Protecting Buffer Op Amp

[N9]

I've been working on a low-power, always-on buffer for my pedalboard. I've got the basic design, but I'm a little puzzled about the reliability aspects, especially when it comes to protecting the op amp.

Here's my circuit:


And here are the main things I'm puzzled about:

- How can I determine if the input protection is sufficient? The OPA703 can withstand 10mA surge currents on its inputs, but I'd rather not rely on that.

- Should the output have more series resistance for current limiting?

Any other comments are very welcome; I'm not terribly familiar with op amp circuits and there are undoubtedly things I've missed!

[R.G.]

And here are the main things I'm puzzled about:
- How can I determine if the input protection is sufficient? The OPA703 can withstand 10mA surge currents on its inputs, but I'd rather not rely on that.

The problem is, you have to rely on something. There are always inputs which can fry the opamp, no matter how it's protected. All you can do is read the specs and take your best shot. With a 2K series resistance and a diode to the power supplies, you're good up to the voltage that would kill the diode. If you have  0V power supply on one side, the voltage that kills a 1N4148 is the max current spec on the diode times the 2K current limiting resistor. I make that 100ma*2K = 200V. Pretty good. But a 240Vac line connected to it would kill the diode, then the opamp. You could put in 1A diodes. Or 10A or 100A diodes. With those, the input currents would fry the power supply, then the opamp. You can only protect so far.

- Should the output have more series resistance for current limiting?

Does the opamp spec say it's internally current limiting?

Any other comments are very welcome; I'm not terribly familiar with op amp circuits and there are undoubtedly things I've missed!

The biggest thing to do is to read the datasheet for the limiting conditions and absolute maximums.

[N9]

The problem is, you have to rely on something. There are always inputs which can fry the opamp, no matter how it's protected. All you can do is read the specs and take your best shot. With a 2K series resistance and a diode to the power supplies, you're good up to the voltage that would kill the diode. If you have  0V power supply on one side, the voltage that kills a 1N4148 is the max current spec on the diode times the 2K current limiting resistor. I make that 100ma*2K = 200V. Pretty good. But a 240Vac line connected to it would kill the diode, then the opamp. You could put in 1A diodes. Or 10A or 100A diodes. With those, the input currents would fry the power supply, then the opamp. You can only protect so far.

I phrased this question poorly; by "rather not rely on," I meant that I don't want to rely purely on the op amp to protect itself against ESD or other ugly inputs. Adding the input resistor and diodes seemed like the best first line of defense...Are there better strategies? ESD, in particular, seems like a tough nut to crack. The voltages and currents can be alarmingly high, even for "small" shocks.

Does the opamp spec say it's internally current limiting?

It can handle "unlimited" output short circuits.
Hmm, perhaps I've been misinterpreting the various schematics I've seen? I commonly see small resistors at the output in op amp circuits; I assumed they were for current limiting. Perhaps they're just isolation resistors for capacitive loading?

Apologies if some of these questions are silly; I may have confused myself on these issues!

[teemuk]

They can be for many purposes, and you'd have to evaluate that based on the circuit.

- They can limit current. Yes, many opamps are internally limited to maybe 20 mA or so but you can take that value and calculate what impedance loads you can drive with that current and what maximum voltage swing the opamp can provide before the current limiting kicks in. Typically an opamp may need at least few hundred ohms of loading so that current draw is not high enough and full output swing (in volts) can still be produced. A very related application is for instance diode clipping. You can't usually connect the diodes straight to output because when diodes forward conduct the opamp would be nearly the sole current limiting element. Usually some series resistance is required in the output.

- Series resistance to capacitive loads. Yes, this is another very typical application for such resistors.

- Any basic function, not neccessarily even related to Opamp itself: "Isolating" circuits with resistance, mixing, resistive dividers, RC filters, etc. A clever mind probably compiles an A4 sheet  -long list of stuff you can do with a resistor.

[merlinb]

Are there better strategies? ESD, in particular, seems like a tough nut to crack. The voltages and currents can be alarmingly high, even for "small" shocks.

