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Will this be safe?

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That still sounds far too high, even for a small transformer.

The SPICE model is still inaccurate, it should have a slightly higher secondary voltage than 9V, probably by about 30% higher. At full load the voltage will be 9V because of the copper losses.
 
I've just tested a cheap transformer.

Specification:
230V to 9V
1.65VA

Measured:
235V to 14.28V
Rpri = 1.42k
Rsec = 6.2R
Imag = 12.1mA

Calculated:
Turns ratio = 14.28/235 = 1:0.06077
Maximum current = 1.65/9 = 183mA

Primary inductance calculation:
[latex]Z = \frac{235}{12.1 \times 10^{-3}} = 19.42 \times 10^{3} \Omega\\
X_L = \sqrt{(19.42\times10^{3})^2 - (1.42\times10^{3})^2}=19.37\times10^{3} \Omega\\
L = \frac{X_L}{2\pi F} = \frac{19.37\times10^{3}}{2 \pi 50}=61.66H\\[/latex]

Secondary inductance:
Ls = 61.66×0.06077² = 227.7mH

Putting it all into LTSpice it looks like it'a going to work.
 

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Swap D3 from 1N4002 to a much faster 1N4148, it's still plenty big enough for that task as the relay back emf won't have much energy. Never use a slow relay diode with a micro circuit! D4 is not needed.

Replace D2 with a big cap, at least 470uF, this will decouple any relay current surge from your main supply AND supply the main current pulse to let the relay pull-in. Then you can increase the value of R3 to reduce relay operating (hold) current to a value less than its pull-in current but safely above its drop-out current. This reduces FET heating AND reduces the back emf energy from relay turn off AND provides greater isolation between relay and PSU/micro in general.

Get rid of the inductor after the regulator (blech!), replace with a small resistor for filtering (assuming you are not using ADC and dont need a perfect 5.0v) drop about 0.1v on the resistor at normal micro operating current and add a 10uF tantalum across C5.

Finally if you are just driving the FET on/off (not pwm) then replace R2 with a safer value like 2k7 which will save your micro if the FET lets go and you get full PSU voltage where its gate used to be you will only get a couple mA into the micro pin internal protection diode.
 
Swap D3 from 1N4002 to a much faster 1N4148, it's still plenty big enough for that task as the relay back emf won't have much energy. Never use a slow relay diode with a micro circuit! D4 is not needed.

I'll do that piece of cake. :)

Replace D2 with a big cap, at least 470uF, this will decouple any relay current surge from your main supply AND supply the main current pulse to let the relay pull-in.

Available board space doesn't allow another big electrolytic cap. :)

Then you can increase the value of R3 to reduce relay operating (hold) current to a value less than its pull-in current but safely above its drop-out current. This reduces FET heating AND reduces the back emf energy from relay turn off AND provides greater isolation between relay and PSU/micro in general.

I won't reduce relay coil voltage (and current) even for holding. The relay used is a power relay switching mains driven loads. With reduced pressure on the contactors the relay will burn up quickly. The FET used in the circuit is capable of 19.4 times overkill (3.5A continuous drain current) It won't probably notice the 180mA relay current. :)

Get rid of the inductor after the regulator (blech!), replace with a small resistor for filtering (assuming you are not using ADC and dont need a perfect 5.0v) drop about 0.1v on the resistor at normal micro operating current and add a 10uF tantalum across C5.

I've got hundreds of those micro-inductors. Using an additional tantalum I'll check if I can squeeze it in.

Finally if you are just driving the FET on/off (not pwm) then replace R2 with a safer value like 2k7 which will save your micro if the FET lets go and you get full PSU voltage where its gate used to be you will only get a couple mA into the micro pin internal protection diode.

The FET is zener protected, so no reason to protect the MCU.

Not the produced car makes the money. The spare parts do. :D

Boncuk

BTW, your post contains an answer to the original problem which was my concern about EMF.
 
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Here's an idea.

Connect a capacitor from the relay to 0V.

When the relay is turned on, the higher voltage and current will causing it to close very quickly.

Once it's closed the higher value resistor will providing the holding current.

Contact life can also be saved by including a resistor in series with the freewheeling diode. This causes the current through the coil to decay faster. As a general rule of thumb it shoud be the same value as the relay coil, you can use a higher value if you like but the back EMF will be greater.
 

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Revised circuit design

Hi,

here is the revised schematic.

Note: No transformer delivers nominal output voltage at idle. Iron core transformers normally have an idle voltage 1.35 to 1.65 above nominal voltage.

Toroidal transformers are driven close to saturation. Therefor their idle voltage is between 1.15 to 1.25 the nominal output voltage.

The values calculated to gain relay supply voltage are based on 11.5VDC after rectification. (The total current with the relay activated is 202mA)

Boncuk
 

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Hi hero999,

I'll try to get the extra resistor on the PCB. The idea sounds good.

Hans
 
Why not set it for landscape? It's much easier to read.

That 200mA fuse might blow with 260mA RMS flowing trough it, I'd go for a 300mA fuse.
 

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Why not set it for landscape? It's much easier to read.

I use PDF-creator directly out of Eagle. Printing landscape the print will be two pages.

Opening the file using Adobe Reader9 you can rotate the picture cw and ccw.
 
Is that better now?

Hi hero999,

I guess I managed to print the way you'll not twist your neck.

Hans
 

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I'm still using Adobe 8, maybe I should consider upgrading.

EDIT:
Thanks for printing it the right way round.:D
 
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Available board space doesn't allow another big electrolytic cap. :)

Yeah I hear that, and they are not cheap either when it comes to production.


I've got hundreds of those micro-inductors. Using an additional tantalum I'll check if I can squeeze it in.

But do you really want all the potential problems of ringing and instability and inductor life (they are notoriously unreliable compared to a resistor) when there is absolutely no need? A 10 ohm resistor would perform so much better than a 10uH choke for suppression there unless you need high Vdd accuracy for ADC use.


The FET is zener protected, so no reason to protect the MCU.

But still, what's your reason for using such a low gate resistor to switch a relay? Again if there is no NEED for a low resistor why not use a higher value and reduce the micro output pin current to drive the gate capacitance? It's a few uS extra to turn the FET on (relay takes 10's of mS anyway) but is many times less current from the micro pin.
 
But do you really want all the potential problems of ringing and instability and inductor life (they are notoriously unreliable compared to a resistor) when there is absolutely no need? A 10 ohm resistor would perform so much better than a 10uH choke for suppression there unless you need high Vdd accuracy for ADC use.

OK, inductor out - resistor in. :)

But still, what's your reason for using such a low gate resistor to switch a relay? Again if there is no NEED for a low resistor why not use a higher value and reduce the micro output pin current to drive the gate capacitance? It's a few uS extra to turn the FET on (relay takes 10's of mS anyway) but is many times less current from the micro pin.

My co-designer. May he has a contract with the MCU to control high power devices directly. :D
 
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