I designed the circuit below after a suggestion that I use two gain stages rather than one large stage with an expensive opamp. This help keep the cost of my circuit down (the same application has been discussed in my previous questions).

My schematic is:

PCB0032 schematic

My thinking when designing the circuit was:

  • U1 buffers the Vcc/2 virtual ground (perhaps unnecessary to buffer it, but not harmful)
  • U3 should match the input impedance of 52 ohms and amplify x10 (OPA355 datasheet)
  • U4 should amplify a further x10

The input signal is 5Mhz and the circuit is on a PCB, seen below.


So for a small input of 10mVpp I expect approximately 1Vpp out. I'll probably reduce the overall gain once I've done some testing but x100 is a good place to start testing.

With a 10mVpp input from a signal generator, checking TP2 with a scope shows virtually the same signal as the input. Testing at TP3 shows a signal x10 larger than the input.

I assume I must have made a very silly basic error. Therefore my question is: can somebody point out what is wrong with the circuit which results in only x10 overall gain.

Note I am not currently testing the comparator (U2), as I haven't yet integrated the data slicer discussed in my previous question. Also I am using the same scope probe for all measurements, so I'm not mistaking x1/x10 settings by accident.

  • \$\begingroup\$ OPA355 has relatively high DC offset voltage. The offset at U3 is +/-9 mV at 25 C, 5 V supply. The circuit is DC-coupled, so this means a 0.9 V error at the input to U2. AC-coupling the output or the in-between stage might be desirable. \$\endgroup\$
    – markrages
    Commented Mar 21, 2014 at 2:46
  • 1
    \$\begingroup\$ @markrages What's the DC gain of the first stage? \$\endgroup\$ Commented Mar 21, 2014 at 2:58
  • \$\begingroup\$ @SpehroPefhany good point, the DC gain is one, so offset voltage shouldn't be an issue in this circuit. \$\endgroup\$
    – markrages
    Commented Mar 25, 2014 at 21:09

2 Answers 2


The impedance of your 100pF coupling capacitor is too high at 5MHz (about 318 ohms). in comparison to the 52 ohm resistor.

The total impedance of the series combination is about 322 ohms (they add in quadrature) so you're getting an ideal gain of 1.58 from the first stage and 10 from the second stage.

  • \$\begingroup\$ I had started writing another question about choosing an appropriate coupling cap and deleted it as I found some existing ones. This is excellent information, I will read up further tomorrow. \$\endgroup\$
    – David
    Commented Mar 20, 2014 at 22:45
  • \$\begingroup\$ @David In the meantime, you can try something that has Xc of ~1/10 of 52 ohms, or more like 10nF. \$\endgroup\$ Commented Mar 20, 2014 at 22:48
  • \$\begingroup\$ Superb! With 10nF I now get approx. x4.5 (13dB) in the first stage and x10 (20dB) in the second. And more importantly I learned something very interesting and useful. Thanks. \$\endgroup\$
    – David
    Commented Mar 21, 2014 at 21:36
  • \$\begingroup\$ @David And why do you think you got a gain of about half in the first stage? Something do with your source impedance maybe? \$\endgroup\$ Commented Mar 21, 2014 at 22:03
  • \$\begingroup\$ Just to clarify that is x4.5 not x0.5 - the source impedance in this case was my function generator which is 50R (no other termination used). \$\endgroup\$
    – David
    Commented Mar 21, 2014 at 22:17

You are using an LM358 as a mid-rail generator - you are hoping it will provide a steady mid-point voltage and be rock-steady in the presence of a 5 MHz signal that it gets inflicted with via the 100kohm resistor R8.

R8 is big in value but your gain is also large - the LM358 has a unity gain bandwidth of 1 MHz and it cannot possibly hope to keep its output stable given that the output is also in series with a 1 kohm resistor - it hasn't the speed to keep up with the tiny (but relevant) fluctuations coming through R8. Read this: -

enter image description here

Replace the mid rail generator with 1k resistors and a 100nF capacitor or, just for fun solder a 100nF cap between TP1 and ground and see what happens.

Even if R9 is shorted you still have the basic problem that your mid-rail is being inflicted with a 5 MHz signal - it's worse though because the resistive mid-rail generator is now an impedance of 100k//100k = 50k.

In fact R8 isn't needed because you have decoupled the input with C1. So, remove R8 and see what happens. Also C1 looks a little too high in impedance even at 5 MHz.

  • \$\begingroup\$ This is a single supply circuit, I assumed I needed R8 and the bias to keep the input within the operating range of U3? \$\endgroup\$
    – David
    Commented Mar 20, 2014 at 22:46
  • 2
    \$\begingroup\$ R8 is not needed - think about it - if you removed R5 (52 ohms) the output would equal the voltage at pin 3. R5 (without R8 being present) cannot remove current and unbias the input because you have C1 (which by the way is definitely too small for 5MHz). \$\endgroup\$
    – Andy aka
    Commented Mar 20, 2014 at 23:41
  • \$\begingroup\$ I agree with Andy. To focus a little: (a) You do not need to set the DC voltage or otherwise "bias" the input to R5. U3 has a negative feeback loop around it, so its output and -ve input will adjust to match the DC value of the +ve input (your Vmid). You don't need to buffer Vmid, because it looks to me like it only needs to go to the (high-Z) +ve inputs of U2, 3, 4. What would be worth doing is bypassing Vmid to ground with a capacitor (to reduce noise on Vmid). And lower-value resistors, as Andy said. \$\endgroup\$
    – gwideman
    Commented Mar 21, 2014 at 9:09
  • \$\begingroup\$ ...cont'd: If you did want to buffer Vmid using U1, then I'm not sure what you wanted to accomplish with R3, but this, combined with the capacitance of whatever load you connect it to, will introduce a delay in the feedback loop, turning U1 into an oscillator. Hence Andy's 100nF to TP1 comment :-) \$\endgroup\$
    – gwideman
    Commented Mar 21, 2014 at 9:10
  • \$\begingroup\$ I have removed R8 and more importantly I spent 20 minutes with the schematic figuring out why my understanding was wrong, thanks Andyaka and @gwideman. I'll fix the mid-rail reference when I start on the comparator stage. \$\endgroup\$
    – David
    Commented Mar 21, 2014 at 21:48

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