Buffer Amp IC suggestions plse

Sep 21, 2005 15 Replies

Hi, Ive been trying to locate a suitable unity gain buffer amplifier IC that can drive a 50 load. Application is to buffer a VCO output at 5Mhz. Needs to be 8pin DIP and operate on a single-ended supply. I had found what I wanted in the CLC109, but cant seem to find a supplier. Any suggestions greatly appreciated regarding type and supplier that is user friendly to international countries. JEFF ZL3JK New Zealand


LTC's LT1206 comes in TO220, miniDIP and soic packages, and Digi-Key has them in stock for $6.25 each...

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Oops, +/-5V supply. What's your signal voltage? And did you want your output to have a 50-ohm source impedance?

Thanks, - Win

Needs to drive into 50 ohms. input level approx 150mV at 5Mhz

An AD817 can deliver +-50 mA and has a gain bandwidth product of 50 MHz.

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Unless you source with 50 ohms (2x signal plus 50-ohm resistor), and terminate with 50-ohms, the issue becomes how much cable capacitance? Either the load looks resistive (100 ohms in the former case) or capacitive.

Thanks, - Win

can

supplier.

user

What you want is a video buffer or video op-amp. Linear Technologies, National Semiconductor, and Analog Devices are among the many companies that make such things.

Note that some of these devices get a bit squirrelly if they see a capacitive load. So I recommend that you do something like this: (use courier or other constant-width font for ASCII art schematic)

|\ 50 in--|+\__+____/\/\____/> to 50 Ohm load +--|-/ | | |/ | | | +--/\/\-+ | Rf \ /Ra \ / | ----- GND

Rf and Ra would be the same value. Use a value recommended by the datasheet, or 1k if the datasheet doesn't recommend a specific value. Follow any other suggestions in the datasheet as well. For example, they may suggest a small capacitor directly from the op-amp output to inverting input.

You probably want to terminate the input with a 50 Ohm resistor, too, but I don't know, so I'll leave it out.

It sounds like you may need AC-coupling, so you can put blocking capacitors on the input and/or output. Oh, if you put a blocking cap on the input, be sure to also provide a DC (resistive) path to GND or to the mid-rail point if you are going single-supply.

Just chose a capacitor big enough so that 1/(2*pi*R*C) is much lower than

5 MHz, your signal of interest. (R in this case is 50 Ohms) There is no problem putting a capacitor like this in the signal path: Since it looks like a short circuit at higher frequencies, it does not capacitively load the amp at frequencies where the amp might be unstable. Also, the series 50 Ohm resistor keeps the overall impedance resistive.

HTH

--Mac

Maybe an LT1191?

digikey #: LT1191CN8-ND (24 in stock at USD 3.25)

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--Mac

Ouch, a 30mA supply current!

Thanks, - Win

Well- why not your super-duper 74HCU04 linear amp or whatever it was?- is that available in Tiny Logic?

can

supplier.

user

Here's a dual op-amp in 8 pin DIP which should do what you want

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Try a SL560

"Jeff" schreef in bericht news:k3kYe.13434$ snipped-for-privacy@news.xtra.co.nz...

can

user

I would say it depends on the cable you choose. If there is a 50 Ohm load and you connect it to your buffer amplifier with 50 Ohm coaxial cable, then the buffer amplifier will see a 50 Ohm resistive load, not capacitive at all.

It is probably still wise to use the gain-of-two amplifier and 50 Ohm source resistor in case the load happens to be a poor match. If you really don't need this feature then I would be tempted to use an emitter follower with just one transistor. (Do they make transistors in DIP-8 packages?)

Chris

I agree, if the load has no capacitance, source terminating with 50-ohms eliminates the cable-capacitance issue, and the driver only sees a 100-ohm load at high frequencies, above those where the open- end reflection has time to return. Although the input to the cable is attenuated by 2x (hence the 2x gain you suggested), the end of the cable gets the 2x back due to the open circuit, so a gain-of-two amplifier shouldn't be needed. Just the 50-ohm sourcing resistor, after the emitter follower.

Thanks, - Win

What I meant to say is that there is no need for a resistor at the amplifier end of the cable, provided we know for certain that the load is resistive and of the right impedance to terminate the cable, i.e. accurately 50 Ohms. If this is the case, then there is no need for an amplifier with voltage gain, which might make thing easier. If the load is not close to 50 Ohms and if we care about the frequency response, then terminating both ends of the cable makes a lot of sense.

There is a more important question which the OP could answer which is: what is this whole thing for? Is the VCO incapable of providing enough current to drive the cable, or is the amplifier there to provide reverse isolation so that the load cannot affect the frequency, or is there another reason why we need this buffer amplifier? Anyway at 5MHz I would probably try to do it with one or two transistors rather than buying a chip, since the continued availability of transistors is more likely.

Chris

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