Control motor speed with transistor, WITHOUT PWM

Thread Starter

Mrdouble

Joined Aug 13, 2012
107
image.jpg
Learning electronics is something fun to do and for the last 4 months I've been designing a circuit that will remove the power from the motor when it is stalled.The criteria for my circuit is completely arbitrary but I set my goals on completely discrete components with no fancy controls like PWM.

Is it possible to contol assorted small DC motors (some low amps /voltage , some high amps/ voltage) with a single pot on the base of a transistor.

The problem I'm having is that I can bounce from motor to motor and it works but the way I have it designed I have to adjust a pot at the collector and a pot at the base. I'd really like to just get rid of resistor at collector and use the transistor as more of a resistor. I have did a fair bit of research and have concluded that transistors don't work well as resistors due to beta moving fall over he place. :/ I could just put a pot on the base but without the resistor on collector the beta would be all over the place, atleast from my limited understanding of beta stability. I was thinking negative feedback (collector - base), would that work? Thanks in advance
 

alfacliff

Joined Dec 13, 2013
2,458
why dont you consider the motor the colloector resistor? the resistor in series with the motor will only reduce the current and limit the maximum current.
 

Thread Starter

Mrdouble

Joined Aug 13, 2012
107
why dont you consider the motor the colloector resistor? the resistor in series with the motor will only reduce the current and limit the maximum current.
Ha. Thought about that after I posted. Thank you though. Atleast it confirms my thoughts
 

ScottWang

Joined Aug 23, 2012
7,397
1. Vcc?
2. The V/I of motor?
If the voltage of motor is equal to the Vcc, and then the e of bjt connected to gnd, the c connecting to the (-)motor(+) to Vcc, using a 5K or 10K pot, and the pin 1 and 3 connecting to the Vcc and gnd, the b of bjt connecting a 1K resistor and connecting to the pin 2 of pot, the npn bjt can be use as 2SC1384 or similar, but it is depends on the current of motor, you will also need a fly wheel diode as 1N4001 to protect the bjt.

You can try to draw the circuit, that is one kind of learning, good luck.

You can find the drive circuit in ScottWang's Blog.
 

AnalogKid

Joined Aug 1, 2013
10,988
Is it possible to contol assorted small DC motors (some low amps /voltage , some high amps/ voltage) with a single pot on the base of a transistor.
Yes.

You are right, when a transistor is running "wide open" (full gain, no feedback, etc.) the gain wanders around with base current, collecgtor current, temperature, phase of the moon, whatever. But there is a solution, called the emitter follower. An emitter follower has all of the current gain, but no voltage gain. With your circuit, a very small voltage change at the base brings large voltage changes at the collector. With the emitter follower, the load (motor) is connected to the emitter. The emitter voltage tracks the base voltage almost exactly, so you have much better control of the motor voltage.

You don't say what the + voltage is, what the pot value is, or what the range of motor currents is. With these details the rest of the circuit is pretty straightforward.

ak
 

MikeML

Joined Oct 2, 2009
5,444
Here is what AnalogKid suggests with one better, namely a Darlington-connected emitter follower.

Everything is plotted vs pot position (in percent %). Note the base voltage V(b) green trace and the voltage across the motor V(m). Finally, note the violet trace, which shows the power dissipation in Q1. You would have to use a small heatsink to get rid of the heat produced there if the motor draws as much current as is implied by a motor resistance of 10Ω with a supply voltage of 6V.

247.gif
 

Thread Starter

Mrdouble

Joined Aug 13, 2012
107
Yes.

You are right, when a transistor is running "wide open" (full gain, no feedback, etc.) the gain wanders around with base current, collecgtor current, temperature, phase of the moon, whatever. But there is a solution, called the emitter follower. An emitter follower has all of the current gain, but no voltage gain. With your circuit, a very small voltage change at the base brings large voltage changes at the collector. With the emitter follower, the load (motor) is connected to the emitter. The emitter voltage tracks the base voltage almost exactly, so you have much better control of the motor voltage.

You don't say what the + voltage is, what the pot value is, or what the range of motor currents is. With these details the rest of the circuit is pretty straightforward.

ak
I was thinking same thing as emitter follower :). The supply is +6v
 

Thread Starter

Mrdouble

Joined Aug 13, 2012
107
1. Vcc?
2. The V/I of motor?
If the voltage of motor is equal to the Vcc, and then the e of bjt connected to gnd, the c connecting to the (-)motor(+) to Vcc, using a 5K or 10K pot, and the pin 1 and 3 connecting to the Vcc and gnd, the b of bjt connecting a 1K resistor and connecting to the pin 2 of pot, the npn bjt can be use as 2SC1384 or similar, but it is depends on the current of motor, you will also need a fly wheel diode as 1N4001 to protect the bjt.

You can try to draw the circuit, that is one kind of learning, good luck.

You can find the drive circuit in ScottWang's Blog.
Great blog. Looks like some good resources.
 

Thread Starter

Mrdouble

Joined Aug 13, 2012
107
Here is what AnalogKid suggests with one better, namely a Darlington-connected emitter follower.

Everything is plotted vs pot position (in percent %). Note the base voltage V(b) green trace and the voltage across the motor V(m). Finally, note the violet trace, which shows the power dissipation in Q1. You would have to use a small heatsink to get rid of the heat produced there if the motor draws as much current as is implied by a motor resistance of 10Ω with a supply voltage of 6V.

View attachment 88729
That's EXACLY what I was looking for. Why the darlington though? I understand they have tons of gain but I'm not seeing the relation.
 

#12

Joined Nov 30, 2010
18,224
That's EXACLY what I was looking for. Why the darlington though? I understand they have tons of gain but I'm not seeing the relation.
There are different ways to do it. Would you rather use 10 times as much current in the pot and drive one transistor?
 

AnalogKid

Joined Aug 1, 2013
10,988
As with your previous scheme, the transistor's power dissipation needs to be watched. While it is natural to assume that a motor running at half its design voltage draws half its normal current, this is a very coarse approximation of reality. Unless you are moving amperes of current, you can add a 1 ohm resistor in series with either the emitter or collector to measure the load current without affecting (much) the available voltage at the max adjustment.

ak
 
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