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Assume I have a 50MHz signal with 300mV pk-pk amplitude and I want to amplify it to 8v pk-pk. For this purpose I selected AD8055. This is a 300MHz VFO that is suitable for this purpose. I selected resistor values according to its datasheet (Table3 page 16). As the maximum gain with that config is 10, I need two op-amps connected in series (schematic below) hoping the first device produces a 10 fold gain and second one a 2.2 .

The first one produces a 3 volt that is OK. but the second one's gain never goes higher than 1 ( the total never goes higher than x10).

This problem is just seen in high frequencies (in lower frequencies as 1MHz it can make a 9v pk-pk output). The supply voltage is +/-5v. The first output is 3v pk-pk that is well below the maximum input (+/-2.5v) for the second one.

I tried using different resistors ( a Pot) between 100ohms- 1k but the second Op-Amp gain never goes higher than 1. I also tried attenuating the input voltage ( reduced it to 100mv). In this case the first output becomes 1v (gain=10) and the second output can exceed unity (it reaches gain=3) but finally the total output is 3v pk-pk yet.

Is there any consideration that I am missing?

enter image description here

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    \$\begingroup\$ The first stage has a gain of 10. The op amp only a gain bandwidth product of 300MHz, so you never get 50 MHz bandwidth out of this (there also needs to be some margin left). Compare also the frequency responses for gain 10 on page 1 in the datasheet. \$\endgroup\$ – Andreas H. Oct 12 '13 at 20:34
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Try two CFOs each with a gain of √(8/0.3) = 5.16.

As you know, VFOs have a gain-bandwidth product, meaning that the bandwidth decreases as the gain increases. If you place two VFOs in a cascade, as you have done in your circuit, the bandwidth of the whole system is limited by the lowest bandwidth of either VFO. Therefore, both VFOs should have as small a gain as possible. This is achieved when the gain of each VFO is the square root of the total gain of the system, in this case √(8Vpp/0.3Vpp) = 5.16. CFOs are very much less susceptible to change in bandwidth as a function of gain, but since the frequency is so high, it may be worth considering.

[Original] You might want to consider using a current feedback amplifier approach. They can be orders of magnitude faster than voltage feedback amplifiers and their gain is much less affected by frequency.

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  • \$\begingroup\$ +1 for good suggestion!. After your suggestion I tried AD811 ( a CFO ) . It got better and I could reach 6vpp but the same scenario happens on this voltage ! I think I never can get it to work! \$\endgroup\$ – Aug Oct 13 '13 at 1:41
  • \$\begingroup\$ Try two CFOs each with a gain of √(8/0.3) = 5.16. If this doesn't work, perhaps you should consider using a RF power transistor for amplification, for example Motorola MRF492. \$\endgroup\$ – Atuos Oct 13 '13 at 8:59
  • \$\begingroup\$ Could you please explain the formula? √(8/.3) \$\endgroup\$ – Aug Oct 13 '13 at 9:33
  • \$\begingroup\$ As you know, VFOs have a gain-bandwidth product, meaning that the bandwidth decreases as the gain increases. If you place two VFOs in a cascade, as you have done in your circuit, the bandwidth of the whole system is limited by the lowest bandwidth of either VFO. Therefore, both VFOs should have as small a gain as possible. This is achieved when the gain of each VFO is the square root of the total gain of the system. CFOs are very much less susceptible to change in bandwidth as a function of gain, but since the frequency is so high, it may be worth considering. \$\endgroup\$ – Atuos Oct 13 '13 at 9:42
  • \$\begingroup\$ Wow! I tried 2 CFO ( AD811) with gain of 5 ( I had to change the RG from 50 ohms to 100 ohms, reducing RF didn't work). Now I can achieve even 20 Vpp! at 50MHz! (Never believed that!) .As this answer will remain for others who may later see it, please add the last comment to your answer, and explain that formula to make a good conclusion and let me know done. I will mark it as correct answer. \$\endgroup\$ – Aug Oct 13 '13 at 9:51
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If you look in the datasheet (page 4, output characteristics ... also figure 30 on page 11) you will see that the AD8055 is not specified for any more than +/-3V output with +/- 5v supply. So, the output is out of dynamic range.

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My guess would be that for such large output swings, you're running into output slew rate limitations, which is spec'd at 750V/µs, or 0.75V/ns for that chip with gain set to 2 and a 4V output step.

I would further note that the 300MHz gain-bandwidth product means that you'll only get (best case) a gain of 6 from each stage at 50 MHz, and this would be assuming no negative feedback. Note that these bandwidth numbers represent -3 dB points, so the actual gain at the corner frequency will be 0.707× the "nominal" value.

But finally, note that the output swing for the chip is only guaranteed to be ±2.9V, with a typical value of ±3.1V.

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  • \$\begingroup\$ I changed the first gain=5x and second gain to 3x ( expecting total=x15). Now I should see an output swinging between -2.25 to +2.25 (5x3x300mV =4.5 pk-pk = -2.25 to +2.25). But result is the same as before (-1.5 to +1.5). The same also with 4x,4x and 5x,2x. \$\endgroup\$ – Aug Oct 12 '13 at 21:36
  • \$\begingroup\$ This is exactly the result I get with BJT ( 2N2219 , BC108, BC109, BC114 !) It seems there is no 50MHz 6vpp in the world!! \$\endgroup\$ – Aug Oct 12 '13 at 21:43
  • \$\begingroup\$ Oh I found it with your guidance. I had not noticed that 300 MHz is "-3dB" Bandwidth product. I should have entered a 0.707 in my calculations: 5 x 3 x 300mV x 0.707= 3.185 and this is exactly I see on my oscope. If this assumption is correct please add it to your answer until it becomes the perfect answer ans I will mark it as the correct. Thanks again. \$\endgroup\$ – Aug Oct 12 '13 at 21:57
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  1. You will need rail-to-rail opamp for 8Vp-p with 5V supply voltage. The normal output stages will eat at least 1.5V with high distortions.

  2. These 300MHz are for K=1; For K=10 you get 30MHz maximal frequency.

  3. 2 stages at 30MHz gives you lower than 30MHz at -3dB level.

So, you need 2 stages with lower gain (but higher BW) - 5 for example. And opamp with Ft at least 500MHz and at least 1000V/us

Something like LT6200-5? (After some check - NO, it has only 200V/us and is very slow for 8V p-p)

BTW, there are so called video amplifiers with fixed/programmable gain. They are not exactly opamps, but are much faster.

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  • \$\begingroup\$ Thanks! I agree with numbers 2,3 but why I can't amplify to 8vpp with +/-5v (= 10v) ? \$\endgroup\$ – Aug Oct 12 '13 at 21:12
  • \$\begingroup\$ @Aug Because 8Vp-p means from -8V to +8V output and this is more then +-5V. \$\endgroup\$ – johnfound Oct 12 '13 at 21:14
  • \$\begingroup\$ But I think Vpp is the "Difference" between max and min. 8vpp means from -4 to +4: en.wikipedia.org/wiki/Amplitude and whatis.techtarget.com/definition/peak-to-peak-pk-pk This is exactly my digital oscilloscope calculates \$\endgroup\$ – Aug Oct 12 '13 at 21:22
  • \$\begingroup\$ @Aug - Oh, my bad. I need some sleep. :) Anyway +-4V is also too big for +-5V power, but possible with rail-to-rail opamp. \$\endgroup\$ – johnfound Oct 12 '13 at 21:29
  • \$\begingroup\$ Yes I should go for 6Vpp , but please note my comment under Dave Tweed's answer. New design is not working yet ! \$\endgroup\$ – Aug Oct 12 '13 at 21:39

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