Hybrid Power Amplifier Design

=Preface= This book is about how to design a power amplifier for audio while using tube input and magic eye indication of both output level and push-pull MOS balance. I think that I have come to the conclusion that the nice sound from tube amplifiers actually doesn't come from the tubes but comes from the output transformer so it seems like tubes are not necessary here and I use a pair of MOS transistors (2SK135) instead which also simplifies the output transformer (OPT).

=Amplifier Block Diagram= The amplifier consists of five different modules which I first wanted to reside on a single PCB but have changed my mind and I will now use five different PCBs. The reason is that the different modules may be used for other applications. The first module (KHR) is the rectifier which rectifies 18VAC and filters it, the second module (KHS) is the heater supply which also steps up DC (with the use of a transformer, observe) for input tube usage, the third module (KHB) biases the 7247 input tube and makes it amplify and split, the fourth module (KHO) only sets bias for the output MOS and receives a split input signal from KHB, the fifth module (KHM) indicates offset balance and output level with the aid of a PM84 Magic Eye. It is not shown in the block diagram but there will be a special kind of output transformer also.

KHR, Hybrid Rectifier
This unit supplies the whole amplifier and it does that with a delay given by the double tube diode V1. The delay is done by the heating time of the tube which smoothly onsets the voltage. It is a well known fact that silicon amplifiers has the risk of harming the tweeters due to the onset voltage spike which happens at normal onset without delay. This could be done by using timers but I think it is more smooth to do it with a single tube.

This circuit works in such a way that when SW1 goes on 2x18V is immediatelly there, their smoothed voltage is supplying the V1, EB91 however takes some 15s before it is ready and is thus our timer. During most of this time T3 and T4 is strangled which means that the power transistors T1 and T2 also are strangled. While EB91 is being heated up a voltage is being built up at the bases of T3 and T4 which makes them conduct in such a way that T1 and T2 also are able to supply current.

Here the currents are not less than 4A, they are less than 5A while the output MOSs draw 7Ap each at maximum output. This 5A is however a RMS from full wave 7Ap (while each transistor draw 7Ap alternatingly building a full wave supply signal). Now this current is a RMS. Normally a capacitor filter gives some DC-current and let's say the DC voltage is stable, the the AC current through the transformer is 1,4 times higher. I'm not sure about RMS but if I suppose the same the current at the transformer may be as high as 7A. I will use 8A rectifiers.

My transformer is however of 160VA only, two windings of 18V makes a maximum current of 4,44A. This is a bit from 7A but I reason like this that firstly I don't need to run the amplifier at maximum output (some 5W is enough) secondly a toroidal transformer does actually not have a fuse so for a short time 7A may actually work.

For 7A, the primary fuse will have to be 1,1AT. I'm not sure how fuses work but i think it is the rms current that blows them so maybe I should use 1AT instead because we don't want our transformer to burn (for normal use the fuse should be 0,7A). For toroidal transformers I have learned the hard way that you have to set a higher fuse than normal probably because the high inrush current.

KHS, Hybrid Supply
This unit supplies the high voltage supply for the input tube and the Magic Eye, it also supplies the heater voltages for all the tubes where a 4/6/12V option is built in. With paralleled sekundaries it looks like output voltage at full load (13mA, estimated only) is around 235Vdc. The PM84 works best at around 250Vdc so this voltage is okey. It is however a bit sad that the voltage is over those 200V my capacitors can withstand (in singular) but on the other hand the voltage before the tubes are heated (and current is drawn) is around 318V.

Some cooler will have to be used, the formula for estimating a cooler is

$$\frac{125}{P}=K+Kiso+Kjc$$

where K is the thermal resistance of the cooler, Kiso is for the insulation, Kjc is for junction to case. I am keen on using a push-on cooler for TO220 of 16K/W and the Kiso is around 0,3 (if used) and the Kjc of TO220 is around 2.

Here we then have the power dissipations IC1: (25-4)*0,3A=6,3W IC2: (25-6)*0,3A=5,7W IC3: (25-12)*0,15A=2W

Using my formula it can be shown that a 16K/W cooler can cool 7W so my approach works. 2W is by the way the limit for a naked TO220 whatever.

KHB, Hybrid Bias
Here I use a double triode to amplify the input signal to the saturation level of the output stage. This level is lower than 10Vp and a Line level is defined as 0,5Vp so if we want to saturate the output stage we need a gain of 20. A ECC82 has a my of 20 but the gain can't be so much, a practical ECC82 gain is about 15 tops. While we need to hit rather exactly 7Vp a ECC82 may actually work but then we have to bias it for a gain of 15. I however am of the opinion that while feedback is kind of magical low a feedback as possible is the best. On the other hand we here have a transformer of rather unlinear properties where we need feedback to linearlize the output transformer. So a gain of 15 will not suffice, we need another tube. A ECC83 could be used but it works poorly as splitter. This is why a 7247 is my choice because it combines ECC83 with ECC82. Here I then have some power to linearlize both the transformer and the tube with feedback.

