equilibrium at its inputs. Of course, for this to work, the opamp positive supply voltage needs to be typically 2 to 4 volts higher than the maximum value for Vin.

If you wanted to be a little more precise about things, the next consideration is that the current through R2 is actually motor current plus the base current and, if the beta of T2 is low you will have a significant error in the simple analysis. Normally you could expect less than a 5% error but, as you turn up Vin this may rise to 10% as beta drops towards 10.

  • \$\begingroup\$ Thanks for your analyses. Does this also imply that my equations are wrong? \$\endgroup\$ – asd Jun 11 '18 at 18:55
  • \$\begingroup\$ Can you tell me that? \$\endgroup\$ – asd Jun 11 '18 at 19:16
  • \$\begingroup\$ Yes your equations are wrong. R2 is simply Vin/Imotor for high beta and it develops into Vin/(Imotor+Ibase) as beta drops. R1 can be a range of values depending on the beta of T2. Have you thought about using a simulation tool to resolve this? \$\endgroup\$ – Andy aka Jun 11 '18 at 20:35
  • \$\begingroup\$ Where does the rule of 2 to 4 volts higher than the input voltage come from? Can you provide me a link so that I can learn more about that? \$\endgroup\$ – asd Jun 12 '18 at 17:01
  • \$\begingroup\$ An op-amp cannot deliver the full rail of its power supply and you maybe loses two volts. So, if Vin is (say) 5 volts, to deliver a Vout (across R2) of 5 volts + 2Vbe (= maybe 7 volts), you need a power rail of about 9 volts. \$\endgroup\$ – Andy aka Jun 12 '18 at 17:53

Your design is sub-optimal. The R1 will get hotter than the motor. All equations are NG.

The proper design ought to measure current in a 50mV shunt using Max. current of V+/{DCR(coil)+RdsOn FET(s)}=Imax using a “FET bridge design” to control acceleration and braking current which may be up to 10x motor rated current. PWM is a more effective and efficient (cool) way to drive motors.

Then 2 half bridge MOSFET pairs are used for bi-directional control . Google “images” has tons of examples.


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