Anything not specifically related to the Speeduino hardware. Eg sensors, bluetooth, displays etc
By androidcho
#5021
Ok, in my project I decided to go with a 3-wire Bosch IAC(I do have a 2-wire Bosch IAC from Siemens ECU) and I searched the net for a resistor value. In all topics people say that you must connect one of the coils to ground through 35Ohm/50W resistor. But here is my question.
When I calculate the power through the resistor i get only 5 watts of dissipated energy which is far from 50 watts. That's without the load which will add a little more resistance.
U=14.2V
R=35Ohm
=>
I=0.40A
P=5.76W

Let's double the resistor wattage and go to 35Ohm/10W resistor. Is that okay? Yes, it'll get a little hot, but it'll be far from burning something. I need to test this.
By edc_atl
#5024
androidcho wrote:Good news, but I think we don't have a software support right now. Another question is what is the current consumption of this valves?
Support for single PWM type devices has been there for some time, but there was no interface per say until now. The valve can use up to 2A when in rapid fire mode, or so we have measured in the past.
By edc_atl
#5025
androidcho wrote:Ok, in my project I decided to go with a 3-wire Bosch IAC(I do have a 2-wire Bosch IAC from Siemens ECU) and I searched the net for a resistor value. In all topics people say that you must connect one of the coils to ground through 35Ohm/50W resistor. But here is my question.
When I calculate the power through the resistor i get only 5 watts of dissipated energy which is far from 50 watts. That's without the load which will add a little more resistance.
U=14.2V
R=35Ohm
=>
I=0.40A
P=5.76W

Let's double the resistor wattage and go to 35Ohm/10W resistor. Is that okay? Yes, it'll get a little hot, but it'll be far from burning something. I need to test this.
It's a 35 ohm / 50W resistor that you use to tie pin 3 of the valve to ground.
By androidcho
#5026
I know what value they recommend, but I'm asking why. Because you can't dissipate more than 5-6 watts through a 35 Ohm load on 14 volts. :shock:
By edc_atl
#5028
androidcho wrote:I know what value they recommend, but I'm asking why. Because you can't dissipate more than 5-6 watts through a 35 Ohm load on 14 volts. :shock:
Let's try this way, measure the resistance between pins 1 and 2, and also between pins 2 and 3. You should have about 12-15 ohms on each of the coils which relates to about 1A per coil at 12Vdc without external load components. I could be wrong but when you apply ohms law you get 35W being dissipated by the 35 Ohm resistor, hence 50W is the ticket. :mrgreen:
By androidcho
#5029
edc_atl wrote:
androidcho wrote:I know what value they recommend, but I'm asking why. Because you can't dissipate more than 5-6 watts through a 35 Ohm load on 14 volts. :shock:
Let's try this way, measure the resistance between pins 1 and 2, and also between pins 2 and 3. You should have about 12-15 ohms on each of the coils which relates to about 1A per coil at 12Vdc without external load components. I could be wrong but when you apply ohms law you get 35W being dissipated by the 35 Ohm resistor, hence 50W is the ticket. :mrgreen:
Ok, so here are my calculations :D
R(valve) = 12 Ohm
R(resistor) = 35 Ohm
They're connected in series, so we get
12 + 35 = 47 Ohm
And from here we know these two
U = 12 V
R = 47 Ohm
From which we get
I = 12 / 47 = 0.255 A
P = 12 * 0.255 = 3.06 W
And for U = 14.2 V
I = 14.2 / 47 = 0.302 A
P = 14.2 * 0.302 = 4.28 W
User avatar
By PSIG
#5033
By comparison, I ran my napkin calculator, and with a worst-case 15 volts and dead short through just the resistor. I only get 0.43A for 6.45 watts:
15 volts / 35 ohms = 0.429 amps
and
15V * 0.43A = 6.45 watts
Done with a 12R motor winding in-series, I get (15V / 47R) * 15V = 4.79W, using 15V (typical very cold charging system voltage) to be sure the extremes are covered.

David
By noisymime
#5041
androidcho wrote:P = 14.2 * 0.302 = 4.28 W
And remember that this is 4.28W dissipated by both the resistor and the coil in the valve!
PSIG wrote:By comparison, I ran my napkin calculator, and with a worst-case 15 volts and dead short through just the resistor. I only get 0.43A for 6.45 watts:
I tried, but couldn't think of any possible way to get a higher number than this.
Short answer, I have no idea why such a massive resistor is recommended other than crazy safety factor. :lol:
By androidcho
#5042
So after a little spam, the myth is BUSTED! You don't even need a 10 watt resistor, but just in case I will grab a 10 watts to be sure. :D

And yes noisy, you're right when you say that not small part of that energy will be dissipated into the coil itself.
By edc_atl
#5067
androidcho wrote:So after a little spam, the myth is BUSTED! You don't even need a 10 watt resistor, but just in case I will grab a 10 watts to be sure. :D

And yes noisy, you're right when you say that not small part of that energy will be dissipated into the coil itself.
Just keep it in a cool place :lol:
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