10kHz to 2MHz , 0.25mW to 150mW amplifier - Page 1
Forum section: electronics / beginners

author: raff5184
author_rank: Regular Contributor
timestamp: April 17, 2018, 04:53:56 pm
content: Hi all,I need the following amplification:INPUT:frequency range = 10 kHz - 2MHzV = 1 V peak-2-peakI = 1-2 mA peak-2-peakOUTPUT:Vo = 12 V peak-2-peakIo = 80-100 mAPower = 30 dBm (about 1W)Does something like this exist that I can buy? Or I can also build one, but what basic component do I need? Would an OpAmp do the trick?Also, if the bandwidth is too large for the amplification that I need I can split the amplification in 2 (max 3) sub-amplifiers, each covering a sub range of frequencies, this is not an issue.Thank you

author: dmills
author_rank: Super Contributor
timestamp: April 17, 2018, 08:24:46 pm
content: ADSL line driver or two plus a transformer or Balun?Linear or TI spring to mind for some reason.Regards, Dan.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 17, 2018, 09:44:20 pm
content: FYI, ADSL drivers are apparently really quirky, don't use them outside the range that ADSL needs (even if the specs say or imply they can run over a normal op-amp sort of range). Maybe okay here.A modestly fast line-driving-capable op-amp or buffer should do fine here. Shouldn't have trouble finding one off the shelf.Possible gotchas: supply voltage range, slew rate, distortion, RMS output, load impedance (resistive? complex? nonlinear?), transient ratings if any, passband flatness, cutoff band slope, etc.Reminds me of something I breadboarded years ago,You might want something more powerful than '390x's for output, for which you might use some SOT-89 BJTs instead, which will get you plenty of dissipation into a copper pour, for 1W output. Or a TO-126 something or other if you prefer THT.A discrete solution should be a rather extreme measure, for something that ought to be commercially available, so this is just FYI.Tim

author: Cerebus
author_rank: Super Contributor
timestamp: April 17, 2018, 10:33:07 pm
content: Tim, nine transistors? Two too many surely?As to the OP's problem, Jim Williams' Linear application note 47 has a couple of handfuls of useful designs that will meet your spec and a lot of relevant tutorial material. Now on Analog's website:http://www.analog.com/media/en/technical-documentation/application-notes/an47fa.pdf.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 17, 2018, 10:41:23 pm
content: Quote from: Cerebus on April 17, 2018, 10:33:07 pmAlas, t'was before 7TL's time (I said it was old).It could... ehhhh... let's see:- The one current mirror could be a resistor and transistor, simple enough. Worse offset, who cares right?- The diff pair tail could be a resistor. This impairs common mode range, but that would be acceptable for modest-gain and inverting applications.Ahh... that feels betterTim

author: edavid
author_rank: Super Contributor
timestamp: April 17, 2018, 11:07:41 pm
content: An HP 465A is pretty close to what you want: 5Hz-1MHz, 0.5WThey are not hard to find on eBay.

author: Zero999
author_rank: Super Contributor
timestamp: April 18, 2018, 07:54:23 am
content: Quote from: raff5184 on April 17, 2018, 04:53:56 pmYou've made a an error with your power calculation.12Vpp = 6V peak.Assuming you meant a current of 100mA peak, not RMS, that's a peak power of 0.1*6 = 0.6W and the RMS power of a sine wave is half the peak, so 0.3WRMS. If you meant 100mARMS, then multiply by 20.5.

author: raff5184
author_rank: Regular Contributor
timestamp: April 18, 2018, 08:40:14 pm
content: thank you all. Sorry there was a mistake in the specifications.It is not 30dBm, I meant that I want a gain of30dB.IN power = 0.125mWOUT power = 150 mWhence, gain = 30dB.

author: raff5184
author_rank: Regular Contributor
timestamp: April 18, 2018, 08:48:28 pm
content: The correct requirements are:INPUT:frequency range = 10 kHz - 2MHzV = 1 V peak-2-peakI = 1-2 mA peak-2-peakIn Power = about 0.250 mWOUTPUT:Vo = 12 V peak-2-peakIo = 80-100 mAOut power = 150mW (or more)amplifier gain = 30dB (or more)

author: Zero999
author_rank: Super Contributor
timestamp: April 19, 2018, 10:59:27 am
content: Your specification is still contradictory.1V peak to peak (not peak-2-peak) which is 0.5V peak, P = (Vp*I)/2 = (0.5*0.002)/2 = 0.0005 = 0.5mW.Are you talking about power or voltage gain?If the input power is 0.5mW and output 150mW, that's a power gain of: A = 150/0.5 = 300 = 10*log(300) = 57dBIf you're talking about voltage gain then: AV= 12/1 = 12 = 20*log(12) = 21.58dBThe voltages are clear, Vinpeak= 0.5V & Voutpeak= 6V. Now what about the input and output impedances?I think a video amplifier IC is more than adequate for this task.

author: dmills
author_rank: Super Contributor
timestamp: April 19, 2018, 03:33:20 pm
content: You sure?0.5mW in 150mW out, so A = 300 times.300 times power in dB is 10 log(300) = ~25dB, I have no idea where you got 57dB from.Yea, current feedback opamp or something of the sort, but notice that the source impedance seems to be quite high, so something fast and voltage feedback may be a better choice. Either way, not hard.Regards, Dan.

author: Zero999
author_rank: Super Contributor
timestamp: April 20, 2018, 08:01:45 am
content: Quote from: dmills on April 19, 2018, 03:33:20 pmMy mistake. I used the wrong log function. I pressed loge, rather than log10.QuoteYes, a high speed op-amp my do, although it may be a bit more expensive, than a video amp IC, but I haven't looked at this for a long time.