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What kind of IC is needed for 12V relay?
Basically I need two IC, (AND groups) one for LOCK, and another group for UNLOCK. I want to make it as simple as possible. If anybody can suggest how to connect it, drawing, power supply is 12V and I output needs to control small 12V relay. What IC is the best for this? Thanks in advance! Table 1. Truth table for relay logic. OP to edit.
What is microcontroller used in tida-01168 DC / DC converter?
The system uses two LM5170-Q1 current controllers and a TMS320F28027F microcontroller (MCU) for the power stage control. LM5170-Q1 subsystems use average current feedback for current control whereas the C2000™ microcontroller provides voltage feedback. Such a control scheme eliminates phase current balancing typical for multiphase converters.
Is the compfuser a 12V or 5V regulator?
The engine management compfuser does have its own internal regulation, usually at 5 Volt to operate itself and all sensors. Leave it, unless sensors do not get stable 5V. And if that happens, the compfuser regulation is faulty, not the “12”V system
How are resistors adjusted to match 12V?
Adjust resistors to match 12V according to IEC61131-2. The concept of the standard is to ensure the input has to sink a minimum amount of current before considering it a ‘1’. The three types specify how much, and are applied based on environmental noise. This prevents glitches from touching it or nearby relays.
Can a MOSFET drive a gate at 12V?
Once you get the logic worked out, if you implement it using CD4xxx chips, they should work at 12V. Then you can drive the gate of a MOSFET to drive your relay. Reading between all the lines, I’m gathering you have four SPDT switches, which may either be in the LOCK or UNLOCK position.
How many diodes are on a 5 V relay?
Adding the Zener diodes of Figure 2 and using popular 5 V relay coils should solve the hold-on problem. With the switches aligned the energised relay receives 12 V – 3 x diode – 1 x Zener = 12 – 3 x 0.7 – 5.1 = 4.8 V or so. With switches unaligned the relays will get 12 − 2 × 5.1 2 = 0.9 V each which should be enough to guarantee drop out.