test - amplifier from the eighties

Measurement of a forest and measdows hifi amplifier from the eighties


peaceful meadow in the Black Forest

I would not like to call the name of the manufacturer of this amplifier, because I had nothing at all of it a product by the cocoa to pull. It is a receiver, it sold in the early eighties, the price lay at that time with approximately 400-500 DM. Such a part is easy to measure, since the usual sources of error in this amplifier step out immediately. It is to serve as a training model for sources of error.I don't want to mention the name of the manufacturer, I don't have any gain for it. It is a receiver sold in the early eighties for a price of 200 - 250 USD. It's easy to measure such an amplifier, the errors appearing quickly. This amplifier should be my demonstration model for typical errors.

Measurement of the Amplitude Response

The measurement of the amplitude response was done with the Measurement System , demonstrated in the categorie Amplifier. This measurement shows the amplitude response under full load.

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Figure 1 shows the amplitude response under a variable input voltage and a 4 Ohm load. the bandwidth is with 35 kHz only less above the audio range.

Figure 2 is a zoom of figure 1. Clearly to see the nonlinear amplitude response within the audio range and the heavy gain variation against the output voltage.

Figure 3&4 showing the inut and output amplitude (true RMS volts)

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Figure 5 shows the amplitude response with a variable input amplitude under a 4 Ohm load. The bandwidth decrease to 25 kHz, almost in the audio range. Figure 6 is a zoom of figure 5. Under a low load decrease the influence of input variations against the gain. Figure 7&8 showing the  input and output amplitude (true RMS volts)

Comments on Figure 1

Bandwidth only 35 kHz

that means an early voltage drop for high signal frequencies, which lie still in the audible range. Result is a smaller amplitude for this signal frequency. We go once of it out of the developers a theoretical reinforcement of 40dB had aimed at, in the range of 20 kHz reach it only 38,2dB. As relationship expressed: 40/20 and this quotient in the exponents to the basis of 10 sets, results in 100. The amplifier strengthens thus maximally with factor 100 in the desire conception. However 1,8dB are missing to it to the ideal value. DB's can be reckoned back again: 1,8dB/20 and this quotient in the exponents to the basis 10 set, there come out a number of 1,23. 100/81.2 = gain error ratio of 1.23

What does it matter 1.23?

it is the gain error ratio. Let's have a look at figure 3, we define the maximum input to exactly 150mV, the output should be 150mV * 100 =15V at 20kHz. Unfortunately there is an error 15V / 1.23 = 12.2V instead of 15V.

Short calculation for influence on power:

Ptarget = 15V*15V / 4 ohm = 56.25 Watt

Pnow = 12.2V *12.2V / 4 ohm = 37.21 Watt

Sorry, 56.25 Watt - 37.21 Watt = 19.04 Watt missing output power at 20kHz.

Very nice?

Comments on Figure 2

Twisted Amplitude Response

the frequency response should be smoothly like a ruler over the entire audio range and sufficient in addition it. Following the writing in the left column, the equations can be applied naturally also 1:1 to the entire amplifier frequency response . Optically clearly recognizable statement, the amplifier is clearly violently stressed in the range upper bass and the mid range. The amount of the deviation is not tolerable. Can this error be corrected by adjusting at the bass and treble control? My answer to it: the sound controllers stood during the measurement accurately in mid range position. Is it now actual the task of the customer to correct by means of complex measuring technique the work of the developer again? Probably hardly. I naturally tried it. The curve can be improved, it becomes bent for it however again with other frequencies.

Reasons for these nonlinearity?

is to be said only with difficulty completely exactly and always "probably" at the end attached. In the completely deep frequency range at the AC signal coupling, which probably begins relatively late from avoidance cost reasons for a large condenser. Reasons of the increased height could be that the Bass/Treble filter is unfavorably dimensioned. The low range; it is missing to the amplifier at sufficient open loop gain with high frequencies.

Comments on Figure 2

Variation of Gain against Oput Power?