A small capacitor from the opamp input to ground is sometimes included (10pF say), which would help reduce the dV/dt of any potential spikes (which is often more damaging than the voltage itself).
I suppose you could go as far as adding a MOV, but really it's overkill for audio applicances like this. Also I'm not sure of the capacitance of a MOV; perhaps it is too large. An alternative would be a neon lamp!
I pair of anti-parallel diodes between the opamp inputs is something else you could add. Oh, and a fuse in series with the input (or maybe a polyfuse/resettable fuse?) But now we're getting silly.
Perhaps look at some DMM schematics and see what protection measures they use.

I commonly see small resistors at the output in op amp circuits; I assumed they were for current limiting. Perhaps they're just isolation resistors for capacitive loading?

Yes, it's usually to prevent oscillation when driving capacitve loads. Most opamps these days are current limited/ proof against output short circuits.

[samhay]

I'm not sure what value R5 is - is to 10 ohms. Either way, what is it doing in the feedback loop of the op-amp?

If you can live with an input impedance of a few hundred k, you could consider using an op-amp with BJT inputs, as I believe they are less susceptible to ESD than those with FET inputs.

[R.G.]

I phrased this question poorly; by "rather not rely on," I meant that I don't want to rely purely on the op amp to protect itself against ESD or other ugly inputs. Adding the input resistor and diodes seemed like the best first line of defense...Are there better strategies? ESD, in particular, seems like a tough nut to crack. The voltages and currents can be alarmingly high, even for "small" shocks.

The datasheet for all modern semiconductors will include the amount of ESD tolerance the manufacturer guarantees, as well.

I take a mixed approach to ESD. I come from a time when ESD was not considered for semiconductors until the introduction of MOSFET transistors and then CMOS logic. It was only then that more attention was paid to ESD. As a practical matter, bipolar and JFET devices are modestly immune - meaning that really big discharges will get them, but normal handling that's NOT in winter-heating dried air and synthetic carpet environments are generally not issues. MOS devices need special handling, such as not wearing shoes (yes, really - this prevents most static buildup unless you're wearing all-synthetic clothes) and touching a ground before touching the part pins. It's not strictly according to ESD handling specs, but I have not lost a MOS part in a decade or two this way. Once MOS parts are in the circuit, they are fairly immune to ESD, excepting when they have inputs going directly off-board.

In that case, I worry a bit about ESD. For CMOS, using a 1K or larger resistor in series seems to be all they need. For MOS input opamps, I'd probably use a series resistor in the K range and maybe protect it with diode clamps to the power supply.

As with most topics on pedal building, there is an article on this at geofex.com; in this case "What are all those parts for?"  
http://www.geofex.com/circuits/what_are_all_those_parts_for.htm

It can handle "unlimited" output short circuits.
Hmm, perhaps I've been misinterpreting the various schematics I've seen? I commonly see small resistors at the output in op amp circuits; I assumed they were for current limiting. Perhaps they're just isolation resistors for capacitive loading?

Yes, as noted above, they're for isolating the output from capacitive loading for stability reasons. The TL072, otherwise a sweetheart of an opamp, is especially prone to going into oscillation from certain capcitive loads.

There is another situation  where series resistance is useful, and that's driving inductive loads. Inductive loads can store current, then force voltage back onto the output pin. In that case, a series resistor and clamp diodes on the output pin can save the device. The resistor absorbs some of the voltage, limits current, and the diodes channel the voltage into the power supplies without exceeding the voltage limits on the output pin.

Apologies if some of these questions are silly; I may have confused myself on these issues!

No, you're just being careful. It takes a long time to pick up a sense for what is wise protection and what is silly overkill. Those two can look remarkably similar.

I'm not sure what value R5 is - is to 10 ohms. Either way, what is it doing in the feedback loop of the op-amp?

It's probably for capacitive isolation, although including it in the feedback loop makes that problematic. 10 ohms is pretty cold comfort for current limiting on an opamp if the opamp won't protect itself.


