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Anonymous Poster

Insufficient current problem

05/16/2010 1:50 AM

I'm a bit of a beginner when it comes to building circuits and I've run into a bit of a problem. I'm running a PICAXE-28X1 microchip powered by 5VDC at 1A, and the output pins are 5VDC at 25mA. The problem is that these output pins need to switch a relay that is rated 5VDC at 1A. I don't have enough current flow to switch the relays. I'm kind of stuck – I can't increase voltage in the circuit because anything over 5V will damage the microchip (supposedly.) Any thoughts/suggestions/ideas?
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Tobugrynbak
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#1

Re: Insufficient current problem

05/16/2010 2:00 AM

Do you need to switch one or more relays?

You could use the output of the picaxe to drive a transistor or a fet. Alternatively if you need to switch several relays then you might like to look at using a ULN2003 IC, this will give you ttl input and high current (around what you need for a relay) output for up to seven devices.

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#2
In reply to #1

Re: Insufficient current problem

05/16/2010 2:05 AM

I'm using two outputs from the PICAXE, each output switches its own separate relay. So would a transistor be the best option to try?
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Tobugrynbak
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#3

Re: Insufficient current problem

05/16/2010 3:10 AM

That would be the way to go, any common transistor like BC108, BC548 or 2N2222 would do.

Plenty of examples around to give you the idea.

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stevem
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#4

Re: Insufficient current problem

05/16/2010 4:11 AM

Whether using a 2003 or 2222 to buffer the output be sure to include a reverse biased zener to clamp the back emf from the relay.

1A into an inductive load can generate destructive reverse voltages.

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JohnDG
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#5
In reply to #4

Re: Insufficient current problem

05/16/2010 5:23 AM

ULN2003A has a built-in freewheeling diodes - just need to remember to connect pin 9 to the power rail.

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Anonymous Poster
#6
In reply to #5

Re: Insufficient current problem

05/17/2010 5:18 PM

Dear Uncle JohnDG,

I like your idea but I am a bit confused. The relay requires 1A and your individual driver can pass maximum 350mA. Besides, the Vce is 1.6V, so only 3.4V are across the relay's coil. What shall we do?

Your prodigy niece

Indela
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JohnDG
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#8
In reply to #6

Re: There are many ways to skin a <strike>cat</strike> relay<sup>[1]</sup>

05/17/2010 6:37 PM

This is from the linked datasheet:

Just parallel up 3 or 4 of the drivers to give the current handling. The VCE(sat) is (realistically) going to be 1V or less - relay coils are pretty tolerant; it should work fine.

As has been said elsewhere (sorry to whoever posted it - I can't remember who it was and I CBA to go back & look - when I've posted I'll review the thread & give you a GA), it must be a dam' big relay to need 1A. If you really need that current, it may be easier to use a small low-power relay beween the driver and the main relay.

Alternatively, get another power supply (say 24VDC) and a relay to suit. You would then have the option of using a relay with a built-in transistor driver which would work directly from the PIC output.

Alternatively, use a HEXFET (fancy MOSFET thing with TTL-compatible input levels and built-in protective diode) or similar.

There are loads of ways around this - just go for it. Read around a bit - it ain't rocket science.

But please first check out the spec of the relay (and post it here).

[1] Del made me say it.

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Tobugrynbak
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#7

Re: Insufficient current problem

05/17/2010 5:45 PM

How smegging big is this relay? Are you perhaps confusing the contact current rating for the current drawn by the solenoid of the relay?

Most common relays will draw around 200 to 250 mA at the rated voltage, smaller signal relays will be much lower. If your relay draws 1A then it must be a rather large one and would be considered a "Contactor" rather than a relay.

I suggest you measure the current drawn by the relay note that you won't be able to measure the inrush current without some sophisticated equiment but you can guesstimate it at 25% over the steady state current. I.E. if the steady state current is 200mA then make an allowance for an extra 50mA. So using a transistor that can sink 300mA or better will give you sufficient headroom (safety margin).

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