The Fundamental Problem
The fundamental and obvious problem to solve with electrostatic amplifiers is the fact that the amp must symmetrically swing very high voltages. To make 1000V peak-to-peak each output must swing to 250V peak. The output device will see 500V from one end of its excursion to the other. There are not that many devices that can handle these kinds of voltages.
The second problem is that both the inputs and outputs of the amp must be at DC ground. To get a 0VDC offset at the output requires a split HV supply which in turn means that the output device that provides the high voltage swing must be located at the bottom rail. That way it can pull the output down to the bottom rail or push it to the top rail while the quiescent operating point remains at 0VDC.
What this means is that the input signal (referenced to ground) must first be translated down to the bottom rail across many hundreds of volts and then the signal must be translated to the top rail by the primary output device. There are not that many good ways to do this at high voltages.
What we mean is this:
You can see that the active devices plus their associated passive components are strung between the rails and have to stand-off the full rail-to-rail voltage.
You may wonder why we translate to the bottom rail first and then up to the top rail. This reason for this is simply that the output device must be able to sustain the entire rail-to-rail voltage. It is easier to find and there is more selection of N devices that can handle these voltages. And N devices have their emitters/sources/cathodes connected to the negative power supply.
One very common way to stand off high voltages is to use cascode type topologies. In a cascode design one device is stacked on top of another. The bottom device, usually, provides the gain and is a low voltage device. The top device may or may not provide gain, but it is the high voltage device. It protects the low voltage device from the HV.
A typical transistor cascode circuit looks like this.
The base of the top device is set above ground so that the collector of the bottom device sees a good operating voltage. With the base voltage held constant the HV device is driven at its emitter. There are mosfet and tube versions of this circuit. In fact, the cascode originated with tubes and the word cascode is shorthand for cascading triode.
Remember that this amp is designed to be SIMPLE, but good. It is not designed to use over-the-top topologies, hard-to-find parts, or to burn megawatts of power. Our goal is, constantly, keep it simple but effective.
In this vein, we really only need two simple building blocks. A stat amp is a differential amp because the two stators are driven out of phase. So, we need a differential input. The input stage can translate the signal to the bottom rail. Then we need an output stage. The output stage can translate from the bottom rail to the top rail and provide the voltage swing needed for the stators.
The Input Stage
One very nice thing about solid state devices is that fact that we have both polarities of device. Thus, the input stage can be P devices and since we want a nice high input impedance we use p-jfets. Then, in keeping with the cascode hv protection scheme we simply cascode these with suitable HV PNP bjts. Like this:
The resistors on the collectors of the BJTs serve to isolate the input (which has no blocking capacitor) and the source from the HV B- in the event that the BJTs short out.
The Output Stage
For the output stage we need to take the signal passed down to the bottom rail and translate it upwards. The simplest way that I know of to do this is with a current mirror. However, a typical current mirror with just two transistors will not work here. This is because to have a good mirror both transistors should be of the same type. And since the output transistor must be a HV device the other transistor must be HV also. We can do this, but there is a better way. It's a Wilson current mirror. A Wilson mirror uses three transistors, the third one in the cascode position to shield the other transistors from the higher voltage. With a Wilson current mirror we can use good, low noise, low voltage transistors for the actual mirror and the HV output transistor for the cascode device. Like this:
A quick look at this circuit tells you that the LV transistors never see more than two diode drops while the HV transistors sees all the rest of the voltage to the top rail.
The Basic Amp
The basic amp consists of putting these two pieces together as a transconductance amplifier. That is, it is a current amplifier not a voltage amplifier.
To make the final amp we will add a few flourishes to help performance. First, we will add emitter followers to the outputs to reduce the Zo and to make it easier for the amp to drive difficult loads such as two or more headphones in parallel. And then some negative feedback to reduce the distortion a little, bring the gain down to about 1000, and increase the bandwidth some.
No Constant Current Sources?
