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MEHA v0.1a
This is the project page for the MEHA v0.1a.
A few prototype boards were sent around, but I have not heard whether they worked, whether people liked them, or what the story is.
For now, I'll leave this page up as a bit of history which might be useful to someone.
Background: This project is intended to extend the low voltage hybrid headphone amplifier designed by Pete Millett. In that project, Pete used a low voltage triode tube (12AE6A) for voltage gain and DC coupled it to a solid state buffer for current gain to drive headphones. While the project is very innovative, and sounds quite good, it requires a coupling capacitor on the output, and because headphones present a low impedance load, that capacitor must be quite large -- upwards of 470uF for 32ohm headphones. The changes of this project over the original are several fold. The first, is that the tube is AC coupled to the buffer. The reason for this is that, hopefully, it allows the use of a high quality film capacitor in between stages instead of the electrolytic on the output. The second change is the replacement of the 12AE6A's with a a dual triode (6GM8/ECC86) which is subjectively of higher quality and, as a dual tube, only requires a single tube. Third, the project uses a buffered ground on the output which has been found, in many cases, to be beneficial for headphone amplifiers. Last, the project is physically smaller than the original, which seems like a good thing. At this point, the project is in the early prototype stages. You can view the schematic here, or the inital board layout here. Prototype boards have been ordered, and we'll see where it goes from there. Keep an eye here or on http://diyforums.org/ and http://headwize.com for updates. You can also send me an email at meha@ecp.cc. Why "MEHA"? There are a whole bunch of low voltage hybrid headphone amplifiers floating around. With the exception of Pete's, these all seem to have names like SOHA, or YAHA, etc. The "ME" here stands for Millett-Esque. I am not sure if the "HA" part is Hybrid Amplifier, or Headphone Amplifier, or both, or neither. Some Gory Details: In order to pick some of the parts, it is important to understand the basics of how a tube works, and how to pick operating points. However, to get a more thorough understanding, it is worth reading some of the many tutorials on the web. Here are some good links: Anyhow, the basics are that a tube utilizes the thermionic effect. That is, a basic tube has two parts or electrodes. A Cathode and an Anode (usually called the plate). If heat is applied to the cathode, this causes electrons to boil off. If a voltage is then applied to the anode, which is some distrance from the cathode, the electrons are attracted to it, and a current is caused to flow with the electrons jumping across the vacuum. (It is worth noting that current flows from a negative to a positive potential. This is the opposite of how people tend to think in solid state circuits.) This flow of current is constant in a two electrode tube so long as the heat and the voltage applied to the anode are constant.However, a third electrode can be added. If a grid of wire is placed between the cathode and the anode and is at a negative voltage to the cathode, it will repel some portion of the jumping electrons. By altering the voltage of the grid up and down, the flow from the cathode to the anode can be increased and decreased. And since music signals do alter in voltage, applying that voltage to the grid causes the flow to increase and decrease. Further, since it is the amount of potential difference between the cathode and the grid (that is, how much the grid's voltage is negative with respect to the cathode) that determinales how much current flows, it is by setting the grid's voltage that a tube's operating point is set. This is generally referred to as biasing the tube. (A note -- on the original Millett amplifier, the bias is measured on the plate, not on the grid. The plate and the grid are proportional to each other, changes in one influencing the other, so measurements can be taken at either. This design has test points at both the plate and the cathode (the grid being connected to ground) so you can convince yourself of this.) There are several methods that can be employed to bias a tube, but the simplest one, and the one used here, is to place a resistor between the cathode and ground. Since current must flow in a complete circuit, the current flowing across the tube must begin at ground. Since, at a given current, a resistor will decrease voltage by a predictable amount (V=IR) adding a resistor in series between the cathode and ground will drop a certain amount of volts as the current flows resulting in the cathode being at a positive potential (voltage). Further, if one references the grid to ground (by connecting the grid to ground with a large resistor), the cathode is effectively "biased" at a positive voltage with regard to the grid. To determine how much current will flow across a tube in a given setup, one must look at the datasheet. Below, the red dot on the graph shows an operating point where there are approximatly 16V on the tube's plate, about 2mA flowing through the tube, and by looking at the curve, the grid is biased at -0.75V with respect to the cathode.
