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Servo control using relays

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jcertain

New Member
I am working on a circuit to drive servos for a small bot I'm working on. I tried using darlingtons to build an h-bridge, but after several failed attempts and after branding a transistor shape on my finger while checking to see if it was hot I decided to use relays instead.

I have changed the reverse option from a 5v output of a microcontroller to a push button to voltage for simplicities sake.

What I am trying to figure out is the necessary resistance for the section of the circuits hooked up to the coils of the relays. Each relay has a 5vDC nominal range with 89.3mA nominal current.

There is also a voltage dropout of .25vDC

The Formula I am using is as follows
R1+R2+Rx=E/I
56+56+Rx=5/.00893
Rx=447.91

Here are my questions:
Based on this info I need a 447Ohm resistor to achieve the correct current?

Does the voltage dropout affect the equation?

Is there anything else I need to be aware of?
ReverseButton-1.jpg


well I found the problem iwth this...... I made a mistake in my calculations.... shoutl be .089 not .0089
 

Klaus

New Member
Looking at your circuit, you appear to have connected the two 5V relays in series via a series resistor to 5V. This way, each relay will get less than 2.5V at its coil, why did you do that?.
Usually a relay with a 5V coil is connected directly to the supply if this is also 5V. There's no need for a dropping resistor, the relay's coil has sufficient resistance if its rated for this voltage.
If you have two relays, which should switch on at the same time as per your motor reversing circuit, just connect the coils in parallel without the resistor.
You could do this reversing with just one relay if you can find one that has DPDT contacts.
Klaus
 

crust

Member
Are you building your own servo or using an off-the-shelf RC servo. It seems from your circuit you are building a reversible motor driver with two relays. Since the relays are wired in series, I dont think either one will get enough current to turn on. At 5V, with RX=0, the total load will be close to 100 ohms. So only about 50mA will flow through each relay coil. This may/may not be enough to trigger the relay.


If you are connecting this circuit to a micro, you definately need some protection diodes in your circuit.


Did you use protection diodes on your transistor h-bridge circuit?


The relay approach will *probably* work if you eliminated Rx and wired both relay coils in parallel.

This circuit will always run the motor unless you cut the power to the +5V terminal. Is this what you want?
 

jcertain

New Member
Well, I reworked the circuit design and I was able to achieve voltages above the minimum for the coils, but it is academic at the point because I have obtained a different relay that has 2 ins and 4 outs with one coil...

anyway here's what I did with the circuit

ReverseButton3-1.jpg


And no I didnt use protection diodes... I'm trying to figure out how they work. Someone Explained how the diode would work on this circuit, but I am not confident about the idea as of yet.... I am doing more research to understand how best to achieve propper protection...

If you know of any sites that would help me with this it would be greatly appreciated. (my "Basic Electronics Course" book is great on math, but very poor on practical application)
 

ivancho

New Member
Check out this article it will clear up some stuff.

Brief H-Bridge Theory of Operation

:!: The diode on the relay...... to understand that you need to know how a relay works. A relay is nothing more than an inductor. Inductors is a wire around a piece of iron (or other core material). When current flows thru a inductor, this builds up magnetic field.... Inductors don't like sudden changes in current so if you remove the current from the relay suddenly, the relay will generate from the magnetic fiel collapsing a huge voltage accross its leads. This will have some nasty effects on the power lines, and will zap your components. To solve this you use a diode in reverse biase capable of handling the normal coil current rating. This diode will start conducting when the power is removed from the coil, before the large voltage can be generated by the collapsing magnetic field. One diode to use is the 1N4004. And you better put it in there :shock: .

:idea: Have you try one of those normal H-bridge IC like the L293? They work well.....

ivancho
 
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