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the point is that when the input voltage is more positive than the +12v (12+ diode forward voltage) then the diode will conduct.
The other diode will work the same way if the input becomes lower than the -12v-Vf.
It is also important to use diodes with low forward voltage drop, 1n4148 has about 0.7v so 12+0.7=12.7v, a Schottky with a Vf of 0.3v (like BAT85) will conduct at a lower voltage.
The diodes act as bypass valves for the incoming signals; they are normally "off" until the voltage gets close to the rail voltages, then they begin to conduct.
Since a 1N4148 has a forward voltage of ~0.7V, they will begin to conduct when the incoming voltage reaches +12.7V or -12.7V (+12+0.7 and -12-0.7). If the incoming signals try to push more current into the op-amp, the extra energy is routed to the rail voltage (power supply), which effectively "clamps" the voltage near the range of Vcc to Vee (+12V and -12V).
when A contact is on we dont have any probelem in prtoection
but when it s off U_ contact = 13,5 V > supply 12 V
so there what is hapening ??**broken link removed**
Your attachment doesn't seem to be working. Try again.
If either of the inputs tries to rise to 13.5 volts, the diodes going to the +12V rail will be conducting a HUGE amount of current, which should quickly discharge the extra energy on the line causing the high voltage. This type of protection is meant to remove short duration, high voltage transients (high current for a very short period of time) or long duration, low overvoltage conditions (which would draw a much lower current). In either case, the diodes will survive.
You cannot analyze this circuit assuming a constant DC 13.5V input, unless you are willing to say that the diodes burn up and fail (which is what would happen in a real circuit). The diodes would be conducting somewhere in the range of 100's of amps, indefinitely, and that's far from realistic.
INPUT PROTECTION
The AD620 safely withstands an input current of ±60 mA for several hours at room temperature. This is true for all gains and power on and off, which is useful if the signal source and amplifier are powered separately. For longer time periods, the input current should not exceed 6 mA.
For input voltages beyond the supplies, a protection resistor should be placed in series with each input to limit the current to 6 mA. These can be the same resistors as those used in the RFI filter. High values of resistance can impact the noise and AC CMRR performance of the system. Low leakage diodes (such as the BAV199) can be placed at the inputs to reduce the required protection resistance.
---------- Post added at 12:14 ---------- Previous post was at 12:09 ----------
so that haw can i choose R2 ,R4 protection resistor ??
i'm using 1N4148 diode
Sheet 12 of the spec sheet shows the effective internal wiring of the device. I see a 400 ohm resistor between the IN's and the rails. The rest is Ohms Law.
If you go to the extremes...
Vsupply = 13.5V
Vinternal = -12V (negative rail)
I_in = 6 mA (for super-long durations)
V=I*R... or R = V/I = (13.5- -12)/0.006 = 4250 ohms (total)
4250 - 400 ohms internal to part = 3850 ohms (minimum)
Using 10k resistors should give you plenty of room.
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