e015008.PDF

AU DIO

PPP HiFi

Valve Power Amplifier

using KT88 valves

Design by G. Haas (Experience Electronics)

Thanks to their pleasant sound, valve-type power ampliﬁers continue to

enjoy uninterrupted popularity. With such an ampliﬁer, you can eliminate

the impression of coldness, sterility and artiﬁciality that many people expe-

rience with CDs.

The design for a power ampliﬁer that is pre-

sented in this article is based on the PPP

principle. PPP stands for ‘Parallel Push-Pull’.

‘Push-pull’ means that the output stage is

composed of two active elements acting in

phase opposition. One of the valves handles

the positive half-cycles, while the other one

handles the negative half-cycles. In

a Parallel Push-Pull conﬁguration, the

valves in the output stage are con-

nected in parallel with respect to the

ac signal. The disadvantage is that

the power efficiency per valve pair is

less than with classical Class AB

push-pull operation. Otherwise the

PPP principle has only advantages.

This output stage conﬁguration was

invented in the early 1950s, and it

was the configuration of choice in

studios. Reduced distortion, good

sound and a wide frequency

Elektor Electronics

5/2001

AU DIO

response were more important in this appli-

cation area than high power efficiency. With

the triumphal march of semiconductor tech-

nology, PPP was unable to retain any signiﬁ-

cant territory, due to its relatively low power

efficiency and relatively high price. However,

with the aid of modern resources it is possi-

ble to construct excellent HiFi PPP power

ampliﬁers at an acceptable price.

AB versus PPP

R g2

In order to enable you to better understand

the PPP technique, we ﬁrst have to delve into

a bit of theory. The schematic diagram of Fig-

ure 1 shows the basic circuit of a classical

Class AB push-pull circuit. Each of the con-

trol grids of the power valves is driven by a

half-wave signal from a paraphase circuit.

This is shown symbolically in the figure; in

practice the two valves would naturally be

driven by positive half-wave signals, since

otherwise the circuit would not work. Here

the left-hand valve is responsible for the pos-

itive half-cycle and the right-hand valve for

the negative half-cycle. The output trans-

former combines the two half-cycles to form

the complete sine wave. The ﬁgure shows the

polarities of the voltages that appear in oppo-

site phases across two halves of the primary

transformer winding (N1 and N2). They gen-

erate a voltage on the secondary side of the

transformer that is reduced by the trans-

former ratio n . From the figure, you can also

see the response of the power stage to hum

components in the supply voltage. If a hum

voltage is superimposed on the supply volt-

age U B , it will be fed into the two windings

N1 and N2 in equal measure. If the circuit on

the primary side of the transformer is com-

pletely symmetric, which means that the

windings, output valves and quiescent cur-

rents are identical, the hum voltages cancel

each other out due to the antiphase feed into

the transformer. Unfortunately, perfect sym-

metry cannot be achieved in practice, so U B

must be adequately ﬁltered.

Figure 2 shows the ac equivalent circuit of

the basic circuit of Figure 1. The power sup-

ply U B represents a short circuit for ac sig-

nals. The ac internal resistance is given by

the series connection of the internal resis-

tances of the two valves and the two wind-

ings N1 and N2 with their equivalent resis-

tances Ra. Both output valves are operated

with a non-zero quiescent current, in order to

avoid cut-off distortion when the sine wave

signal passes through zero.

The situation with a PPP output stage is dif-

ferent, as can be seen in Figure 3 . You will

notice that there are two supply voltage

sources and that the valves are connected to

U B

000118 - 11

Figure 1. Basic circuit of a conventional Class AB push-pull output stage.

R a

R i

U B

000118 - 12

R I1 = 2·R i + 2·R a = 2·(R i +R a )

Figure 2. The ac equivalent circuit of Figure 1.

U B

000118 - 13

R I2 = R i /2 + R a /2 = 1/2 · (R i + R a )

R I2 = 1/4 · R I1 (!)

Figure 3. In a PPP output stage, the signal current ﬂows through the entire primary

winding of the transformer.

