# circuit/IC required

Discussion in 'The Projects Forum' started by jayanthyk192, Mar 15, 2010.

1. ### jayanthyk192 Thread Starter Member

Jan 23, 2010
80
0
hello everyone,

i am need of a circuit/IC for one of my projects.the requirements are

1) it should have 2 switches when one is switched on the output should be say +5V and if the other switch is ON the polarity should be -5V
2)the output current must be around 700mA

thank you.

2. ### Bernard AAC Fanatic!

Aug 7, 2008
4,952
551
If you can use SPDT switches:

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3. ### jayanthyk192 Thread Starter Member

Jan 23, 2010
80
0
i got the basic idea about the switch.But i am using a battery for my project and i don't know how to give a -5v.please help.

thank you.

4. ### SgtWookie Expert

Jul 17, 2007
22,202
1,793
This will allow you to reverse the current flow direction:

If both switches are down, the difference between Out1 and Out2 is 0v.
If both switches are up, the difference between Out1 and Out2 is 0v.
If S1 is up and S2 is down, Out1 is 5v more positive than Out2.
If S1 is down and S2 is up, Out1 is 5v more negative than Out2.
However, you have not said anything about your battery voltage or type. If it is not 5v, this will not regulate the voltage down to 5v, or do anything to limit current.

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5. ### SgtWookie Expert

Jul 17, 2007
22,202
1,793
If you need to regulate voltage AND limit current, you will need something like this:

Note that the battery voltage has increased.

S3 was added so that the battery can be disconnected; otherwise the idle current of the 7805 will eventually drain the battery.

If you want 5v out of a 7805 regulator, you will need at least 7v in. If you're going to approach 700mA out, you'll need more like 7.5v in. See a datasheet.

The LM317 when used as a current regulator will drop a minimum of 3v across itself. At higher currents, it will drop more than 3v across itself. See a datasheet.

So, 7.5v + 4v = 11.5v.

The 1.8 Ohm resistor causes the LM317 to output a maximum of around 700mA.

C1 and C2 are required for the stability of the 7805 regulator. See the datasheet.

You will need heat sinks on both regulators, or they will overheat.

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6. ### jayanthyk192 Thread Starter Member

Jan 23, 2010
80
0

actually this does'nt work for my project.

i'm sorry its my fault i had'nt given the complete details.so here they are,
actually i'm building a device which maintains a horizontal position.for this i'm using 4 switches,2 for right-left and 2for front-back.the switches are hand made(by myself) in such a way that if mercury flows over the open terminals of switch1 the motor turns in one direction,and in the other direction if the mercury flows over the other switch.(the motor is connected by gears to the platform which has to be maintained horizontal)if the platform is tilted then the mercury flows over the switches to make the platform horizontal.the potential given to the motor can be in the range of 5-8v.

because of the above reasons i cannot use the switch you earlier suggested.i cannot ground the terminals while reversing polarities.

thank you.

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7. ### SgtWookie Expert

Jul 17, 2007
22,202
1,793
Have your switches energize the coils of single-pole double-throw relays.

Mercury is a deadly poison. Be careful to not get any of it on your skin.

See the attached. S1 and S2 are your mercury switches. S3 and S4 are limit switches in case your motor runs out of control.

• ###### MotorReverseLimitSwitch2.PNG
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Last edited: Mar 18, 2010
jayanthyk192 likes this.
8. ### jayanthyk192 Thread Starter Member

Jan 23, 2010
80
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hi,
i used your circuit for my project and it worked well.instead of the switch you told i used two solenoids from an old printer,modified the circuit a bit to use the two solenoids for the front-back as well as right-left and it worked pretty well.
i had new idea which could save space.the circuit diagram is attached.
if the mercury fows over to left side then the flip flop is powered by the battery and s=1 is given by the continuity,then q=1 and q'=0 then the motor turns in one direction.
if the mercury flows to the other side then again the circuit is activated and this time r=1 is given and the output isq=0 and q'=1 so that the motor spins in opposite direction.

i want to know if the circuit works and please tell me if i have to any more components before giving the output to the motor.

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9. ### SgtWookie Expert

Jul 17, 2007
22,202
1,793
Flip-flops are not designed to supply very much current. Unless your motor required less than around 20mA, the flip flop would probably burn up.

Your circuit has no "off" state. The motor would always have current flowing through it in one direction or the other.

Look up a datasheet for an L272 or L2720 power opamp. It can be used to drive motors forward and reverse in an H-bridge configuration.

10. ### jayanthyk192 Thread Starter Member

Jan 23, 2010
80
0
i think the circuit actually works because when the mercury flows over one switch it activates the flipflop an then say s=1,then the motor turns to bring the platform(explained earlier) to horizontal position and the mercur flows away from the switch and thus switching off the flipflop itself so that the flipflop is itself off.the same thing happens in the other switch also.it had a mistake and i rectified it(circuit attached).please tell me if this circuit works.if it works then i can use mosfets to increase current to output.

i saw the datasheet of l272,i could understand the circuit's logic but i cannot build the complete circuit myself,so could you plase help me with the resistors that have to be conntected.

• ###### jajaj.JPG
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Last edited: Mar 25, 2010