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I'm currently designing an active filter which has to pass an analog signal between two cables (PLC). I'd however like to not let it pass in case the signal is already present (up to a certain amplitude) on the second cable, to avoid harmonic distortions and presence of multiple frequencies.

I was thinking of using a CMOS analog switch (as indicated here for instance Switching op-amp gain resistances - how?) at the output of the OAP which I'll turn on when I need it.

The biggest problem is that: the resistance is highly non-linear (because in the circuit I have I can have down to 2 Ohms of load) and I may have to drive up to 1A (the output amplifier is a power amplifier).

Another solution I though was simply using an inverting amplifier with two feedback resistances: one very small (no transfer) and one equal to the resistance put in front of the inverting input (unit gain).

The bandwidth is about 100kHz. No problems concerning gain, since I already amplify the signal before (using a static gain, or even a programmable gain operational amplifier). The problem is how to connect the output only when I need it. Another potential problem might be the small amplitude of the signal: 2V_RMS (a sine wave at about 100kHz).

Switching has to be relatively fast: less than 1us.

I was even thinking of using static relays but I'm not really sure either. Normal or even reed relays seem to be quite slow and deteriorate with time.

Basically I was looking for a switch with: low R_ON, high R_OFF (>> 1MegaOhm), quite fast, bidirectional (supporting ANALOG signals in sine form of low amplitudes), low THD, and possibly easy to control. If there can be 3,6 or more switches inside a SOIC (or other IC) package it would be very good, since I have very limited space.

I'm hoping to see if there are other (better) solutions available.

EDIT: I though it would not work, but what about using an OAP with a switch to control power supply ? I did a simulation with PSPICE and it seems to work. It only produces a small voltage offset output. This is not a problem since there is a decoupling capacitor in series at the exit, so DC voltage/current doesn't pass through. Could this actually work ?

OAP power off http://img140.imageshack.us/img140/7462/opampturnedoff.png

Relays should be able to take as much as 230V_AC when turned off (open circuit): there is a decoupling capacitor at output, but its impedance is pretty low at 50Hz/60Hz (200kOhm) compared to that of the open relay. I haven't found those with a low ESR yet.

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user51166
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3 Answers3

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The Analog Devices ADG812YRUZ has 4 SPST switches, 0.65 ohm typical on resistance, and comes in a 16-pin TSSOP package. It switches in < 30 ns, has a THD of 0.02%, and a -3 dB bandwidth of 90 MHz. However it can handle only 300 mA continuous. (I assume you could parallel them for higher current.)

tcrosley
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  • Thank you. Not really a problem the 300mA continuous limit: the signal is transmitted over something like 20ms, then there will be a larger period of pause (I don't know exactly the length of this pause). The real problem is the peak value: it should be able to jump up to 1A-2A (on a 1-2 Ohm load), down to something like 5mA. It all depends on the load, which is variable due to external conditions. I really like the R_ON characteristic which is pretty flat. Any other analog switch with a little bit more of amps ? I don't really like putting 3-4 of those in parallel because it takes much space – user51166 May 05 '12 at 19:43
  • If you know other solutions as well, please suggest them. – user51166 May 05 '12 at 19:44
  • What about static relay ? Something like http://www.farnell.com/datasheets/358214.pdf support up to 2.5A. I just cannot figure if it's only for logic signals or if it works for analog ones as well. Any ideas ? Thank you. – user51166 May 05 '12 at 19:51
  • peak current for the ADG812 is only 500 ma. the problem with the relay (which would otherwise be a better choice) is that the switching time is a few ms, and you stated < 1 us. – tcrosley May 05 '12 at 21:02
  • hmm, another option might be discrete mosfets. have to think about that and get back to you. – tcrosley May 05 '12 at 21:05
  • Thank you tcrosley. What about the solution I posted in my original (edited) question below "EDIT" ? Is it just the simulator or powering off the OAP might actually work ? – user51166 May 05 '12 at 21:21
  • Not too crazy about turning the op-amp on and off. First of all, I would think there would be a period of time when the input-output would be non-linear as it powers up, plus I don't think there is even a spec on turn-on/turn-off times. – tcrosley May 05 '12 at 21:30
  • Technically the transient response shouldn't be an issue. Using FSK modulation with something like 30 period of sine wave per bit. 2 periods are already out for processing and turning switch on. If another 1-2 periods are missed, it wouldn't be too bad. – user51166 May 05 '12 at 21:36
  • I'll let someone else with more analog experience (I'm mostly a digital guy) come in and comment on alternate solutions. – tcrosley May 05 '12 at 21:40
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You seem to be thinking about switching the output of a opamp, but consider instead having a power amp that always drives this line and switch its input signals. Now you don't need really low on resistance and can use more ordinary analog switches. The input of the power amp wouldn't need particularly high current.

Olin Lathrop
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  • You mean like a switch that switch from GND (if I don't want output) to the input signal (if I want to output) of the OP-amp ? That migh actually work, since the R_ON becomes very small compared to the input impedance of the amplifier. – user51166 May 06 '12 at 07:51
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Out of what you proposed, an analog switch is probably the best bet. First let's look at your other two proposals:

simply using an inverting amplifier with two feedback resistances: one very small (no transfer) and one equal to the resistance put in front of the inverting input (unit gain).