The gain for the first stage (V1:b) may be written

$$Av=\frac{\mu Ra}{rp+Ra+(\mu+1)Rk)}=\frac{100 \cdot 470k}{100k+470k+101\cdot 470}=76$$

here 470*101 is only 47k which is less than 10% of the denominator so we could have skipped that, but this is the correct expression, on the other hand none of this is that exact.

The gain of the splitter (V1:a) is very close to 1 at both ends which means that our sensitivity is around 7Vp/76 or approximatelly 0,1Vp. Line level is said to be 0,5Vp which means that we have a margin of 5 for feedback. The only thing to then consider is that the output impedance of the anode is quite high (approximatelly Ra and thus 44k) which means that we can not load this stage so much, somewhere around 470k is perfect which I also do.

Actually I don't want to use R3 but it is needed for the possibility of feedback. Here we then have a curious situation because while my estimated anode current is 400uA the R3 voltage is only some 0,2V. An output of my hopeful 35W means an output voltage of 12Vrms (or 17Vp) in a 4 Ohm speaker, if feedback of 5 is supposed to be relayed back to R3 the voltage will be around 3V which overshoots 0,2V. Here one may suspect hard distorsion (which would happen in a solid state amplifier) but I have noticed that this high feedback voltage actually can be handled! The only problem is that DC-bias of the amplifying tube (V1:b) drifts somewhat which also is the reason why the splitter is capacitivelly coupled from the amplifying tube.

The input impedance to V1:a is not trivial but according to Morgan Jones

$$Rin_{eff}=\frac{Rin}{1-\beta Av}$$

here beta is

$$\frac{R8b}{R8b+R8a}=\frac{39k}{39k+3,9k}=0,91$$

and the gain is

$$Av=\frac{\mu Rk_{tot}}{rp+(\mu +1)Rk_{tot}}=\frac{20\cdot 42,9k}{7k+(20+1)42,9k}=0,95$$

while Rin=R9=1M, this sums up to

$$Rin_{eff}=\frac{1M}{1-0,91\cdot 0,95}=7,38M$$

Here we see that the input resistor is amplified, in other words the input resistance is not the input resistor but some 7 times higher! This is actually rather important because feedback may sound simple but it "draws" the zeroes together making the response being unstable if care is not taken. I have learned that for two zeroes it is a good approach to keep them BF apart so if we want the feedback (BF) to be 5 times, we should keep them 5 times apart. If they are the same there will be a rather large peak low in frequency. This peak may not be heard because outside of the audible spectra but to me this is rather irritating, I want a smooth response. I think this also will alter the step response, maybe to instability. I have measured the input resistance of the Edison biased V1:b tube being some 800k.

KHO, Hybrid Output
Here I regulate the bias for the MOSs. I think it is important to regulate the bias especially when running Class B which is my plan. The regulator voltages used are also so much lower than the supply voltage so any spike will not interfere with the bias. Class A generates no spikes but Class B does because current is supplied one half-cycle at the time so this current jerks in the supply. Input signal is capacitivelly coupled to the trimmer and the trimmer sets bias. Depending on bias the zero will differ because the impedance the capacitor sees is different but I have set minimum zero to an acceptable frequency level.

For JFET/MOS transistors and pentodes the gain formula is rather simple

$$Av=gm\cdot Ra$$

this while transistors behave like current generators, triodes however behave like voltage generators where there is a not so negligable bleeding current. Here the 2SK135 gm is rather exactly 1S so the gain is Ra so to speak. I use a reflected Ra of some 2,8Ohms so the gain is 2,8.

We have above calculated the driving circuit gain of some 76, total gain to the drains is thus 213. When the MOS are loaded with 2,8Ohm each, their swing capacity at 25V supply is around 20V so we have a sensitivity of 20/213=90mV for full power output. At the same time the gate doesn't have to swing more than some 7Vp and this is what we have to hit as driving voltage.

Swinging 7V at the gates gives a peak drain current of some 7A (gm=1S). Running Class B means that this peak current is supplied to one of the MOS over one half of the cycle. The rms current through this MOS is then

$$Irms[HW]=\frac{Ip}{2}$$

so the rms current through this transistor is 3,5A (and of course also through the other transistor). The net full input current is 7Ap (5Ae), I then use a transformer of the turn ratio 150/250 which makes the secondary current be 250/150*7Ap=11,7Ap, rms this means (while it is full wave) 8,2Ae. So it seems like the different primary windings work with 3,5Ae while the secondary works with 8,2Ae.