Resistors inside the feedback loop simply add to the internal impedance of the opamp open loop, and are reduced by feedback. They do act like internal current limits at the ends of the voltage range or whenever clipping otherwise prevents feedback from hiding them.

[N9]

Many thanks for the replies!

R5 is included because the datasheet calls for it when driving capacitive loads:



Placing the resistor like that seemed unusual to me as well; perhaps it's a quirk of this particular op amp?

[slacker]

I think the reason for putting it in the feedback loop is so that the resistance doesn't affect the output voltage, I think that's what they mean by DC accuracy.
If you put the resistor after the opamp then it forms a voltage divider with RL, so Vout will be Vin * RL/(RL+Rs), for most of what we do that doesn't matter because RL will be very large compared to Rs, so there will be no audible difference.
With Rs in the feedback loop, the Rs/RL voltage divider tries to pull Vout down but that lowers the voltage on the inverting input, which is connected to Vout. The opamp wants to make both inputs the same voltage so the output of the opamp will go up until Vout = Vin. This basically cancels out the voltage divider effect.

[samhay]

cool - I learnt something today. Thanks N9 and Ian.

[PRR]

> bipolar and JFET devices are modestly immune

His OPA703 says it is "CMOS".

Perhaps not as fragile as the earliest stuff.

I take the 10mA input current at face value. I would bet more than a nickel that there are actual diodes inside, and that they (and their bond wires) will take 10mA all day and all of the night.

If we take the input pin to -1V or +10V, "infinite" current would flow.

You have 2K in series to the Big Bad World. So -21V or +30V will be 10mA and will be safe.

A loudspeaker line can exceed 21V peak.

But why 2K? Why not more?

Yes, 2 Meg might get you into hiss trouble (but be safe for 20,000V!). What is the hiss of the OPA703? 50nV/rt.Hz or 7uV in 20KHz. (Compare TL072 at 2uV.) Going 250K series resistor raises hiss 3dB, so you want to be well below 250K.

And study your roots. Nearly ALL late-1950s-1960s Fender amps have 34K input series resistors. This is mostly to cut radio, but also limits abusive damage to delicate grid wires. Taking the same 10mA, the input is protected to 340V. And if you have >340V laying loose on a stage, it isn't the 3-buck chip I would worry about.

The output-- yes the OPA703 is self-protected. Bottom of page 7 shows the output acts like about 100 ohms dead resistance, which is reasonable for CMOS. Another 10 ohms inside the loop does no good but little harm. Top of page 8 shows significant overshoot for capacitances typical of stage-cords (30pFd per foot, 300pFd-1000pFd for 10 and 30 foot cables). This alone may cause no real trouble in a bandwidth-limited system (guitar amp), but it may not be as simple as it looks and I'd stick some resistance between NFB loop and output jack. Less than 10K (so you could drive 30 feet cable to 17KHz) but more than 100 ohms (so the OPA703 can still find its NFB when jack is shorted by capacitance or bad cable). 1K is a nice round number.

Your 27K+27K bias network draws 1/10 as much current as the OPA703. That's fine, but it is likely you could go a lot lower current and still bias-up fine. Bias current is 10pA max, so go >>100pA, say 1nA. The "27K" could be 4 Gig-Ohms!! Well, actually you have more than 10pA of stray leakage across your PCP, and in your bypass and coupling caps. But 1Meg seems a fine number. And now your C2 10uFd can be like 0.01uFd which is a film or ceramic with low-low leakage and infinite life.

[PRR]

> Placing the resistor like that seemed unusual to me as well;

I've never seen that. I see what it does. Quirk or not, dunno.

The output of the opamp is inductive. Put a cap on it, we have a series resonant circuit. Usually there's enough stray resistance to keep it mild. But I see a funny local NFB loop in the output stage, which may really be making a good low-loss inductance. Some resistance damps it. Inside the loop, the overall gain is unaffected by loading.

I say dumb resistor outside the loop. You are going into fairly high impedances, it can be large. We don't really need gain of 0.9999, gain of 0.99 or 0.95 is all the same to the ear.