5/2001

Elektor Electronics

AU DIO

it is connected to the mains with no

secondary load. The PPP power

stage can be seen as a predecessor

of the standard transistor push-pull

output stage. As we have seen, a

whole series of indisputable beneﬁts

are associated with this design, to

wit low internal resistance and tol-

erance of short-circuit and open-cir-

cuit loads, as well as an inherently

good frequency response.

nection, the input resistance drops

to 8.5 k

, but the output stage can

be fully driven with only 0.75 V. Suit-

able wire bridges are provided on

the circuit board. If the amplifier is

wired with a Cinch socket, the input

sensitivity remains at 1.5 V.

At the input to the ampliﬁer, the 2.2-

µF bipolar electrolytic capacitor C1

ensures low-impedance coupling

and at the same time blocks any dc

component of the signal. Resistors

R3 and R4 block coupled-in RF inter-

ference. Valve V1a provides the main

amplification, with RF oscillations

being suppressed by C2. Valve V1b

is wired as an impedance converter,

and it serves as a low-impedance

source for the signal applied to para-

phase circuit valve V2. The grid of

valve V2a is driven directly without

a coupling capacitor, since the dc

potential has been brought to the

proper level by suitable selection of

component values. The control grid

of V2b is connected to the same dc

potential as the grid of V2a via R14.

Capacitor C10 provides a short-cir-

cuit path to ground for any ac com-

ponents that may be present at the

grid of valve V2b, which is driven by

the ac signal voltage via its cathode.

Valve V2 must provide the full ampli-

tude of the output voltage, since the

output valves V3 and V4 only pro-

vide the signal with the current

needed to achieve the desired

power. The latter two valves act as

cathode followers, in the same way

as transistors are used as emitter fol-

lowers in the output stage of a tran-

sistor ampliﬁer.

This also gives an answer to the fre-

quently hotly debated question,

‘How do different types of valves

sound?’ Since the output valves of a

PPP design do not contribute any-

U B

R L

U B

A power ampliﬁer in

three building blocks

Figure 5 shows the complete

schematic diagram of a monophonic

power ampliﬁer. In consists of three

blocks, and is consequently divided

into three circuit boards. This allows

the ampliﬁer to be built such that the

valves show to best advantage, as

can be seen from the photograph.

The transformers and the large elec-

trolytic capacitors, in contrast to

many other amplifier designs, are

‘embedded’ in the equipment.

Block 1 is the ampliﬁer circuit, block

2 is the power supply and block 3 is

the switch-on current limiter. At the

input to the amplifier, provision is

made for an E-1220 audio trans-

former. If you want to have a ﬂoating

symmetric input, you can use the

transformer together with an XLR

socket; otherwise, a Cinch socket is

adequate. Thanks to the transformer,

it is possible to use floating sym-

metric cabling, as is common prac-

tice in studios. This avoids earth

loops and quality losses due to the

cabling. Furthermore, the trans-

former can be connected either 1:1 or

1:2. In the 1:1 configuration, the

input resistance is around 34 kΩ ,

with an input sensitivity of 1.5 V for

full output power. With the 1:2 con-

000118 - 14

Figure 4. The ac equivalent circuit of Figure 3.

the transformer in a completely different man-

ner. The ac signal current flows through the

complete primary winding of the transformer,

which is similar to the situation with a mains

transformer. Here the valves are connected in

parallel for ac signals, which reduces the

internal resistance of the PPP circuit by a fac-

tor of four. This has various beneﬁts. First of

all, the transformer may have a smaller trans-

former ratio. This reduces the effects of stray

inductance and winding capacitance, which

yields an improved frequency response.

Besides this, a genuine ac current flows

through the complete primary winding,

instead of one half-cycle per half winding as in

the classic Class AB push-pull circuit. This

avoids the much-feared ‘flyback’ effect that

occurs with Class AB push-pull output

stages, in which a half-wave voltage across

one half of the primary winding generates a

voltage that is twice as large on the plate of

the opposite valve, due to the 2:1 ratio of the

autotransformer formed by the primary wind-

ing. If the transformer core is magnetised and

the stored energy is not fully consumed on

the secondary side, the voltage on the pri-

mary side rises sharply, and to make matters

worse it is again multiplied by the same 2:1

factor. The end result is a voltage breakdown

at the weakest point — valve, socket or trans-

former. This may be a desirable effect in an

automobile ignition coil, but in an ampliﬁer it

is devastating. In the event of a short circuit,

a Class AB push-pull stage will become hot,

but it will not be destroyed. Such a valve out-

put stage is thus by nature short-circuit proof,

but not open-circuit proof.

