(last update: 01 Feb 2012)

The final amplifier has two stages, and should not require more than about 200 mW of input power to achieve a full 100 watts of output power, into 50 Ω.

This implies that a peak voltage of about 4.5 volts at the output port 59 of the *LOW LEVEL AMP 81037* would be required. Transistor Q1 already produces about 2.5 volts of amplitude, measured at about 10.0 MHz.

All that Q2 needs to do is provide another 2 volts or so of voltage, and enough power to drive the final amplifier's input impedance, of about 50 Ω.

The Emitter resistor R9 has 1.126 volts across it at 0 drive, transmit mode.

The Emitter resistor R9 has 3.445 volts-peak across it at FS drive, transmit mode. This is a bit deceptive, since the input AC signal excursion is -2.72 and 2.32 volts.

The Emitter resistor R9 has 0.0 volts across it in receive mode.

Unfortunately the manual does not provide much of a circuit description. Nor does it provide a detailed list of bias voltages, and currents. So we are left with the laborious task of backing such things out of the schematic, and visual inspection of the amp itself.

Since there are no obvious failed components, some detail will need to be garnered as to the operation of this circuit, and why it might cause problems for the output transistor of the *LOW LEVEL AMP 81037*.

T4 is on the left. The input transformer T2 is on the right.

The power amplifier circuit is similar to the Motorola application notes: AN758, EB63, and EB27A. The Corsair 560 design shares many aspects of each.

in the Ten-Tec Corsair 560 HF transceiver manual.

The feedback network is similar to that of EB27A, which is a 300 watt class B design.

The final power transistor DC power delivery is similar to that of EB63, which is a 140 watt class B design.

The final power transistor biasing is similar to that of AN758 (and others), which uses a voltage regulator to set the transistor Base voltage. Once again in a class B design. In such designs both transistors receive the same Base bias voltage.

This implies that both transistors have similar characteristics, so as to minimize distortion in the output. Two such transistors would be a so called "matched pair". Finding such a matched pair can be tedious, and/or expensive.

In some cases manufacturers supply two transistors on the same die, in a single package. This is an optimal solution to the "matched pair" problem.

There is no complex impedance output matching section in the Corsair 560 design. This is also true, more or less, of the Motorola app. note designs. This strongly suggests that the design has a natural power transistor output impedance, with low enough reactance, such that it can be matched to a 50 Ω load via the output transformer's turns ratio alone.

OK... so... If the transistor output impedance has little reactance, how does that come about, and what must the predominant real resistance be?

The winding ratio of the output transformer is 4:1. However, the primary winding is center tapped. So the drive current for each half of the primary circuit only flows through one half of the primary winding, and core. Also note that the "core" isn't a single core. It is two side by side single cores, which do NOT share a common field. This is evident in the photo of the Final AMP, as it was built, as seen in Image 2. The two side by side cores, of the output transformer T4, are prominent on the left side of the photo. Notice in the schematic (upper right corner) T4 appears as a single core, center tapped, transformer.

In any case the turns ratio yields an impedance transfer characteristic of 64:1.

\label{equ:SecondaryTurns}
N2 = 4

\label{equ:PrimaryTurns}
N1 = 1

\label{equ:TransformerRatio}
ratio = \frac{N2}{N1} = \frac{4}{1} = 4 : 1

\label{equ:ImpedanceRatio}
ratio_{Z} = \bigg( \frac{N2}{N1}\bigg) ^2 = \frac{4^2}{1^2} = 16 : 1

However, the primary turns are center tapped so,

\label{equ:PrimaryTurnsReal}
N1 = 0.5

thus,

\label{equ:ImpedanceRatioReal}
ratio_{Z} = \frac{4^2}{0.5^2} = \frac{16}{0.25} = \frac{16}{1} \cdot \frac{4}{1} = 64 : 1

Again, this transformer does not have an AC center tap per-se. This is because the transformer cores do not share the same field. The field in each of these side by side cores is due to the one half of the windings that pass through it. The cores are not a single piece of material, carrying the flux of all the windings in the transformer.

If this 64:1 characteristic is applied to the suspected transistor impedances, the result should be close to a good match to a 50 Ω transmission line.

Naturally this hypothesis should be checked... so let us check...

The first place to check the hypothesis is the MRF458 datasheet.

The datasheet reports,

\label{equ:ZparalleloutL}
Z_{pL} = 2.0~\Omega - 8.842j~\Omega

equation \(\ref{equ:ZparalleloutL}\) for the low end of the Amateur band (1.5 MHz), and

\label{equ:ZparalleloutH}
Z_{pH} = 1.5~\Omega - 2.65j~\Omega

equation \(\ref{equ:ZparalleloutH}\) for the high end of the Amateur band (30 MHz).

Unfortunately these two impedances are quite high. When multiplied by the transformer ratio (64:1) the resulting impedances are a poor match to 50 Ω.

