Phono Preamp Project

The Novo kit demonstrated that discrete circuitry could be very problematic to the inexperienced: too many things to get the wrong way round, and in the case of a transistor - the rest (or most of them) blow with it.

Therefore, for the Genera phono preamp it has to be an op-amp (operational amplifier chip), and just one at that!. But is that a dual op-amp, or a single?

In the Genera phono preamp it's going to be two singles, one per channel. The reason why will become apparent as the design progresses.

What performance can you expect? I'm looking for another winner here, so this is not going to be second-fiddle to a Gram Amp 2 phono preamp: I want it to do better or at least equal.

So how do I go about chosing an op-amp?

First we have to take a look at the RIAA reproduction curve.



The chart [1] above shows the output from a RIAA record according to a magnetic (moving magnet or moving coil) cartridge.

To end up with a flat frequency response, the phono preamp frequency response has to be the inverse - the chart upside down.



The above chart [2] shows the curve from the first chart, upside down - the black curve. It is part of the blue-grey curve which shows the extended frequency response we can expect.

This is the RIAA reproduction curve of our proposed phono preamp.

The red curve is the gain bandwidth product of an op-amp called the LF356, and below [3] is the same taken from the data sheet: the curve has the callout LF156, which is the family name covering the LF356.



The gain is inside the curve, that is to say to the left and below the curve. At approx. 5MHz the curve intersects 0dB on the x axis. Therefore the Gain Bandwidth Product is 5MHz. In actuality it is stated on the data sheet as 4.5MHz.

Where the gain is maximum at about 105dB - a gain of approx. 178,000, the bandwidth read from the graph is only about 30Hz. In fact, if we know the Gain Bandwidth Product is 4.5MHz then at 105dB gain the bandwidth is 4,500,000/178,000 = 25Hz, which correlates with our estimations off the graph.

To ensure an undistorted phono preamp using this op-amp, the phono preamp gain and bandwidth needs to be well within the op-amps curve - a rule of thumb being a factor of ten. For a 41dB gain (just over 100, and suitable for a moving magnet pick up) as shown in the second [2] graph, the rule of thumb exists frequency wise. In the direction of gain we have over 40dB spare, which more than covers the rule of thumb... 40dB being 100.



Edited by Graham Slee - 17 Jan 2010 at 6:36pm

Following this with interest. 

SLEW RATE

From previous post: "Notice C2 in the voltage amp differential. I'll be talking about that next."

Notice C2 is the same as C in the diagram below...



The above diagram taken from the Art of Electronics, Horowitz and Hill, as featured on Google Books (skip back a page)

Here (fig. A) we have the innards of the opamp represented by bipolar transistors, but you may be able to see the similarities (because they are there) between it and the innards of the LF356 shown last post.

Slew rate is one of those really hard to grasp properties in electronics. Here I will do my best to explain so you all understand it a bit better.

Capacitors store charge. To obtain the charge you'd place a capacitor across a voltage. That would cause a current to flow from the voltage source for a period of time required for the capacitor to adopt that charge - with me so far?

So we have five properties here: charge (Q) in coulombs; capacitance (C) in farads; voltage (V) in volts (of course); current (I) in amps; and time (T) in seconds.

Now here is a (dreaded maths again!) formula...

CV = IT (which means: C x V = I x T )

which is a great way for simple folk like me to remember what I'm supposed to be doing...

Also, Q (charge) = CV, so because CV = IT, also Q (charge) = IT  (yes?)

Slew rate is volts per micro-second, and so we can use the V and the T like this: V/T (Voltage divided by Time), which represents volts per time (got it?).

So we want to know V/T, and by rearranging CV=IT we get...

V/T = I/C

So volts per time = current per capacitance

Great!

From the data sheet we know the slew rate is 12V per micro-second which can be put as...

12/0.000,001.T (which means 0.000,001 x T, which means micro-second), or put it all on the top line (by doing dv/dt), which is 12,000,000.

From the data sheet we know C = 10pf.
10pf is 0.000,000,000,01 Farads - a bit cumbersome, but never mind (the scientific among us will be making use of that calculator "x10x" key)

So we now have, 12,000,000 = I/C = I/0.000,0000,000,01

So by rearranging the formula so that I is on the left (I = dv/dt x C) the current required to charge the capacitor that quickly is

I = 12,000,000 x 0.000,000,000,01 = 0.00012 A = 120uA (120 micro-amps)

which suggests that the current called Ie in the above diagram is 120uA, or that the current flowing in the differential input stage of the LF356 is 120uA ???

