Virtual ground and stability driving a capacitive load

[Kevin.D]

It took you all of 3 minutes to answer you must'nt have thought about your reply much

I think you are forgetting that the high gain op-amp with negative feedback lowers the effective output impedance (look where the negative feedback is coming from - it's after the 100 ohm resistors).

 It's irrelevent where the opamp feedback is coming from .The transistor has to source current via it's emitter (it's the emitters resistance that the source current has to come through that is Re+RE where re is intrinsic emiter resitance(small) and RE is any external emitter resistor ) . Think of it like this,Re is  small but even if your bjt turns hard on then max current it can source is now limited by RE .  thus any load cap can only be charged at at time constant of RE*Cload

Ideal caps with zero ESR isn't the issue here. It's a widely understood fact that op-amps have stability issues when driving large capacitive loads.

then  widely misunderstood it seems:) .

The problem isn't 'large caps' it's actually 'small caps' like ceramics with low esr's (Electrolytics with there large esr's are nearly always stable ,) the cap forms a pole you see with the output impedance Rout (in your case the 100 ohm resitor you put in the emitter) . this cap creates a pole  at 1/2 pi C Rout+ESR so you see the bigger Rout or C then the lower frequency (bad) this pole goes . BUT any resistor in series with the capacitor  forms a zero or 'anti pole' which cancels the first pole .This anti pole is at 1/2 pi C esr , so you see if ESR OR C is bigger this zero also moves down in frequency moving down with the pole and this is why you can put massive electrolytics with aparantly little esr on your power supply and it will be stable .Since if we increase capacitance but esr stays more or less the same  the anti pole  still goes down in frequency with the pole.

I couldn't disagree more. Since the circuit is designed to sit exactly midway between the two input rails, without the biasing scheme the transistors would continually be sitting right in the dead zone. This definitely has a detrimental effect on stability and output ripple of the circuit.

Did you even read anything what i typed before you answered ? .The time it takes for the opamp to turn on the complement bjt is small compared to how long a 10 uF capacitor can source  current to the load for  ,thus you may hardly wont even notice it .THATS WHY I suggested you do some load steps with lt spice and  posted a bode lot so you could see a real plot  of the loop response speed  and a 10uS recovery
'you strongly disagree with that' ? 
Let me ask you what you think the GBW of your is ? (with some real capcitive loads on it).but I think you dont need evidence you just to know yeh

[motocoder]

It took you all of 3 minutes to answer you must'nt have thought about your reply much

Turning the other cheek on that comment...

It's irrelevent where the opamp feedback is coming from .The transistor has to source current via it's emitter (it's the emitters resistance that the source current has to come through that is Re+RE where re is intrinsic emiter resitance(small) and RE is any external emitter resistor ) . Think of it like this,Re is  small but even if your bjt turns hard on then max current it can source is now limited by RE .  thus any load cap can only be charged at at time constant of RE*Cload

No, it's not irrelevant. Let's just look at one transistor, the NPN. It's connected to 24V. Assuming 1V drop across the transistor at saturation, it can source (24 - 1 - 12)/100 = 110mA and still keep the output rail at a stable 12V.  That said, I just checked my original circuit, and it IS a typo (should be 10 ohms). Doesn't change my point.

then  widely misunderstood it seems:) .

Ok, I guess the EE community at large is all wrong on this issue, and Kevin is correct. Duly noted.

The problem isn't 'large caps' it's actually 'small caps' like ceramics with low esr's (Electrolytics with there large esr's are nearly always stable ,) the cap forms a pole you see with the output impedance Rout (in your case the 100 ohm resitor you put in the emitter) . this cap creates a pole  at 1/2 pi C Rout+ESR so you see the bigger Rout or C then the lower frequency (bad) this pole goes . BUT any resistor in series with the capacitor  forms a zero or 'anti pole' which cancels the first pole .This anti pole is at 1/2 pi C esr , so you see if ESR OR C is bigger this zero also moves down in frequency moving down with the pole and this is why you can put massive electrolytics with aparantly little esr on your power supply and it will be stable .Since if we increase capacitance but esr stays more or less the same  the anti pole  still goes down in frequency with the pole.