A PPP output stage, by contrast, is safe under

both extreme operating conditions. Since the

transformer is connected as previously

described, the output stage does suffer from

‘flyback’. It can be compared to a mains

transformer, which also suffers no ill effects if

Elektor Electronics

5/2001

AU DIO

3p3

R10

220k

V1 = ECC83

V2 = ECC81

2k7

R11

47k

R15

R27

56V

47µ

450V

47µ

450V

220n

1 00V

R23

10k

V2a

630V

KT88

R19

1 00V

R25

R17

22k

V1b

see text

1 00V

220n

voir texte

C11

C13

25k

630V

R21

1N4007

47µ

100V

47µ

100V

Tr2

8 Ω

V1a

50V

Tr1

E1220

10p

4 Ω

R13

2k2

2 Ω

2µ2

R22

50V

R14

1N4007

C12

C14

AP-234

47µ

100V

47µ

100V

100p

25k

R26

R18

22k

R20

C10

1N4007

R24

10k

KT88

1000µ

V2b

100 Ω

22µ

100V

220n

630V

40V

R16

R28

Re1

R12

47k

1N4007

1000µ

40V

1000µ

40V

R29

100 Ω

6V3

100 Ω

Tr3

100 Ω

R30

100 Ω

Si1

100 Ω

200mA T

100n

1000V

GL1

470µ

450V

100µ

450V

Si3

B500C1500

2A T

50V

230V

50V

Si4

B500C1500

2A T

GL2

470µ

450V

100µ

450V

Si2

100n

1000V

200mA T

000118 - 15

Figure 5. A circuit according to the building-block principle: the equipment can be conﬁgured as a monobloc or stereo ampliﬁer, with

different types of output valves.

5/2001

Elektor Electronics

AU DIO

COMPONENTS LIST

R13 = 2k

7, MO, 2W

C15 = 1nF, MKH, lead pitch 7.5mm (see

text)

R14 = 33k Ω

R15,R16 = 10k

, MO, 2W

Power Output Stage

with KT 88 or 6550A

Semiconductors:

D1 = zener diode 56V 1.3W

D2-D4 = zener diode 110V 1.3 W

D5,D6 = 1N4007

R17,R18 = 10k Ω

R19,R20 = 220k

Resistors:

R21,R22 = 33k Ω

R23,R24 = 3k

R25,R26 = 10 Ω , MO, 2W

R27,R28 = 270

(unless otherwise stated, use metal ﬁlm

types, 0.7W, 1% tolerance;

MO = Metal oxide, 5% tolerance)

Miscellaneous:

V1 (Rö1) = ECC83

V2 (Rö2) = ECC81

V4 (Rö4),V5 (Rö5) = KT88 or 6550 A

Tr1 (Ü1) = E-1220 (if necessary)

Tr2 (Ü2) = AP-234

2 off ‘Noval’ (9-way) valve socket, PCB

mount

2 off ‘Octal’ valve socket, PCB mount

Tr1 = Mains transformer NTR-P\7 (Mono)

or NTR-P\5 (Stereo)

, MO, 2W

R29,R30 = 47 Ω

P1,P2 = preset 25k

R1,R2 = 68k

Capacitors:

C1 = 2

R3 = 2k Ω 2

R4 = 150k

F2 50V bipolar

C2 = 10pFceramic

C3 = 100pF ceramic

C4,C5 = 47 µ F 450V axial

C6 = 10-33pF (ﬁt only if parasitic oscilla-

tion is noticed)

C7,C8,C9 = 0

, MO, 2W

R5 = 2k Ω

7, MO 2W

R6 = 2k

R7 = 1M Ω

R8 = 2k

R9 = 22k Ω , MO, 2W

R10 = 390k

F22 630V, MKS 4

C10 = 10 µ F 100V, lead pitch 5mm

C11-C14 = 47

Ω

R11,R12 = 47k Ω , MO, 2W

F 100V, lead pitch 5mm

Elektor Electronics

5/2001

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