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Circuit Optimization

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kbs88

New Member
Hey guys,

I have a working circuit that flashes a few car lights and is controlled by a PICAXE 18x chip. Im kinda new at this, but there must be a way to cut down the number of components I'm using. Im using a high side switching method to switch the lights on and off because a lot of the light share a common ground and didnt want to have to run new ground lines.

Any thoughts??
 

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Nigel Goodwin

Super Moderator
Most Helpful Member
Then don't use FET's - check my PIC tutorial 'Hardware Extras' for high side switching using an NPN and PNP transistor.

You could also use relays, in keeping with car technology.
 

Electronworks

New Member
you are driving N-Channel MOSFETs on the high side. The component you are using generates a voltage higher than the supply so you can drive the gate above the supply voltage.

To keep things simpler, you need to use P channel MOSFETs. You pull the gate low with respect to the source to switch the device on and pass the current. However, your source will be at the input voltage (12V) so you need to translate the 0-5V PIC signal to 0-12V. You can do this with a simple transistor inverter stage.

Drive your PIC into the base of the transistor via a 1k resistor. Put a 10k in the collector (to get a 0-12V signal from the transistor) and drive the gate of the P-FET from the collector.

You should not have to invert the drive signal from the micro either. Switching the PIC o/p high switches on the transistor which pulls the collector low (and the gate) and activates the FET - phasing is maintained.

Good luck:D
 

kbs88

New Member
Hey guys,

Thanks for the advice. I took what you both said (it sounded to me like what you were saying was very similar) and put together what I though you were saying. But in my animation the lamp doesn't go out completely. I'm assuming that one of my transistors is not turning off completely. How can I fix this? Or am I not making my connections correctly.

Thanks again
 

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KMoffett

Well-Known Member
You don't need R2. It will limit the gate voltage on Q1 to ~1/2 of your supply voltage...~6.5V.

Ken
 

KMoffett

Well-Known Member
If you open the circuit where R2 was, does the lamp go out? That is, disconnect the Q2's drain from Q1's gate.

ken
 
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kbs88

New Member
If I remove open the circuit at that point I get a faint dim out of the lamp. But it does not turn fully on at all. I get a constant dim.

EDIT:

I inserted a diode between my lamp and the ground connection. This seemed to work and allowed the light to fully illuminate and go out. Voltage from Q1 source pin must have been leaking into my lamp thus not letting it go out fully. I have attached a final scheme for reference.
 

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crutschow

Well-Known Member
Most Helpful Member
You shouldn't need the diode.

If the circuit just has R3 connected between the gate of Q1 and its source, and no other gate connection, then Q1 should be fully off. If the lamp still glows, then something else is not connected properly.
 

kbs88

New Member
Carl,

In the last attachment (working circuit) if I remove the diode and just have a wire the lamp glows. Now this is in an animation (Proteus), i haven't actually built the circuit yet. But its usually pretty accurate. I put a virtual volt meter across my lamp, it gave a reading of 1.1volt when it should be 0 in theory. and a full 13.3v when it is on.

Any suggestions?
 

KMoffett

Well-Known Member
Is the voltage across R3 "0.0" when Q2's gate is low?

Ken
 

Roff

Well-Known Member
Pick a PMOSFET with higher source-drain breakdown voltage. The IRF7410 is 12V max.
 
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