It is indeed a pole in the overall transfer function created by the cap.

Did you even read anything what i typed before you answered ? .The time it takes for the opamp to turn on the complement bjt is small compared to how long a 10 uF capacitor can source  current to the load for  ,thus you may hardly wont even notice it .THATS WHY I suggested you do some load steps with lt spice and  posted a bode lot so you could see a real plot  of the loop response speed  and a 10uS recovery
'you strongly disagree with that' ? 
Let me ask you what you think the GBW of your is ? (with some real capcitive loads on it).but I think you dont need evidence you just to know yeh

I did read what you said. I just disagree. By not biasing those two transistors, the op-amp has to slew rapidly from above to below the midpoint (12V) as the load changes and pulls the signal above or below the 12V mark. So instead of reaching a steady state value with minor output changes as the load varies, the op-amp is swinging wildly over an approximate 1.4V (2x one Vbe drop) range. That has all kinds of potential to inject noise and instability into the system.

[rs20]

I think there's confusion here around what "source impedance" means in the context of a circuit like this. You're both right in some sense.

A. For small changes in output current, the op-amp enforces that the Vgnd voltage stay very constant, so for small currents (and sufficient low frequencies) the output appears to be an extremely low-impedance source, 1 ohm or better.

B. If you're only interested in the ability of the circuit to provide a bulk amount of current, then yes, the 100 (or whatever) ohm resistors make the source reach a limit at roughly 11.xV / 100 ohms. Still performs better than 100 or even 50 ohms tacked on an ideal voltage source, but nowhere near 1 ohm in this sense.

So just saying that "the circuit has an output impedance of x" seems like a vague oversimplification.

[motocoder]

I think there's confusion here around what "source impedance" means in the context of a circuit like this. You're both right in some sense.

A. For small changes in output current, the op-amp enforces that the Vgnd voltage stay very constant, so for small currents (and sufficient low frequencies) the output appears to be an extremely low-impedance source, 1 ohm or better.

B. If you're only interested in the ability of the circuit to provide a bulk amount of current, then yes, the 100 (or whatever) ohm resistors make the source reach a limit at roughly 11.xV / 100 ohms. Still performs better than 100 or even 50 ohms tacked on an ideal voltage source, but nowhere near 1 ohm in this sense.

So just saying that "the circuit has an output impedance of x" seems like a vague oversimplification.

There's no question that the 100 ohms was a mistake. If the circuit is sourcing 100mA, that's 1 watt being dissipated in that resistor. I've since corrected the model to use 2.2ohm emitter resistors (which I think is just enough to help with temperature stability), to add some realistic ESR to those outpuit caps, and to move the series isolation resistor that is used to grab the AC feedback to the output of the op-amp before the transistors. At least in simulation, it's looking very good.

[dom0]

Actually the open-loop output resistance of the circuit is that one thing (okay, together with the o.l. gain/phase response of the op amp, i.e. where it's dominant pole is) that determines stability when driving capacitive loads, since that's whats creating the phase shift of the feedback in the first place.

[motocoder]

Actually the open-loop output resistance of the circuit is that one thing (okay, together with the o.l. gain/phase response of the op amp, i.e. where it's dominant pole is) that determines stability when driving capacitive loads, since that's whats creating the phase shift of the feedback in the first place.

That is true, and I definitely saw that in the sim when I lowered the value on those two emitter resistors.

[motocoder]

Have some updates on this based on building some of the circuits discussed in this thread. Yes, they do oscillate as shown in the simulation

[motocoder]

I ended up building the circuit using the push-pull discrete transistors and putting it in the Keithley Current Source for which it was intended. The Keithley is well within specs, but it bothered me that I could see a few millivolts of noise on the output. This seemed to be originating from my +/-12V (+24 V with virtual ground) supply.

So I kept tinkering with things, and end up with a different circuit. Along the way, I discovered that the primary source of the noise was actually the LM317 I was using to generate the 24VDC from the 31V rectified AC. There's a simple fix for that shown in the TI data sheet. It involves adding a 10uF capacitor from the adjust pin to the negative supply, and a diode to discharge this cap if the output gets shorted to ground.