A second problem is that they are parallel impedances. Not series impedances.

The circuits in which the MRF458 devices are used are series circuits. The datasheet parameters were found as parallel elements. The two should not be considered interchangeable. Using parallel characteristics, in a series circuit, can cause serious design errors. It is often best to transform one into the other before proceeding with a design or an analysis of a design.

Series parameters are not provided in the datasheet. However, the parallel parameters may be converted mathematically to their series equivalent. Equations to do this are listed in Motorola app. note AN282A.

Converting parallel resistance ($R_p$) to series resistance ($R_s$),

\label{equ:Res_pToRes_s}
R_{s} = \frac{R_p}{1 + \Big( \frac{R_p}{X_p} \Big)^2 }

Converting parallel reactance ($X_p$) to series reactance ($X_s$),

\label{equ:X_pToX_s}
X_{s} = \frac{R_s \cdot R_p}{X_p}

To go the other way,

\label{equ:Res_sToRes_p}
R_{p} = R_s \Bigg[ 1 +\bigg( \frac{X_s}{R_s} \bigg)^2 \Bigg]

\label{equ:X_sToX_p}
X_{p} = \frac{R_p \cdot R_s}{X_s}

All parameters retain the ± sign, of Complex Impedance notation, and are in ohms.

Armed with equ. \(\ref{equ:Res_pToRes_s}\) and \(\ref{equ:X_pToX_s}\) the series representation may be calculated, and the analysis simplified.

Plugging in the numbers for the low end of the band (1.5 MHz),

\label{equ:Res_pToRes_s_numericL}
R_{sL} = \frac{2.0 \Omega}{1 + \Big( \frac{2.0 \Omega}{-8.842j \Omega} \Big)^2 }

\label{equ:X_pToX_s_numericL}
X_{sL} = \frac{2.108 \Omega \cdot 2.0 \Omega}{-8.842j \Omega}

\label{equ:ZsLnumeric}
Z_{ssL} = 2.108 + 0.477j \Omega

and for the high end of the band (30 MHz),

\label{equ:Res_pToRes_s_numericH}
R_{sH} = \frac{1.5 \Omega}{1 + \Big( \frac{1.5 \Omega}{-2.65j \Omega} \Big)^2 }

\label{equ:X_pToX_s_numericH}
X_{sH} = \frac{2.207 \Omega \cdot 1.5 \Omega}{-2.65j \Omega}

\label{equ:ZsHnumeric}
Z_{ssH} = 2.07 + 1.249j \Omega

The impedances $Z_{ssL}$, and $Z_{ssH}$ can now be used with some confidence that the analysis will yield reasonable results, in a simplified circuit.

Which begs the question, what does that circuit look like? Here's one way to look at it...

amplifier circuit of the Ten-Tec Corsair 560 HF transceiver.

In Schematic 2 only one side of the final drive is shown. This is underscored by the fact that one side of T4's primary is shown without any connections.

In addition, all of the biasing for each output MRF458 transistor is also omitted. Other compensation components have also be left out. Such components improve the stability and/or frequency response of the circuit. What is left behind in this simple schematic is the equivalent elements that deliver RF power, and serve to illustrate the operation of the circuit.

The dashed box of Schematic 2 encloses the equivalent circuit of one of the MRF458 transistors (Q5 in this case). This is a large signal model of the device when biased with 100 mA of current at the Collector, at 12.5 volts, and RF power output of 70 watts into a matched load. All of this is stipulated in the Motorola MRF458 datasheet.

Note that the datasheet's stipulations suggest that an equal amount of power is being dissipated in the transistor itself, since it is specifying the OUTPUT power. This is consistent with the Collector Efficiency listed in the datasheet as 50%. There are 70 W of power dissipated in the load, and 70 W of power dissipated in the transistor. Cycle to cycle, a DC power supply will need to source at least 140 W of power to satisfy the circuit. Seventy watts is one half (50%) of the total power (140 watts) used. Thus an efficiency of 50% at the Collector. However, don't confuse this transistor parameter with the overall efficiency of the amplifier. The two are not interchangable, although they are often related.

Another thing to note is that the transistor dissipation is well within the 80 watt limit of the device.

But to get a better feel for where the power goes, it may be better look at it this way,

class B final amplifier circuit of the Ten-Tec Corsair 560

HF transceiver, redrawn for power analysis.

In Schematic 3 the circuit has been arranged so that the power distribution is a bit more clear. As current flows out of Vs. It passes through ZssH (impedance, series, source, High band) and then through the load (shown as a resistor), reflected through the output power transformer T4.

In this way, an current "produced" by the transistor dissipates power in itself, and the load. The total power, if the load dissipates 70 watts, will be 140 watts. The transistor ZssH dissipates 70 watts, and RL dissipates 70 watts. The RF voltage across RL is one half Vs.

