11-19-2009 02:59 PM
Greetings,
I am running an NI based engine control unit. For my spark, I have a TTL module that works beautifully with the stock smart-coil system. However, I needed to switch to a higher-energy spark system that uses a 12VDC ground trigger instead of the 5V TTL. I have not been able to find a relay that can operate fast enough, so I am looking for any suggestions. I have attached a schematic of what I have. When I measure the amperage going from the 12VDC to ground it is at 130 mA. I am looking for something to physically replace the "green box". I need turn-on and turn-off time to be on the order of 1 microsecond, and able to handle cycling at 50Hz or better. Any suggestions?
11-20-2009 02:06 AM
11-20-2009 01:41 PM
Building on KC's suggestion, optical isolation and surge suppression may be wise to add. A solid state relay would be another device to use.
-AK2DM
11-20-2009 01:43 PM
11-21-2009 02:17 PM
What is a 4505 isolation chip Can you more like brand and complete type number ?
11-22-2009 05:24 PM
11-26-2009 03:44 AM
Electromagnetic relays typically have switching times in the milli-sec range. So you could switch around 100 times a second ... but the problem is contact bounce, the contacts usually bounce for at least 1 or 2 milli-seconds, and this spurious contact can be very annoying. - I agree that a PhotoMOS relay might work best in this application, they provide galvanic isolation and usually can handle up to 300mA or so. Contact bouncing is no problem, and switching delay is in the micro-sec range. You can calculate the series resistor quite easily: Rv = (Voh-Vf)/If, where Rv = series resistor, Voh= minimum output voltage at logic high level, Vf = forward voltage of input LED, If = forward current of the input LED. Since the input LED usually needs several (up to 10) mA current, only TTL buffer outputs can drive a PhotoMOS relay directly, check if the output you use can handle this. Most TTL outputs can handle somewhat more current in sink mode, i.e. with the load connected between +5V supply and the output. The logic state of the output is inverted, i.e. if the output is in logic low state, current will flow through the input LED. Also, the calculation of Rv is somewhat different: Rv = (Vol-Vf)/If, where Vol = maximum output voltage at logic low level (output saturation voltage at low level).