No. Here's why. In a low voltage amplifier where there is not much voltage drop between the rail and the active device and in certain HV situations CCSs are really useful circuits. But, remember, we are going for simple but good.
Look at the input differential pair. The tail of this pair is connected to the HV rail through a resistor. The HV will be on the order of 300V. If only a few mA are flowing this will be a pretty big resistor. For our purposes a big resistor is very nearly as good as a CCS. The difference in performance is negligible in a transconductance amp.
There are three places in this topology where most every designer would want to put a CCS:
- The tail of the input stage.
- The load for the HV output transistors.
- The load for the followers (not shown yet).
In fact to get really good performance we don't need any of these CCSs. However, in the prototype (designed to make about 700-800Vpp) some buiders experienced clipping due to current starvation in the output followers. The final amp, therefore, has CCSs on the output followers.
The Sand Amp
To complete the design of the solid state version of the amp we need to set a few more requirements.
One of the requirements was no exotic transformers. Transformers start to get exotic, more expensive, and bigger when their secondaries exceed 240V (generally twice the line voltage in the US). If we stick with 240VAC-250VAC secondaries there is a small selection of relatively inexpensive transformers to choose from. 240VAC-250VAC means a rectified voltage of about 338VD-350VDC which, with drops in the filter/regulator, sets our rails at +/-300V. An amp with these rails can make somewhere between 800Vpp to 900Vpp. This is simple, but good.
No exotic parts. We are tempted to use the Toshiba 2SJ74, low noise high transconductance jfet. But these are out of production and will get harder to find. Instead we'll use the easily obtainable and in current production J271. Its tranconductance is lower, but there are lots of them around.
We'll use the Fairchild KSC5042 HV transistor for the output devices. It has a max of 900V. Just what we need and lots of them around.
With these design choices in mind and with a few other obvious choices for components the basic amp is this:
There are a few things to note about this design:
- There are no exotic parts. Just simple resistors and transistors (and one zener diode).
- This is not entirely a perfect transconductance amplifier since the jfets are voltage driven devices. However, the voltage at the input is converted to current in the jfets at which point the amp is entirely a current amplifier. In the simplest sense the jfets, through the chain of devices down and up again, drive the 100k load resistors. Thus, the voltage gain of the amp is totally dependent on the transconductance of the jfets. This will be important later.
- To relieve some of the current swing in the jfets to reduce distortion the current mirror has a current gain of 3. The jfets have a current excursion 3 times less that what is flowing in the load. However, current mirrors amplify both DC and AC hence we cannot arbitrarily increase the gain of the mirror because as we do, due to the DC current gain, we must decrease the size of the load resistors. The values chosen are a good compromise giving the amp plenty of bandwidth.
- The followers are CCS loaded (as per the design discussion).
- The offset and balance are adjusted at the tail of the differential amplifier. To get a 0VDC offset at the output the 100k resistors must drop 300V which means they need about 3mA. With a 3:1 current ratio back through the mirrors the jfets must each have 1mA. Thus, the tail resistor must pass 2mA (1mA for each jfet). Hence the resistor selection.
- A small amount of NFB is applied to the sources of the diff pair. However, the NFB is small enough that it does not entirely control the gain of the amp, which is still primarily in hands of the jfets. The values chosen are an average value to bring the average gain down to about 1000. Jfets with high transconductance will cause the amp to have higher gain and vice versa.
- The offset/balance trimpots control the output through a long chain of solid state devices, each of which will undergo temperature shifts. The offset/balance will wander some as a result of this. But, in keeping with simple but good, we accept a few volts of wander in an amp that is driving devices that want hundreds of differential volts. In the prototypes, it has been possible to set the voltages within less than 1V of 0V and see them wander less than 1V after warmup. This is good enough and saves much expense on more complicated designs.
- However, because of the balance/bias arrangement, the jfets will have to be Idss matched within 10%. It is easy to do this and, for the prototype builds, it has been easy to find a matched quad of J271s in a batch of 20 devices. This is a small price to pay for a much simpler design.