This is a evil bad card for this amplifier. Take a loo at figure 2 on frequency 350 Hz. with the minimum adjusted load the gain reachs 40.19 dB (cyan 17%). The darkblue colored curve under full load (100%) the gain decrease to 40.06 dB. A loss of 0.13 dB.

What's the problem with a loss of only 0.13 dB?

Divide 0.13 by 20, take it to power of 10 = 1.015. A changed gain of 1.5% agianst the input voltage. What are the consequences?

That is not beautiful result, why? The amplifier represents 350 cycles per second the loud tones with the frequency around 1.015% more quietly than it to be should, if the quiet tones are consulted as reference for a correct volume. The result are not linear distortions, speak distortion factor. Mathematically regarded, here a multiplication of the not constant amplification factor with the input voltage takes place. The communications technology calls this happening "mixture" or "amplitude modulation". With an ideal amplifier this "constant" amplification factor is in an XY coordinate system (gain vs. input amplitude) regards a parallel to the axle of the input voltage. With this amplifier it is a straight line with more easily negative upward gradient, which is curved very probably still additionally to something. The result of such a multiplication are mixing products, which are reflected on the right of and left at the carrier frequency (here the signal frequency 350Hz) in an integral and not integral relationship.

Figures: left from the 350 Hz a frequency 350Hz - 350Hz = 0Hz = DC component. Right from the 350Hz a mixing product of 700Hz, often called K2. The next overtone can be found at 1050Hz called as K3, and so on.

The answer to the question: which amplitude and phase have these new frequencies? On it, "like much" and with "which form" these amplifiers straight line depends is bent. At the form of the "buckling" decides whether for example K2, K3 or also K4 etc. dominates. Mathematically to prove this leaves itself for example by an approximation with a Taylor polynomial to these "nearly straight lines". The individual amplitudes and phase shifts of the mixing products DC, K2, K3, K4.... Kn can be computed in such a way. The mathematical expenditure for this is not at all once as enormously as one assuming could. It is somewhat more near described in the report Distortion factor forest and meadows amplifiers.

The serious problem for such a computation lies in it the actual accurate behavior this "amplifier straight line" to get. It is in particular with a very good amplifier an instrumentation and after today's conditions of the measuring technique a nearly impossible task, which excellent extremely good and calibrated AC test equipment would require. The work expended would not be small. Fortunately there are still simpler methods to receive around the distortion factor - measure. In order to come now on the point, much can be computed, the correct parameters too gotten is hell. Where would the use lie? Nowhere to only satisfy in order any theoretician? A half year later is noticed that the parameters of the praised calculation were wrong (the type makes then a stupid face and was not not debt). Often in it also a deficiency of simulated circuits, mathematically first-class, in the parameters description lies poorly. I take the liberty to state these damage mechanisms to have understood. It does not correspond to my philosophy most complicated at the symptoms to tinker, it is better the effects develop not to be let at all. In the plain language spoken, rather a good amplifier develop, than a less good amplifier to talk beautifully with rhetorical garb and much twaddle. To perhaps even still state, these instrumentation bad characteristics are in music insignificant. Which is then importantly? Is an amplifier, which corresponds less to the theoretical ideal for instance a better than that, which comes the ideal more near?

Where now does the cause for the variation of the reinforcement lie as a function of the output voltage and power output? 

A transistor or a tube is a strengthening element. Stupid way is this gain not so strongly and also not as constantly as one it gladly would have. Would like that be, then it would be possible to built from only one transistor or a tube an amplifier. A transistor behaves in such a way: if no collector current flows, then it has also no gain called beta. Only a very small collector current flows in such a way has a little gain. Flows somewhat more, the gain rises. Sometime it reached one point, to which it exhibits a high gain to a still comparatively small current. Starting from there history turns, the amplification factor beta becomes ever worse with increasing collector current. As next miserably characteristic is added, this gain sinks with rising frequency. Next miserable characteristic is the temperature dependence of the gain. In addition comes still the dependence on the collector voltage.