As long as ZssH = RL, the rated output power will appear in the load. However, if RL is larger than ZssH, the current through the circuit drops. The source Vs is fixed, and as the current drops so too does the power. As RL goes to infinity the current goes to zero, and no power is dissipated at all.

Alternatively, if RL goes to zero ohms, the only thing dissipating power will be the transistor. This is because Vs is fixed, and the maximum current it can supply is limited by ZssH. If RL gets smaller and smaller, the RF output voltage will get lower and lower. When the output voltage is zero, RL is a short, and no power gets dissipated in the load.

So... The only way to get the maximum power out of the circuit is to match ZssH and RL.

There are other issues when ZssH and RL are not matched. However, for the time being, let us get back to the how the match may be achieved in the circuit in question.

The impedance $Z_{ssL}$ of equ.\(\ref{equ:ZsLnumeric}\), and $Z_{ssH}$ of equ.\(\ref{equ:ZsHnumeric}\) are $2.108 + 0.477j \Omega$, and $2.07 + 1.249j \Omega$ respectively.

Our desired output impedance is 50 Ω.

Somehow $Z_{ssL}$, and $Z_{ssH}$ need to be transformed into 50 Ω. The only way that can happen, given our circuit, is through transformer T4. There are no other components, resonant or otherwise, that could do the job.

The transformation of the impedances will be dictated by the square of the turns ratio of the transformer. Given our circuit, the ratio is that of equation \(\ref{equ:ImpedanceRatioReal}\) or 64:1.

Unfortunately this yields the following,

\label{equ:ZssHnumeric}
64 \cdot Z_{ssH} = 141.259 + 79.958j \Omega

and,

\label{equ:ZssLnumeric}
64 \cdot Z_{ssL} = 134.902 + 30.514j \Omega

well that didn't work out too well... Neither of these impedances is remotely close to 50 Ω.

The key here is the factor of 64. It is a big factor, as far as transformer impedance ratios go. Such high ratios are often difficult to work with in practice. Ordinarily designers shy away from such high ratios. However, the folks at Ten-Tec didn't, so we just have to work with it.

One good thing about such a high ratio is that small changes on the low impedance side make big changes on the high impedance side. With that in mind a change of,

\label{equ:ZbpcLnumeric}
Z_{bpcL} = -1.3 + 0.277j \Omega

and,

\label{equ:ZbpcHnumeric}
Z_{bpcH} = -1.4 + 1.0j \Omega

yield the following final impedances,

\label{equ:ZssHnumericAdjusted}
Z_{ssH} = 51.659 + 15.958j \Omega

and,

\label{equ:ZssLnumericAdjusted}
Z_{ssL} = 51.702 + 12.786j \Omega

The impedances of equations \(\ref{equ:ZssHnumericAdjusted}\) and \(\ref{equ:ZssLnumericAdjusted}\) are now good matches to 50 Ω. This is especially true because the reactive components of these complex impedances are now quite low.

The funny part of this is the negative real resistances in \(\ref{equ:ZbpcLnumeric}\) and \(\ref{equ:ZbpcHnumeric}\). What is meant by those? This is where a change in bias current comes in.

The MRF458 datasheet specifies a bias current of 100 mA. An increase in the bias current will lower the impedance of the device. How much? Hard to say without a graph of impedance vs. bias level. Unfortunately this is not offered in the datasheet. But a drop of 1.3 to 1.4 ohms is not unseasonable. Especially given the fact that at 100 mA the real part of the devices impedance is about 2.0 ohms.

In the Corsair 560 design, The MRF458 is suggested to be biased at about 250 mA, in each transistor. It is reasonable that increasing the bias current by a factor of two or so, will decrease this real resistance by about half or so.

It is harder to argue what will happen with the reactive components. But similar changes may be reasonable. In any case the numbers here yield impedance magnitudes of,

\label{equ:ZssHnumericMagnitude}
|Z_{ssH}| = 54.068 \Omega

and,

\label{equ:ZssLnumericMagnitude}
|Z_{ssL}| = 53.26 \Omega

These impedances aren't too bad. Some power is lost in the reactance. However, careful bias tuning may reduce the power dissipation of the reactance inside the transistor. The reactance likely will not be eliminated by bias adjustments alone. It will likely be present at some level. This may be the price paid for such a simple class B design.

But if it works well enough, there may not be much point in throwing more components, and circuit complexity at it.

If this analysis is correct, we now have a good idea of how the final stage of the Final Amp in the Corsair 560 works. This will help to determine how the earlier stages might be expected to function.

A plausible model of how this stage of the amplifier works also provides a basis on which to determine if this stage is working properly. This can be important, when there isn't any obvious physical indication of failure. Indications such as burnt components or obviously wrong voltages during operation.

Checking out other sections of the Ten-Tec Corsair 560 "FINAL AMP 80565"

Image thanks to ARRL