- Since the overall gain depends on the transconductance of the jfets selecting jfets with high Idss is important. Although Idss is not a measure of transconductance, it tends to correlate with it.
The Hybrid Amp
It's easy to turn the solid state amp into a hybrid simply by substituting a triode for the HV output device. But, since a tube is depletion type device we can't use it in Wilson current mirror arrangement. Instead we fix its grid voltage and drive its cathode with conventional two-transistor mirror. Like this:
In this case the mirror transistor will see 30-40V and so we choose higher voltage (but easy to get) transistors for the mirror devices. And, in keeping with nothing exotic, we use the plentiful workhorse 2N5550.
Suprisingly, at first glance one might think that using triodes would reduce the bandwith of the amp, but in fact, the bandwidth increases because the triodes offer less capacitance to drive at their cathodes than the HV BJTs do at their emitters.
The Power Supply
We need a +/-300V split supply. With a 250-0-250 secondary we can expect somewhere between 320-345VDC at the filter capacitors. The supply must eliminate the ripple and give good regulation while dropping only 20-35V.
Fortunately, we have one really important feature working for us with this amplifier. Each channel is a differential amp. This means that as one side of the channel sinks current the other side sources it. In the same amount. If all the devices are perfectly matched the net dynamic current draw from each channel will be zero. Devices are not perfectly matched, but the dynamic draw will still be small, probably less than a few mA.
The most efficient way to eliminate the ripple without requiring big capacitors is with a constant current source driven shunt regulator. Shunt regulators are easy to make work at low voltages but high voltages present problems because even small currents in the shunt devices add up to large power dissipations. For example a single shunt transistor across 300V shunting 10mA will burn 3W. In our case, however, since the dynamic current draw is small we don't need much shunt current. And we will use another trick to reduce the power dissipation even further.
The basic CCS shunt regulator looks like this:
In this diagram the zener represents any kind of shunt regulator. However, if we were to use a simple zener we would have two problems: The regulation would be poor and the power dissipation would require 5W zeners at least. We can solve this problem by adding a simple BJT booster to the zener:
If you follow the zener trail you'll see that is passes through the BE junction of the BJT to the top rail. Thus, the rail voltage is shunted at the zener voltage plus one diode drop. In addition, if the voltage tries to rise the current in the resistor will increase increasing the conduction of the BJT pulling the voltage down again. And vice versa. This type of boosted regulator is common and we're not ashamed to use it.
However, the BJT shunt device will see 300V. There are not many, easily obtainable, BJTs that can sustain 300V and, particularly, there are not many PNP BJTs. There are good 300V devices (such as MPSA42/MPSA92) but these would be operating right at their Vceo limit. But these transistors are so plentiful we should use them if we can. And we can if we, once again, use a cascoded design so that two transistors span the 300V putting just 150V on each device. Like this:
The two resistors that set the base voltage of the cascode BJT are the same value to cut the voltage in half.
The CCS can be made as a ring-of-two type CCS. However, when the PS is powered up the load side of the CCS will be at 0V and the line side will come to full voltage very quickly. The primary pass device in the CCS will see over 300V. We have to use a HV device and, in this case, we can use 500V MOSFETs. The BJT will not see high voltage because of the zener diode protecting the GS junction of the MOSFET. But, to keep parts simplicity, we can use the same BJTs that are used in the shunt regulator.
The adjustment is very simple. The trimpot plus the fixed resistor are in parallel with the BE junction of the BJT. They will see approximately 600mV. The current that is set by the CCS will be 600mV divided by the value of the two resistors in series.
Putting the two pieces together we get this simplified CCS fed shunt regulator:
Remember that our goal is simple but good. The actual regulator is a bit more complicated than this. In the final design we'll add another stage of gain to improve the performance of the regulator, but we won't use a differential opamp type circuit that is more commonly used because we don't need that level of performance.
The negative regulator is the exact opposite of the positive regulator.