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Topic: what is this part called?(shematic from playground) (Read 978 times) previous topic - next topic

Kinsi

hey,

im planing to buy all the parts to build 2 relayboards for my arduino. found this nice shematic in playground:
http://www.arduino.cc/playground/uploads/Learning/relays.pdf
now my problem is that i found all the parts, but i dont know what K1 is. hope someone can help me ^^

greetz

AWOL

#1
Jul 01, 2011, 11:00 pm Last Edit: Jul 01, 2011, 11:02 pm by AWOL Reason: 1
A relay.
In this case, a single pole, double throw (SPDT) relay.
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.

Kinsi

#2
Jul 01, 2011, 11:27 pm Last Edit: Jul 01, 2011, 11:36 pm by Kinsi Reason: 1
okay i failed, so this is the relay i want to power ^^
i thought relay power + and gnd go to the relay and wondered why the diode is on the K1 part ^^
another question, can i just connect the 5V rail from the arduino to the positive relay power connection in the shematic?

retrolefty


okay i failed, so this is the relay i want to power ^^
another question, can i just connect the 5V rail from the arduino to the positive relay power connection in the shematic?


Yes, with the caveat of not knowing how much total +5vdc current your project is consuming because of other components, and keeping in mind the 500ma total +5vdc current limit (while on USB power).

Lefty

Kinsi

ill make the relayboards with an extra port for the relay power, ill make some measurements how much ma they are need. i selected 5v relays that are working from 4 - 11V so 5V will be enough. ill reply when i got & soldered all parts ;)

thank

MarkT

Quote
wondered why the diode is on the K1 part


It stops the transistor being destroyed by the inductor.  A relay coil is a large-value inductor that will generate hundreds or thousands of volts if it it switched off fast - the diode allows the current to die down slowly and prevent that inductive spike from destroying other components in the circuit.

A bit more theory - inductors generate a voltage proportional to the rate-of-change of current.  Trying to interrupt the current through an inductor causes a large rate-of-change of current, thus a high voltage (which always acts to reduce the rate of change).
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Kinsi

yes, i know that the diode is there because of the coil that generates voltage by turning off, but i wondered why it was on the K part because i thought the relay will go to relay power +/gnd

retrolefty

Quote
but i wondered why it was on the K part because i thought the relay will go to relay power +/gnd


The transient protection offered by the diode is most effective if wired right across the relay coils terminals. It works like this, when the magnetic field collapses at turn off, the resulting voltage spike developed from the collapsing field is of a reverse voltage of the original supplied voltage. This negative voltage will therefore allow the normally reversed biased diode to conduct and clamp the negative voltage spike.

MarkT

Pedants only past this point: ;)  It's not really clamping the voltage spike, its providing a path for the current to continue flowing (an inductor is a bit like a constant current source with unlimited voltage compliance).  There is simply no voltage spike when the diode is there, its the lack of a current path that 'generates' the voltage spike because the inductor continues to push electrons round the circuit at the same rate.

When the diode is there the current will take a lot longer to die down, so the spike becomes a pulse of much longer duration, an effect that isn't really explained by 'clamping'.

An analogy:  with the diode its a bit like stopping a runaway lorry with a gravel-trap.  Without the diode the lorry hits a concrete barrier when the switch opens.  Substitute velocity for current, force for voltage and mass for inductance.  The forward-voltage drop of the diode plus the IR voltage in the coil provide a small stopping voltage (force), so  things slow down smoothly.  With no current path the road ends in a concrete wall and the stopping voltage/force is extremely large.  If you want to extend the analogy then the IR voltage drop across the winding is the lorry's rolling resistance, and the supply voltage is like the engine's driving force (which normally balances the rolling resistance).

Thus when switching on there is no big spike since all that happens is the engine starts and the lorry pulls away smoothly.  If the fuel runs out that's like the supply voltage falling to zero (not the same as switching the transistor off as current can continue to flow).  Opening the circuit is like the road suddenly ending in a barrier (in fact the barrier is the transistor, and if its not strong enough it gets damaged in the crash).

When there is no inductor it's like having a massless lorry (lets say its an inflatable lorry!) - it's easy and safe to stop with a barrier since there's no real momentum to dissipate. 

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jackrae

and that's why they are often referred to as "flywheel" diodes

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