Not having the "confidential" information from National Semiconductor to confirm this, I can only assume that it is the case from what the maths say.

But is 120uA is quite typical for this sort of J-fet input configuration? I would have thought it would be more, especially as the input noise voltage is pretty good, and that usually suggests more FET drain current than 120uA (60uA per device).

And if the current were higher, then the slew rate would be higher too. But it isn't.

The limiting factor here is input stage linearity. It's far better than that of a non-emitter degenerated bipolar transistor, but nothing electronic is perfect.

But a transconductance input stage results in a slew rate 0.3 x bandwidth, which for 5MHz would only be 1.5V/uS. The LF356 is stated as 12V/uS which can be taken to mean that it is 12/1.5 = 8 times better linearity.

So now you know where slew rate comes from, and how it can be related to bandwidth to reveal an op-amp's linearity, we can now proceed to design the RIAA filter (next)


Edited by Graham Slee - 03 Jan 2010 at 4:08pm

Get the paracetamol out! Here comes the RIAA bit!

Here is a simplified active RIAA filter courtesy of one of the best guys in the industry: Walt Jung, from an Audio Engineering Society pre-print.



Now guess what we're going to discuss first? Yes, slew rate! (What? Again?)

We need to know if the slew rate is sufficient for this RIAA phono preamp, and how the op-amp slew rate affects the choice of the RIAA replay filter component values.

Just look at what's hanging off the op-amp output! A large value of capacitance in series with a small value of resistance (C1 in series with C2, and R3 in series with that). The op-amp has to drive that at 12V per micro-second. If it cannot drive it, it will become unstable and ring, if not oscillate. We could over compensate the op-amp, but that would be a waste.

To answer if and what the LF356 output can drive we need a formula...

And here again it's Q = CV = IT

The first answer we need is to know if 12V/uS is sufficient, and a derivation of the formula gives

SR (slew rate dv/dt) = 2pi x V(peak) x (1/T)

(derived from from "Audio/Radio Handbook", National Semiconductor, 1980)

We need to know the maximum output voltage (V), the desired frequency response to derive time (T), and later, the op-amp's output current (I), to find what capacitance it can drive.

Now, the datasheet tells us it can drive some real heavy capacitive loads, but it won't be performing all that well while doing that: we want the op-amp to be lightly loaded to give off its best.

So, we now need to gather the terms V, T and I.

Voltage

"V" is going to depend on a couple of considerations: 1] the power supply voltage, and 2] more importantly, how much the LF356 can take without its input leaving its linear (or near linear region). This will, in turn, determine the overload margin or headroom. When an input leaves its input linear region it still works, and it'll probably measure quite well on THD (total harmonic distortion), but on the inside it is under undesirable stresses if quality audio is what we're after.

We found from the previous post, that the LF356 is 8 times more linear than a transconductance input op-amp. The maximum (peak) linear input for a transconductance input is about 50mV (before things get forced), so here it's going to be 50 x 8 = 400mV.

That should be more than plenty, but hold on! The RIAA curve tells us that the output of a magnetic cartridge rises with frequency - how high?

Musical instrument's fundamental frequencies don't go as high as you'd imagine, say 8kHz (often much less with traditional instruments). But perhaps the early harmonics have good output, so let's allow 20kHz. At that frequency the cartridge output is ten times higher than it is at 1kHz. Taking a healthy output of 5mV, used on a "hot pressing" (because the young are into vinyl too...), its output is going to be the same as a 10mV output cartridge (ref 1kHz), and so the input should be able to handle 100mV. Now that's RMS, so peak will be 140mV, which means we have headroom of nearly 3 (our 400mV divided by 140mV). Studio PPM meters redline at +8dB which is about 2.5. So on most high energy peaks this input will not distort as badly as a transconductance input (this is pereceived distortion, not steady state THD as based on a sine wave). That covers the high frequency requirements.

Now, the rest (the lower frequencies which are lower output from the magnetic cartridge) are going to depend on what the preamp output can swing.

Some hi-fi magazines state maximum input as being where the output of the phono preamp hits 1% THD.

To determine what the output can swing before clipping (hitting the supply rail), we need to set out the gain we're looking for and determine the supply voltage...

What about gain? I don't think it reasonable to expect a high performance phono preamp to directly drive a power amp. I'd be happy if it did a gain of 100 (40dB) which would give an output of 500mV for a 5mV sensitivity moving magnet cartridge.

