# 555 Timer Circuit Q's

Discussion in 'Electronic Basics' started by Jon Hoyle, Jan 10, 2004.

1. ### Jon HoyleGuest

Hi there,

Basically I'm trying to drive a DC motor (3vdc 110ma) at varying speeds.
There are loads of circuits out there, but I am interested in this one, for
simplicity; http://www.eleinmec.com/figures/003_03.gif

For 'C' I intend to use a 0.1uf cap, and R1 a 1k resistor. Could I replace
R2 with a 100k pot in series with another 1k resistor to give a range of
speeds? The range is not important, even if I get a duty cycle af 25% - 75%
I'll be happy, but the more the better I suppose )

Another thing, I need to drive 2 motors, and I expect I will need to
duplicate this circuit to drive the second one?

Many Thanks,

Jon Hoyle

2. ### John PopelishGuest

That motor current (110 ma) is going to heat the 555 up pretty badly.
added between the 555 and the load. The transistor will produce a lot
less heat than he 555, because it can be configured to switch on so
hard that it drops only a fraction of a volt, while the 55 will drop
something like 1.5 or 2 volts with a 110 ma load. This lower drop
also allows a higher top speed. If you want the two motors to go at
the same speed, you could use a bigger transistor and parallel them.
You also need a diode across the motor(s) to carry their inductive
current during the time between pulses. This would take a transistor
rated for something like 1 ampere (to get good saturated turn on at
220 ma). A TIP29 would be fine and a TIP31 even more rugged. You
might find these at Radio shack for under a dollar.

As to timing, I think you should try putting a pot resistance between
two fixed resistors, with the wiper going to pin 7, and the other ends
of the fixed resistors going to the cap and the positive supply. That
way, both the on and off time can be varied, to get a very wide range
of average voltage out. 1 .01 uf cap and a 100k pot in series with a
pair of 10k resistors should give you about 1000 pulses per second.

Tie the output (pin 3) to an NPN transistor base through a current
limiting resistor (with a value of say, (supply voltage times 20)/
motor current in amps). Connect the emitter to the negative rail, the
collector to the motor/s and the other side of the motor/s to the
positive rail. Diode across the motor/s with cathode to positive
supply. It is also a good idea to connect a bypass capacitor (say,
100 to 1000 uf rated at more than the supply voltage) across the rails
at the point there the emitter and motor attach, to carry a lot of the
pulsation that would otherwise have to pass along all the supply
lines. That would be the 220 uf cap at the left side of your diagram.