In order to control all these negative characteristics, many elements and transistors are needed, in order to reach something property. That is not done, in order to put to the signal as much as possible transistors into the signal path. Rather so many elements are needed, in order to eliminate with many the bad characteristics of a particular. The maintained remark often belonged: "the signal path of an amplifier must consist that of as few an elements as possible, is doubtful in my opinion, except regarding probability of failure. To state such a thing is much too overall, what counts is the final result and not how. Sometimes many construction units can be unfortunately attained necessarily around the wishing.

How now does the open loop gain (open loop) of an amplifier build itself up?

 e.g. the amplifier consists of two transistor stages, transistor 1 has beta of 100, transistor 2 one of 200. The two current amplification factors can after mathematics of control engineering with one another be multiplied and results in those entirely current gain 100*200=20000 to correspond 86dB open loop gain.

Phase Respone Measurement

Normally it's easy to do phase shift measurements. Take an simple analyzer e.g. HP3575A, a networkanalyzer with high input impedance, a two channel FFT analyzer or special Precision Phase Meter. All devices has one admits common: the two input channels have a common ground. Unfortunately the forest and meadows amplifiers do not have common ground, exactly that are the problem at this measurement. This amplifier works in a bridge connection, so that the negative pole at the output is not connected with the input ground. If the measuring instruments are attached, thereby a connection between input and output develops. That does not like the amplifier naturally at all and "spits violently". Most amplifiers however have fortunately a common ground, but phase measurement with a bridge amplifier, directly annoyingly.

What could you do against: 

  • try my luck using a differential amplifier in the scope, two Tektronix 7A13, than input and output will appears with correct phase on the scope screen. This method works with a phase about 2 degrees or more, below it's almost impossible even with a wide openened time base deflection factor. The method is good to get a first glance of the quality of the phase. To to do a plot much work always calculating by scope.

  • take two fast differential amplifier, build of very fast video operational amplifier. They have almost no phase in the audio range. Both buffer amplifier sharing the same common ground, easy to measure. Used resistor dividers should be feqrequency corrected. Better ideas are welcome.

What I've done:

  • nothing, I was too lazy for a forest and meadows amplifier to do that much work. This amplifier it is not worth, it's easy to see in the amplitude response that the phase response having the same bad performance. A first scope measurement shows me a phase response starting already at low frequencies. 

  • Most good amplifiers working with circiut design with a common ground for input- and output signal, than phase response measurements are easy.

Slew Rate measurement

Slew Rate is measured by applying a symetrical square inut voltage. The slew rate is defined as the slope between 10% and 90% of the signal. It can be measured either on the rising or the falling edge. Important, the square edge must be much more faster than the excepted amplifier output response. Best choose a square wave frequency still in the working range of the amplifier. For audio amplifier, for example a 1 kHz with a output amplitude of 10 volt under a less inductive 4 ohm load.



Figure 9 shows the inputsignal, a square wave signal with an amplitude of 1V and a period of 1ms, 1 kHz.

Figure 10 shows the output signal.. Even in this small thumbnail picture easy to see, there is a very high overshooting.

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Figure 11 shows the zoomed inputsignal using a second timebase 7B92A, the squarewave generator is specified with a rise time of 20ns under a 50 ohm termination. This rising edge is fast enough to do the slew rate measurement. The scope shows not any overshooting for the generator, only the the corner could be a little more faster.

Figure 12 outputsignal as in figure 10 with a vertical deflection factor of 5V/DIV. This should be a square wave signal with a 10V amplitude. Overshooting up to 20 volt and needs a settling time of 100µs to reach the final value. Bad results. Measurement done with a compensated scope probe.

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Figure 13 rising edge of the outputsignal as in figure 12. The timebase is setted to a small horizontal deflection factor. The measurement shows a slew rate of 18V/6µs = 3V/µs. Too slow for an audio amplifier.

Figure 14 falling edge of the outputsignal as in figure 12. The timebase is setted to a small horizontal deflection factor. The measurement shows a slew rate of -15V/5µs = -3V/µs. Nice, the positive slew rate is the same as the negative.



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