And if using the OPA703 for its low supply current--- it can dump 50mA peak into a short. So a shorted output cable with any decent signal will suck >20mA from your battery (you were expecting 0.2mA). With a 1K out the worst-case battery drain is 2mA; still a drain but 10X less than the no-resistor drain.

[N9]

And if using the OPA703 for its low supply current---

Yeah, I chose the OPA703 because it's a low-current, rail-to-rail op amp with reasonable bias current, slew rate, etc. I figured that a rail-to-rail device would be desirable for a buffer, since it's important for it not to clip. Choosing op amps is quite a puzzle in and of itself.

it can dump 50mA peak into a short. So a shorted output cable with any decent signal will suck >20mA from your battery (you were expecting 0.2mA). With a 1K out the worst-case battery drain is 2mA; still a drain but 10X less than the no-resistor drain.

I hadn't considered this! A 1K or 2K resistor seems like a good idea.

You have 2K in series to the Big Bad World. So -21V or +30V will be 10mA and will be safe.
A loudspeaker line can exceed 21V peak.
But why 2K? Why not more?

I figured I might as well start with a low value, since most guitar buffer circuits omit that resistor entirely. You're right though, a value of 20K-50K would work better.

Your 27K+27K bias network draws 1/10 as much current as the OPA703. That's fine, but it is likely you could go a lot lower current and still bias-up fine. Bias current is 10pA max, so go >>100pA, say 1nA. The "27K" could be 4 Gig-Ohms!! Well, actually you have more than 10pA of stray leakage across your PCP, and in your bypass and coupling caps. But 1Meg seems a fine number. And now your C2 10uFd can be like 0.01uFd which is a film or ceramic with low-low leakage and infinite life.

A very good point. I should read up more on virtual grounds. Every circuit I've seen uses pretty small resistors for the biasing network, even when using op amps with low bias currents. I wonder why?

Another thought: wouldn't large-value resistors in the bias network make it much noisier?

[merlinb]

Another thought: wouldn't large-value resistors in the bias network make it much noisier?

No, because it is in parallel with the input signal, rather than in series. (Noise from the divider itself is shunted away by C2 anyway).

[R.G.]

A very good point. I should read up more on virtual grounds. Every circuit I've seen uses pretty small resistors for the biasing network, even when using op amps with low bias currents. I wonder why?

I think that's largely because very few people designing effects understand bias networks very well. Mostly, people have seen 10K/10K networks in other effects and don't think any further - which is usually OK, because that does work for most things. There is a partial description of the what and how of bias networks at geofex:
http://www.geofex.com/circuits/biasnet.htm
The fish hook hidden in there is that you have to know what currents are coming out of the bias network before you can do a good job. For your case, with a single device sipping bias current and nearly none of that, it's easy.

Another thought: wouldn't large-value resistors in the bias network make it much noisier?

Yes, and no. As Merlin says, a cap to ground from the bias network shunts the thermal noise of the bias resistors themselves to ground. A series resistor to an input still has thermal noise, but no voltage induced excess noise and only the current noise caused by whatever bias current actually flows. So if you must have a high input impedance, and must provide it with resistances, the way to do it with least noise is with a series bias voltage generator string, decoupled to ground with capacitance, and a series resistor to the input being biased. That's about as low a noise contribution as you can get with simple resistor biasing.

You can do better in terms of input impedance by bootstrapping techniques to raise the input impedance with signal from the output, and split the incoming resistances from the bias network into smaller resistors, which lowers their thermal noise contribution. The bootstrapping gets the high impedance back for you while letting you use smaller resistors for lower thermal noise.

[PRR]

> wouldn't large-value resistors in the bias network make it much noisier?

Condenser mikes (where hiss is easily evaluated) use 200 Meg for tubes, higher (a Gig) for JFETs.

You want to be *higher* than the source impedance, as high as possible (without getting into other trouble).

Then the source-resistance sucks-up most of the bias resistance's hiss.

Consider 200K source, a rounded approximation of a guitar pickup and cord around its ~~3KHz resonance. While the resonance is narrow, the relative effects are the same if we pretend this is a 0-20KHz bandwidth (which I happen to know).