Below I have estimated Rp/2 to 1,7Ohm and Rs to 1,4Ohm, running these currents through them gives a copper loss of 21W and 94W respectively which is kind of rediculous. So my transformer winding resistances are too high, I need much thicker threads because here I have a copper loss of some 2x21W+94W=136W and my transformer will burn up!

The primary current density is around 14A/mm^2 and the secondary current density is around 21A/mm^2 this while I know that professional transformer manufacturers run 3A/mm^2. So the different wires are too thin. I have calculated that the overall use of 1mm wire will give 4,4A/mm^2 and 10A/mm^2 respectively. This will probably lessen the copper loss from 136W to around 50W which may be acceptable(?).

In any case, there is no need to run maximum output power (which including copper losses seems to be around 42W). If maximum output power of the amplifier is 42W and you normally play at 12 o'clock you play at some tenth of the maximum output power and thus some 5W. So this fantastic amplifier may work anyway.

KHM, Hybrid Magic Eye
This unit sniffs MOS bias unbalance and output level and shows them with the help of a Magic Eye. The bias offset will be very low due to Class B (read almost no bias current) so in this case it only sniffs output level.

This is a rather special circuit, I have designed it to be able to sniff tube power amplifiers too which mean around 500V of supply (B+). T5 needs some voltage over it, in the tube case this is not a problem while output transformers have copper losses. In my case the difference between B+ and the drain potentials are however not that much, on the other hand I have attenuated the input with the aid of for instance R1/R16 which means that the input level at Class B (which means that input voltage is the same as B+) is attenuated 2M2/(1M+2M2)=0,69. At an Class B input of 25V, the voltage at input will be 17V and this is 8V below rail so this circuit may actually work even for Class B.

The DM formula for the gain in this special case is

$$Av_{DM}/2=\frac{\beta Rc}{Rs+hie+Rc+(\beta+1)Re}$$

here we see that if

$$Rs+hie+Rc<<\beta Re$$

and beta>>1, which is the normal case, we have

$$Av_{DM}/2=\frac{Rc}{Re}$$

so for little higher Re the gain is determined by the Rc/Re ratio only. I have set Re to 2k7 and Rc to 3k9 to give the amplifier a gain of slightly more than 1. Let's say the gain is one, then each input gives a replica at the the collector resistance Rc. D3+C3+R12 is a peak detector (with a 50ms time constant to be able to sniff changes). Into this peak detector thus comes the (varying) drain potential. The peak value controls the opening of the Magic Eye (ME). Now we have two scenarios, we wish to sniff balance (DC) as well as VU (AC). The dynamic limitation of the ME is some 11V so the input voltage has to stay within 11V. For KHP the maximum input swing is around 20V so here we would want a attenuation, what's worse is if we connect this to a tube power amplifier we will have a input swing of around 300V. This can't be handled by the ME so in the case of AC-signals I have attenuated the signal some 20dB using for instance R1 and C1. At DC we then have the sensitivity of 11V, at AC the maximum level we can detect is around 100V which is rather appropriate while we often play at 12 o'clock (-10dB i.e Pmax/10).

KHT, Hybrid Output Transformer
In my book OPT Design almost everything about OPT design is explained. If high bandwidth is required it is a sport to design a good output transformer. I will here add actual dimensions of the used LL-core and the nice LL-coils which Lundahl Transformers so thankfully have sold and manufactured for me.

The turn ratio is 500/300 totally. The primary is of 0,56mm diameter copper wire, the secondary is of 0,71mm. Both primary and secondary consist of two coils of half the total turns to fit on the LL C-Core as push-pull coils.

Transformer Loss Model
I have the coils but have not measured their copper resistances. I have however estimated them. The full primary is 3,4Ohm (Rp), the full secondary is 1,4Ohm (Rs) and using the model we see that the attenuation is

$$\frac{Uo}{Us}=\frac{RL}{RL+Rs+n^2Rp}$$

which for a speaker (RL) of 4Ohms gives an attenuation of -4,4dB (60%).

The 2SK135 MOS at my bias gives a maximum output power of some 70W, 70W-4,4dB is then 42W and around that is probably what I with a 4 Ohm speaker (RL) can get.

For curiosity I did not get more than 11W with my Tube Power Amplifier (KPA) using the fantastic tube 807 in push-pull.