What is going to be the largest signal the phono cartridge is going to output? Consider something few others consider - the test tracks on a HFS75 test record (or similar) used to optimise cartridge set-up. Band C tracking test is +18dB and 18dB is times 8. So with a 5mV output cartridge its (normalized) output is going to be 40mV giving rise to a 4 volt output - can we accomodate that with a voltage supply suitable for the novice DIY enthusiast?

One of the highest and easiest to obtain power supply voltages is the CPC 24 volt plug top "wall-wart", which also suggests this design needs to be single supply (and why not keep things simple?).

However, that 24 volt is not regulated, so we need to choose a regulator that can do a slightly lower voltage, say 18V.

On an 18 volt supply the maximum peak to peak swing is 18V if the op-amp can output rail to rail. Most cannot, and can only get within +/- 1.5 volts of the rail. So the maximum peak to peak is 15V. We are interested in the RMS voltage which is 10/28th of the peak to peak output, or 5.35V. So with a gain of 100, the maximum input is going to be around 53mV. That's +14.5dB for a 5mV cartridge playing a "hot pressing" and +20.5dB for the same cartridge playing band C of HFS75's worst tracking test.... good enough?

So, as far as voltage is concerned, it's 5.35V RMS (7.5V peak).

Time

1/T is frequency in the case of the first question we need to answer. It's the highest frequency we want to reproduce accurately. Often, in the past, designers simply plicked 20kHz out of the air, but is 20kHz really the highest frequency we want to reproduce accurately?

Often, older phono preamps emphasised clicks and pops - well that was my earliest experience of phono preamps. Read about clicks and pops and you'll find that vinyl clicks and pops are all in the audible range. True, because they can be heard, so I guess that's how they justify that, but what about the leading edge of a scratch? Could a scratch be chissel shaped? I reckon so. It will have steep sides. So if we imagine the stylus is moving across the record (that's an easy one - for years man thought the Sun revolved around the Earth) at speed, then it finds it has taken off across this ravine, and then hits the cliff on the other side - ouch! It is not trying to reproduce the full width of the scratch, but the "cliff-edge" it just hit. That could be just ten percent of the width?

So will 20kHz cover that? I don't think so. Maybe 200kHz? So let's see if 200kHz will fit our 12V/uS slew rate...

Is the slew rate sufficient?

From SR (slew rate dv/dt) = 2pi x V(peak) x (1/T)

SR should equal or be higher than 2pi x 7.5V x 200kHz... is it?

The answer is 9,424,777.961 V/S which is 9.4 V/uS, so 12 V/uS is sufficient!

Can 12V/uS drive the RIAA filter?

Here we need the op-amp's output current (I). When we find that we can use this derivation C = IT/V or I x 1/(SR x 1,000,000) (as you can see, this is derived from Q = CV = IT) to find what capacitance the op-amp will drive.

So let's take a look at the data sheet (pdf)

And guess what? It doesn't say!

Often, output current is not specified in op-amp data sheets, so how are we going to find it out? One way is to look at the output voltage swing which is usually referenced to a particular resistive load. Here two loads are shown: 10 kOhms and 2 kOhms. Into 10 k it swings +/-13V and into 2 k it swings +/-12V. By the looks of it, 2 kOhms seems to be the limit (or corner) where its output swing starts to drop with increasing load (maximum power theorem), so I think we can assume that Ohms law will tell us its output current.

I = V/R

12V/2,000 Ohms gives 6mA. Huh!

That's where the output stage similarity with the NE5534 (which will do 16mA) ends!

But still, the LF356 has got all the other features desirable for a phono preamp, so let's carry on.

So by dividing 6mA by the inverse of and normalized slew rate we will see just what capacitance its output can drive whilst maintaining both stability and slew rate.

from C = IT/V we get 0.006 x 1/12,000,000 = 0.000,000,000,5 Farads or 500pF!

Now, luckily, that 500pF is the smaller of the two capacitors required (C2 in the above diagram) - the one that dominates at higher frequencies. But will 500pF (470pF being the nearest "prefered" or available value) result in unusable values for the rest of the filter? The only way to find out is to try it! This is where phono preamp design starts to become an empirical excercise.

I'm going to finish this posting here to allow a page break (hopefully), and continue it in the next reply.



Edited by Graham Slee - 04 Jan 2010 at 12:18pm

This is fascinating, thank you very much for sharing your knowledge on this project.