This was just selected as an arbitrary point, and you can pick a different one as well. The first thing one must do is to determine what value of cathode bias resistor will yield this operating point. This is a simple calculation, and plugging in for V=IR , one gets 0.75 = .002 * R or 375 ohms. This is not a standard value for a resistor, but 374 ohm is which is closer than you will usually get. It is also worth mentioning that one should determine how much heat that this resistor will need to dissipate. In small signal tubes this is not usually an issue, but with power tubes it can be. Resistor power is determined by the current multiplied by the voltage, so here we only need to dissipate (0.002)(0.75) or 0.0015W, so any resistor will do. This flow of current, however, is not that useful on its own. However, if some method is employed to convert the current, and more specifically the changes in current, into a voltage, then the tube becomes a voltage amplifier. The most common method of doing this is to add a resistor to the plate. The resistor resists the flow of current by allowng the voltage to swing up and down. This is just a simple application of the common V=IR where voltage and current are proportional. Thus the next step is to use this operating point and a chosen power supply to determine the load line and the necessary plate resistor. Note, you can do this in any order you want. That is, if you know your power supply already, or you know what plate load resistor you want to use, you can work from that. The easiest thing to do here is to draw a line through the operating point that contacts the X and Y axis of the graph. In the figure below, this line passes through our operating point, and contacts the X axis at ~23V (which is the plate voltage at 0mA of current) and the Y axis at 6mA (which is the plate current at 0 ohms).
To determine what value of plate loading resistor will cause such a load line, you simply calculate the change in current for the change in voltage, and use this to determine the size of the plate resistor. Here, current change is 6mA and voltage change is 23V. So, again using V=IR, the resistor should be 3833 ohms. 3830 ohms is a standard resistor value, so again we are pretty spot on. Now, as a general rule of thumb, the plate loading resistor should be 5x the tube's internal plate resistance. The datasheet is not clear what the plate resistance is, however, so we'll use this value for now. There are two things to note here. The first is that the load line contacts the X axis at ~23V. I am assuming that we are using a 24V power supply. The 23V is to allow for the fact that that the cathode is biased up about a volt. That is, the plate voltage is with respect to the cathode voltage, not to ground necessarilly. Second, the signifigance of the load line is that this is where the tube can operate. There is a big graph, but with our power supply, plate load, and cathode resistor, the tube is restrained to positions on our line. With the above information, you could build this amplifier with a resistor plate load. However, as you may be aware, the design calls for a Constant Current Source (CCS) to be used on the plate. The reason this is used is that it forces the tube to operate in a more linear way (meaning with less distortion). The CCS does two things. The first is that it controls the current flowing through the tube rather than allowng the cathode bias resistor to do so. The second is that is presents an ultra high impedance as a plate load. That is, it holds I constant as well as R such that when converting current to voltage (through V=IR) it will operate more efficiently as only voltage can swing. The long and short of this is less distortion (i.e., cleaner sound) and higher gain. To see how this works, once again one must consult the plate curves. Here, however, instead of drawing a load line that slopes downward, the load line is nearly horizontal as the current is constant.
With this horizontal load line, adjusting the cathode bias resistor simply moves the operating point left and right across this line. So, for instance, with a CCS operating at 2mA, a 375 ohm resistor will bring us to the same operating point as above. Increasing the resistance will move the point to the right, decreasing it will move to to the left. One additional important point must be made. Since the CCS holds I constant, it must have room to allow the voltage to swing up and down. That is, the B+ voltage must be higher than the plate voltage or else you will get horrible clipping. And, since this is a hybrid design even without the buffers (the I/V is solid state), that clipping is not nice sounding. Too much voltage, however, and the CCS can burn up by trying to dissipate too much heat.
Parts List
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