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I'd like to build a preamplifier for measuring the noise of very quiet linear regulators using a baseband spectrum analyzer. I found this page, which presents a few designs for this and looks promising. However, I'm having some trouble understanding how the charging transient protection part of the circuitry provides any protection at all. I've copied the schematic here for convenience.

enter image description here

The basic operation of the preamplifier seems fairly straightforward. U101 and U102 act as inverting amplifiers, each providing -31x gain (collectively +60 dB). The output is taken from U102's output. Since U101's inverting input presents a virtual ground, the input blocking caps with R101 form an RC highpass filter with 3dB cutoff frequency of 1.45 Hz.

I don't really understand the purpose of U103. In theory it's function also seems straightforward. First, the purpose is presumably to limit the duration of the high voltage at U101's inverting input. When a DC voltage is applied to the input, that will initially pass right through the highpass filter to U101's inverting input until it dies away as the blocking caps charge through R101. U103 is also configured as an inverting amplifier with large (negative) gain at low frequencies and decreasing gain at higher frequencies. Presumably, it "protects" U101 by sinking current from the initial DC voltage to limit the time that high voltage is applied to U101. However, because R103 is 10M it won't be able to sink much and therefore won't really limit the duration of the high voltage applied to U101. For example, with +/-15V rails, the current it will sink is in the uA range, whereas the current through the highpass filter is initially in the mA range (using a 20V input). I could decrease R103's resistance, but at some point I'll start to hurt the low frequency response. Another easy thing to do would be to increase C112 to 22u, for instance.

Am I right in thinking U103 doesn't provide much protection? Have I missed something?

FYI I didn't choose version 1 of this because I think it's better than 2 or 3 (I'm more inclined to built v2). But, it was sufficient to discuss the protection function.

MattHusz
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  • Having made a rather similar circuit recently, let me say that R101 ought to be much smaller. As it is, it is the largest noise producer in the circuit. If you change to even lower noise opamps (or transistors), one should make sure that capacitor ESR + R101 has negligible thermal noise. And yes, then you have to really think about charging transient protection, as those transient can reach Amps that will surely throw most op-amps into a latch-up state. – tobalt Jul 07 '22 at 05:51
  • What sort of target do you have in mind? Will the capacitor ESR (in mOhms) not be a negligible contributer in any case? Two simple options come to mind: v2 of this circuit splits R101 and adds limiter diodes. I could make the resistors as small as possible for those diodes. Another option would be to precharge caps through a larger resistance and then physically switch over to the normal position, which could use a much smaller resistance. Or I could think of a scheme to do that automatically. If you have any thoughts here would love to hear them. – MattHusz Jul 07 '22 at 11:48
  • The other downside of a lower resistance is a higher cutoff frequency unless I correspondingly increase the blocking capacitance. – MattHusz Jul 07 '22 at 11:49
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    The capacitor mOhms are indeed usually negligible for noise, but electrolytic caps with several Ohms of ESR shouldn't be used, and certainly 1 kOhm is series resistance usually isn't negligible. I tried to stay well below 1 Ohm series resistance for my amp. Your ideas are good options. The diodes should be Schottkies to conduct before the opamp input does. – tobalt Jul 07 '22 at 13:27
  • How did you get the capacitance high enough? Rseries=1 requires hundreds of mF to get a cutoff around 1Hz. That puts you outside the realm of plastic film and into aluminum electrolytic territory (which as you mentioned typically has higher ESR). Another option would be to use batteries for a DC bias (also discussed by the linked page) and remove the blocking cap altogether. – MattHusz Jul 07 '22 at 13:51
  • The cutoff frequency is not determined by the ESR. It is given by C * Rshunt, where Rshunt is a resistor going to a DC reference point, in your case R103. – tobalt Jul 07 '22 at 14:10
  • In a spice simulation, changing R101 to 1 and R102 to 31 increases the low-frequency cutoff by a factor of 1000 (ie by the same factor we decreased R101), so it looks like C10x + R101 are setting the cutoff, not R103. Changing R103 doesn't change the cutoff much. This is exactly what I'd expect given the gain equation for an inverting amp and the fact that the inverting input acts as a virtual ground. – MattHusz Jul 07 '22 at 14:38
  • tbh I am not sure about your circuit with this strange integrator. If you need lower cutoff, use a non-inverting stage first, and put 10 M to GND. – tobalt Jul 07 '22 at 14:44

2 Answers2

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There are antiparallel diodes between IN+ and IN- inside the OpAmp. Since IN+ is connected to GND, the voltage at IN- of U101 is clamped to around +/- 0.6V. This is at least some kind of protection and it accelerates charging of the input capacitors until the circuit is in the linear operating range.

The feedback from the integrator U103 is a DC offset cancel feature. The offset error of U101 is amplified by U102 and will be significant. U103 forms a slowly changing compensation voltage and feeds this back to U101. The remaining offset error is that of U103.

Jens
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  • Can you elaborate on this?: “ The feedback from the integrator U103 is a DC offset cancel feature.” – MattHusz Jul 07 '22 at 02:07
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Adding to what Jens said about U103 being a DC offset cancel feature, U103, along with 106 & C112, form an integrator which is part of a control loop to reduce the DC offset. Having high gain will amplify any DC offsets in the first two stages. The integrator feeds back a near DC correction voltage which causes the output of U102 to be very close to zero volts.

enter image description here

The upper part of the simulation is just a straight amplifier with the resultant green trace with a DC offset of about 405mV.

The lower part of the simulation incorporates the integrator with the resultant red trace. Note that the DC offset is nearly zero.

Note, I cheated a bit and applied a 100uV DC offset voltage to V1 to simulate the offset voltage one may encounter in a typical opamp circuit. The opamp models lacked DC offset issues (i.e., a bit too perfect).

qrk
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