Hi Kaveh,
Just make note that the contact resistance of most reed relays is in the range of 150mΩ - 200mΩ. With a 200mA current, that will mean a 30mV - 40mV drop across the contacts. That was why I leaned away from those options, but if you’re OK with that, then they are a good solution, as OFF leakage current is essentially zero.
Thanks David. Reed Relays seems to be an expensive option for my application specially for the smaller SMD package.
Yes, you definitely pay for their specific benefits. There’s always a trade-off. Good luck to you!
Hi David,
Assuming I use TLP3107A(TP,F in my design and forward bias the diode between pins 1,2 with 10mA current.
Is there any voltage drop between pins 4,6? I can see there are two diodes between these pins in datasheet page 1.
Please let me know,
Thanks
Hi Kaveh,
First of all, let’s look at the internal structure of the TLP3107A(TP,F. On the output side, the Sources of two N-channel MOSFETs are tied together, and that node is brought out on pin 5. The diodes you see are the intrinsic diodes of the MOSFETs and are not added as extra components. Because of these intrinsic diodes, current can always flow from Source to Drain of an N-channel MOSFET, but by placing two of them with such a back-to-back arrangement, one of the two diodes will always be able to block reverse current flow when the MOSFETs are OFF.
This back-to-back arrangement of two MOSFETs in series allows the user to block or pass both DC and AC loads through the device IF connected such that the load current passes through pins 4 and 6. The drawback of connecting it this way is that the resistance seen by the load is the SUM of the resistances of each MOSFET. According to the datasheet, that resistance measured in series might be as high as 40mΩ at 25°C, which means the resistance of each MOSFET could be as high as 20mΩ.
However, if, rather than passing the load current from pin 6 to pin 4, you connect the load in such a way that current passes into BOTH pins 4 and 6, and out through pin 5 (this only works for DC loads because the diodes would no longer block the negative half of the AC current), then the current passes through the two MOSFETs in parallel rather than in series. Therefore, the resistance would only be 1/4 of the series resistance, or no more than 10mΩ at 25°C.
Connection “C” from page 6 of datasheet:
At your stated max current of 200mA, this would mean only a 2mV drop.
Thanks. I am applying 4V DC at pin 4 and my load is connected to pin 6.
When LED of pin 1,2 is on, the current flows from drain of NMOS at pin 4 to its source. Then to pin 6 through body diode of the top NMOS? In this case is there a diode voltage drop?
Hi Kaveh,
When the IR LED is on, it will turn on both MOSFETs, and the current will take the path of least resistance, which is through both MOSFETs (when ON, the MOSFET will appear as a purely resistive device and allow current to pass from either drain to source or source to drain).
Therefore, the body diode will be bypassed, and the only voltage drop will be across the rds-on of the two MOSFETs.