200 Ohms makes about 0.15uV hiss in 20KHz. So 200K makes about 4.7uV hiss.

Grab a 2 Meg resistor. Its hiss is 15uV.

And its hiss comes out of a 2 Meg source.

Connect the 200K across the 2Meg. That 15uV is divided by the ratio 200K/2.2Meg or 1/11. So the 2 Meg is contributing 1.36uV.

The 200K contributes 4.7uV.

The RMS sum of 1.36uV and 4.7uV is 4.9uV.

Meanwhile say we have a signal in the 200K. Say 1mV.

The 2 meg shunt on 200K is again a voltage divider, the signal output is 10/11 or 0.909mV.

No load: 1mV signal, 4.7uV hiss 
2 meg load: 0.9V signal, 4.9uV hiss 

We have lost 1.24dB Noise Figure by loading with a 10X resistor.

With an infinite load we'd have a 1.24dB better signal/hiss ratio.

A 1X load (200K) gives a 3dB NF. (6dB loss of level but 3dB lower hiss.)

100X load is about 0.1dB NF.

A couple dB Noise Figure is usually excellent. (Wavely may disagree but he does something with star-whispers.) 10X even 5X loading is common. 200 Ohm mikes are conventionally loaded with 2K, and this is rarely worth questioning.

This is all "ideal" resistances. Carbon-film is pretty close to ideal. Carbon-Composition is notoriously much-much worse than ideal. Full Metal resistance is a touch-stone but hard to find in Megs. Metal-film is no longer grossly expensive, but the slim data I have seen show carbon-film hiss comparable to metal-film and nearly ideal.

[N9]

Ah, I think I get it now! Thanks guys!

I've been continuing to read about ESD protection, and have a new question: would a dedicated transient voltage suppression diode be inappropriate for this application? The 1N6270A (http://www.onsemi.com/pub_link/Collateral/1N6267A-D.PDF)
seems like it could work; it clamps at 9.1V. I can't find any capacitance information on the datasheet, however. In any case, it certainly looks like an interesting option.

[PRR]

> dedicated transient voltage suppression diode

1500 watts peak, 5W steady-state??

Anyway.... either the diode is scaled lower than the rails, limiting headroom and making r2r chip moot; or the diode threshold is above the rails and the CHIP's diodes conduct first.

Also: if that diode is hit from a zero impedance source it WILL burst. You need some series resistance.

Put 33K in there. It is dang hard to damage things with 33K series.

Remember that millions of $39 pedals are sold with much less attention to ESD, and most of them work forever (dying of crap pots, crap jacks, crap soldering, etc).

[N9]

Anyway.... either the diode is scaled lower than the rails, limiting headroom and making r2r chip moot; or the diode threshold is above the rails and the CHIP's diodes conduct first.

OH! I forgot about the on-chip diodes. Whoops.


Based on the feedback from this thread and other articles I've read, I've made the first revision to the circuit:



Having all those diodes at the input will certainly protect the op amp, but they present a fair amount of capacitance. Is this an issue?

I had another thought while revising the circuit: Hypothetically, would putting R4 before C3 make any difference?
I realized that the vast majority of circuits I've seen place the input resistor in front of the input cap, but I've never heard a rationale for it.

[R.G.]

Having all those diodes at the input will certainly protect the op amp, but they present a fair amount of capacitance. Is this an issue?

Maybe. What is the capacitance of the diodes? The datasheet should have it. The opamp input itself has some stray capacitance, also on the datasheet. The capacitive losses only count with the 2.2M resistor, as the opamp input is probably much higher.

The frequency rolloff of the input is then F = 1/(2*pi*2.2M * (sum of the capacitors)). Maybe it matters, maybe not. Have to do the math.

I had another thought while revising the circuit: Hypothetically, would putting R4 before C3 make any difference?
I realized that the vast majority of circuits I've seen place the input resistor in front of the input cap, but I've never heard a rationale for it.

Makes no difference. The order of items in serial connections makes no difference to the end point voltages or current through them, as long as you're not looking at or loading any of the mid points.