This doesn't disconnect your filter output from the output cable. It just applies a (near) 0 V signal to the output. If there is another source trying to drive the cable, there will be contention. One source will be trying to drive 0 V, and the other source will be trying to drive a different value. One or the other of the two sources is likely to be damaged by this eventually.

what about using an OAP with a switch to control power supply ?

Again, this probably doesn't do what you want, but the reasons are more subtle. When you shut off power to your op-amp, it won't try to drive any voltage onto the output cable, which sounds like what you want. But, that doesn't tell the whole story. Most likely, your op-amp has diodes connected between its outputs and the power supply in the direction so that normally they'd be reverse biased. These diodes are there to prevent the op-amp being damaged under certain circumstances.

Now when the op-amp is powered off, these diodes connect the output to ground. So if your other source on the cable tries to drive a voltage above 0.5 V or so, the op-amp output diode will try to clip it to ground. In the case of a source capable of driving 1 A, again something is likely to be damaged in this case.

Now, what about using an analog switch, and what kind to use?

A simple MOSFET? The bulk diode will prevent a single MOSFET from blocking a bidirectional signal.

A solid-state relay? The SSR's I'm familiar with have switching time measured in ms, 1000x your requirement of 1 microsecond.

A reed relay? Reed relays able to switch 1 A are bulky and expensive.

An armature relay? Even bulkier and more expensive.

If you can't adjust your requirements, I'm guessing you will have to design something like an SSR (two back-to-back FETs) with your own driver circuit. Here's an example of an SSR schematic to show the arrangement of the FETs:

enter image description here

(image from a Vishay data sheet).

You'll eliminate the opto isolation (which you haven't said you need), and be able to reduce the switching time in proportion to your ability to provide gate drive current. Your gate driver will need to drive the gates several V above your maximum signal level.

Your problem will be how to fit this thing into your available space (which you've said is small, but not how small). Chips able to switch 1 A are typically not especially small, and even if they are they're likely to need additional board area for heat-sinking to avoid self-destruction.

The Photon
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  • I said space was only "small" because I fear you'd said it was impossible otherwise. Currently it should fit in 2cm x 2 cm x 2cm (can do 2 PCBs, but even with that it's a tight fit). I think I can go up to 4cm x 4cm x something but even like this I'm not sure it's enough. – user51166 May 06 '12 at 07:57
  • So basically there is no simple way to disconnect the output of an op-amp from the cable. That's quite a letdown. I'm not really used to FETs (a bit more with MOSFETs) so I'm not sure this might actually work. Plus I need 3 switches, 3 power amplifiers, etc. That's going to be very bulky. – user51166 May 06 '12 at 07:59
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    It's not that difficult to disconnect the output of an op-amp. It's difficult because you need bidirectional signals, high currents, fast switching time, *and* small space. If you can find a way to back off any one of those requirements, you can probably find a solution. – The Photon May 06 '12 at 15:06
  • Just found the ADG451BNZ (<5 Ohm). If I go with a 2V_RMS signal, max current will be like 400mA. Add another load and that will fall to 200mA or the likes, just right to avoid problems. Though not ideal, it isn't that bad either. If I have a 2 Ohm load (minimum), the transfer is not excellent (only 1/4 of the signal is measured at reception, but that's FAR ENOUGH for the purpose). Plus these switches are well packaged (4-8 switches inside a package). Its not that bad (unfortunately I have to buy a DIP package for testing, but SMD mounting solutions have even lower R_ON). – user51166 May 06 '12 at 15:52
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    Not sure what you mean by "add a load". Normally adding a load (at the far end of the cable?) would increase the current. Do you mean add an additional series device at the source? Because if that will work, why not just use a driver circuit with a limited output current? – The Photon May 06 '12 at 16:44
  • Well if at the output of the operational amplifier I have a 5 Ohm resistance due to the switch as well as 2 Ohms of additional load, technically I'd have 7 Ohms in total, assuming load is resistive. Therefore the current will be lower. Not sure I need a driver circuit, but it may be a possibility. – user51166 May 06 '12 at 16:47
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    On ADG451, I'm not sure what your power supplies are, but note figure 5 in the datasheet: 4 Ohm Rds(on) depends on having +/- 15 V supplies. With +/- 5 V supplies, Rds(on) increases to near 7 Ohms, presumably also reducing current handling capability. – The Photon May 06 '12 at 16:57
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    On your later comment, I'd call that "adding a series current-limitting device to the source". The load is whatever you're trying to deliver power or signal to. – The Photon May 06 '12 at 17:02
  • Power supply should be +12V since I already one +12V for the power amp. Just need to add another one for negative voltage. Maybe I expressed myself in a confusing way regarding load. Whatever I think that with such a relatively high R_ON I won't need to drive up to 1 Amp. Probably other devices like that will work. I just felt like it wasn't really useful to amplify a signal if some part would be lost in the switch itself. But again, in some conditions that will be the case (but still work good), while in other cases it will work very good. I'll order one and see how it works. – user51166 May 06 '12 at 18:44