I also got rid of the discrete transistors, and substituted an LT1010 buffer. This unity-gain buffer has no trouble driving capacitive loads, and has a pretty good amount of output drive. I am still using an op-amp with the feedback connected to the output of the LT1010. This lowers the effective output resistance and is necessary to keep the output within specs as the load changes. I have no issues with noise or stability with this circuit.

I can post a schematic if anyone is interested.

[helius]

Count me as interested. One of my projects is building a buffered virtual ground for a modular analog synthesizer that is made up of NJM45580 op amps and NJM13600 OTAs. Each module has ~30 uF of capacitance on each rail and there are 5 modules.

[motocoder]

Here's a rough schematic. I am using one half of a TL082, not an LT1056 as the schematic shows, as that is what I had in my parts bin. Also, the feedback resistor is a 4.7K, not 10K. I tested it driving about 50uF of capacitance. It may do better - didn't check.



I am using an LM317 to turn the unregulated 31V into regulated 24V. The calibration procedure for this current source has a fine adjustment of the -12V supply, which is the voltage from the virtual ground to the negative rail of the input to the LM317. That's why I'm using an LM317 versus a fixed 24V regulator. The LM317 is configured as shown in figure 13 in this data sheet: http://www.ti.com/lit/ds/symlink/lm317.pdf

[motocoder]

If you looked at that circuit and thought, "wow, that's just trivially simple", you're right. Basically
I just added an LT1010 to the output of an op-amp buffering a voltage divider. I just installed this circuit, plus the LM317 24V regulator with the improved noise supression cap, in my Keithley 225, and it's working much better than the previous circuit. All the issues I had with noise when trying to calibrate the thing are gone. Performs better than the original now.

[jimon]

Beginner question, What does capacitor to ground on voltage divider do ? (C3 on last diagram, or C1 in first post in this thread)
If it's just power rail filtering, then why not from rail-to-ground, but voltage-divider-to-ground ? Or it's just there are more 12v caps than 24v caps ?

[T3sl4co1l]

If noise could be induced to the voltage divider (say from nearby electrostatic sources), the capacitor will tend to keep the AC voltage ground-referenced instead.

Note that this is undesirable for a virtual grounding system, because the virtual ground itself is intended to be the reference for the system!  All inputs, outputs and bypasses should be referenced to the virtual ground instead, so that it is treated as ground (as odd as that might sound ).

If symmetrical PSRR is desirable, then a cap divider should be used in parallel with the resistor divider.  Using large enough caps, this can suffice for many purposes, including power amplifiers.

It is possible to use virtual grounds in a common-negative (or "single supply") system, but those always suffer from startup or shutdown transients, because coupling capacitors are necessary to isolate the -V referenced connections against the amplifier bias voltage (which is around virtual ground potential).

Tim

[motocoder]

If noise could be induced to the voltage divider (say from nearby electrostatic sources), the capacitor will tend to keep the AC voltage ground-referenced instead.

Note that this is undesirable for a virtual grounding system, because the virtual ground itself is intended to be the reference for the system!  All inputs, outputs and bypasses should be referenced to the virtual ground instead, so that it is treated as ground (as odd as that might sound ).

If symmetrical PSRR is desirable, then a cap divider should be used in parallel with the resistor divider.  Using large enough caps, this can suffice for many purposes, including power amplifiers.

It is possible to use virtual grounds in a common-negative (or "single supply") system, but those always suffer from startup or shutdown transients, because coupling capacitors are necessary to isolate the -V referenced connections against the amplifier bias voltage (which is around virtual ground potential).

Tim

What Tim says is correct, but let me add one data point that may not be obvious. For this supply, which is for a Keithley 225, virtual ground noise relative to the "-12V" rail is what is important. For example, you adjust the LM317 supply until the virtual ground is exactly 12V above this rail. I won't go into detail, but read the Keithley 225 manual for details. That's why there's a cap to the lower rail but not to the upper rail.