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I am building an experimental percussion instrument. I have two circuits in it which are not connected to each other. One is a small linear fan speed regulator controlling a small motor, which requires between 1.5 and 4.5 volts.

The other is a small pre-amplifier connected to a contact microphone which runs on a 9v battery.

I would like to be able to power both of these circuits from a single 9v battery. What is the most efficient way of doing this? I have several 78L05 voltage regulators which to be honest I don't know very much about. Could I use one of those at the start of the motor circuit? It's my understanding that these take the voltage down to 5v from a higher amount (in this case 9v) but correct me if I'm wrong.

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  • \$\begingroup\$ What about power requirements (i.e. consumption)? 78L05 IC gives an output of 5V DC. I'm not sure, but fan speed regulator may get damaged due to relatively high (> 4.5V DC) voltage. It would be better if you have L78L33 ICs which is a 3.3V DC regulator, but, the advantage is surely depends on the current consumption of fan speed regulator. \$\endgroup\$ – Rohat Kılıç Oct 10 '16 at 14:25
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    \$\begingroup\$ "motor" and "9V battery" (if you mean PP3 type) are not a good combination if you want reasonable battery life. \$\endgroup\$ – Brian Drummond Oct 10 '16 at 14:31
  • \$\begingroup\$ The fan controller is this type. 2.bp.blogspot.com/-KvaOVAid5Oo/UvJa6s3NLFI/AAAAAAAAA20/… \$\endgroup\$ – TCassa Oct 10 '16 at 14:35
  • \$\begingroup\$ @BrianDrummond I believe the circuit is a PWM circuit which should take the stress off the battery a little - can anyone verify? \$\endgroup\$ – TCassa Oct 10 '16 at 14:38
  • \$\begingroup\$ It says "Linear regulator" so ... not PWM. How much current does the motor draw in operation? \$\endgroup\$ – Brian Drummond Oct 10 '16 at 14:48
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Assuming that the 9v battery is PP3 size, there isn't any good way to do what you want with the components you have. The motor will probably draw at least 150mA once it gets up to speed, and up to 0.5A under load. At startup it could draw over 2A. The battery will not like this at all.

Here are some typical curves for a PP3 Alkaline battery discharged at 0.5A:-

enter image description here

After 10 minutes at 0.5A the voltage has dropped below 7V (the minimum voltage required for a 78L05 to regulate) which is probably too low for the mic preamp. Furthermore, the high motor motor current could induce noise into the power supply and upset the preamp.

Even if the battery could supply sufficient current to run the motor, the 78L05 regulator can't. You could use an LDO (Low Drop Out) regulator such as the LM2940, but it won't stop the battery from being stressed.

Your motor speed controller is a linear type, which due to its particular design is inefficient at all speeds because the TIP29C transistor is never fully turned on. Whether you connect the speed controller directly to the 9V battery or through a linear regulator the result is the same - most of the limited power available will be wasted.

So how can you get more current at the lower voltage the motor needs, without stressing the battery? The answer is to use a DC/DC converter, which transforms the power from high voltage at low current to low voltage at higher current. Watts = Volts x Amps. A well designed DC/DC converter may be 90% efficient, so if the battery can deliver 100mA at 8V (0.8W) you could get 480mA at 1.5V (0.72W) which might be enough to run your motor and still get reasonable life out of the battery.

Motor speed could be controlled by using a DC/DC converter with variable output voltage. Or you could use your linear speed controller with R2 connected to a higher voltage instead of Vin, which will improve its efficiency because the higher voltage on the transistor's Base will turn it it on harder and deliver full Vin to the motor.

As an alternative to a DC/DC converter you could use another battery (eg. 1 or 2 AA cells providing 1.5 or 3V) to power the motor. This has several advantages:- more power available to the motor, longer run time, and keeps motor noise away from the sensitive microphone amp. You might not like the idea of having to use two batteries, but it is probably the best way to use the parts you have.

Here is an example circuit, which is a modified version of your linear speed controller:-

schematic

simulate this circuit – Schematic created using CircuitLab

U1 supplies regulated 5V to voltage divider R2/VR1/R3, which biases Q1. Since the transistor has a current gain of about 50, bias current drawn from the 9V battery will normally be less than 10mA, rising to a maximum of ~40mA when the motor is starting up (limited by R2).

R2 and R3 set the maximum and minimum bias voltages which determine the motor speed range. The values I chose should give a full range from zero to full speed. If you want a more limited range then you can increase their values. You could use a single 1.5V cell if it provides sufficient motor speed, in which case you should increase R2 to 220Ω or higher to reduce the maximum bias voltage.

A separate power switch is not required for the motor battery because when the 9V battery is switched off Q1 won't get any bias voltage, so it will also be turned off.

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  • \$\begingroup\$ While some good points are made here, the circuit given is probably not worth building compared to a low-side MOSFET based alternative. \$\endgroup\$ – Chris Stratton Oct 11 '16 at 18:22
  • \$\begingroup\$ A MOSFET PWM circuit would be more electrically efficient for sure, but also much more complicated and use many parts the OP doesn't have. I wanted to show a simple circuit that was close to the original design and easily understandable, and given the low expected motor power (0.75~1.5W) I don't think a PWM controller is necessary. 'More efficient' could also mean 'less use of resources' (mine included!). \$\endgroup\$ – Bruce Abbott Oct 11 '16 at 19:29

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