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Last update: 15/09/01 22.14.00

Hallicrafters SX28 - LF Amplifier & Power Supply

This page is dedicated to describe and understand the SX-28's LF-amplifier circuit.

Having a look at the original diagram from Hallicrafters, it was not so clear to me what kind of circuit it is. So I have re-drawn it in a different and more understandable way.

This is the original Hallicrafters diagram   and this my redraw of it:   

The two triodes within V12 are configured as voltage amplifiers providing two phase opposite input signals to the push-pull power tubes.

The preamplifier (and the inverter)

Let’s examine the first stage working conditions. The following table compares the conditions as required by the 6SC7 specs and the value required by the Hallicrafters manual. By the diagram we can see as the voltage supplied to the triode circuit is  limited and stabilized to 200Volts by R38 and C44.

Parameter

6SC7 specs

Required by SX-28 manual

My rig

Plate current (mA)

2 (max)

0.7 (read text)

Not measured

Cathode bias (Volts)

-2

-2

-1.6

Plate voltage (Volts)

250 (max)

130 

127

Control grid (Volts)

0

0

0.01

Plate load line:

Let’s take the 6SC7 characteristics and let’s trace the 100K load line for a B+ voltage of 200Volts. This line crosses the points 200V, 0mA on the horizontal axe and 0Volts, 2mA on vertical axe. Using the voltages indicated by the Hallicrafters manual (TM 11-874 @ page 34) we find that the no signal working point should belong to the –2Volts characteristic and at 130 Volts. Unfortunately this point doesn’t belong to the 100K load line. The condition seems impossible (see pic). Is anybody able to explain me why ?
Looking at the plate characteristics, you can easily verify that the load line point corresponding to 130Volts plate voltage stays on the 1,5Volts characteristic line. Under this condition the plate current results 0,7mA that seems reasonable. Here are the correct working conditions. Measures in my rig confirmed the values graphically derived (127 Volts on plate and a bias of 1,6Volts); applying an input signal of 2Volt p.p. the plate voltage ranges from 95 and 165 volts, in line with the graph; this happened only at very low frequencies; as soon as the frequency increased over the 50Hz value, the voltage range dropped significantly. It remains to be verified if the dropping is compatible with the bandwidth diagram given in TM 11-874 manual….
The bass control circuit remains to be clarified. It look like there is an error in the Hallicrafters diagram. I am submitting questions about this topic to relevant mailing lists and I ‘all report foundings as son as they’ll be available.

The power amplifier

The push-pull working condition is class AB1. The working condition chosen by the designers is the one highlighted  on this 6V6GT specifications. Those values exactly match the SX-28 specifications: the output choke specs as well as the voltages required by the SX-28 maintenance manual.

Here are the figures:

Parameter

Required

Measured my rig

Plate current

35 mA

32 mA

Cathode bias

15 Volts

14.5 Volts

Plate voltage (Vs Cathode, not GND!)

250 Volts

250 Volts

Screen grid

250 Volts

264Volts

Control grid

0 Volts

0.014 Volts

load dynamic line:

In tube power amplifiers with an output transformer, two different working conditions have to be considered. The first being the one with no signal, the second including signal. This is due to the fact that the amplifier load is constituted by the primary of the output transformer and thus inductive, meaning that the load dynamically changes with the frequency or without. This means that you don’t have a single load lines, but two.

Let’s find the no signal working point. With no signal the tubes have no load (I consider the primary resistance equal to 0) so the entire voltage coming out from the power supplier is applied to the plate. The load line for load=0 is a vertical line passing through the power supply voltage B+=250Volts. The point we are looking for is at the intersection between this line and the tube characteristic of the specified bias, -15Volts in our amplifier. The load line we are looking for has to pass through this point. But with which inclination?? (see pic)

With a dynamic load impedance of 10K plate-to-plate each tube sees a 5Kohm load resistance. To find the line we can temporarily assume that the 5 Kohms load of the single tube was purely resistive. In that case the load line should have been  the one crossing the points 250V, 0mA on the horizontal axe and 0Volts, 50mA on vertical axe. This line is not the one we are looking for as it doesn’t pass through the resting point, but it has the correct inclination. So the dynamic load line is the only line passing through the resting point and parallel to the 5K line we just plotted down.
Don’t get surprised that with that line it seems the power supplier supplies more voltage than it is expected to do. In fact it supplies the correct voltage, but the inductance effect of the choke’s primary increases the plate voltage that way. Think of it has an attempt of the inductor to compensate for the current reduction during the falling part of the signal slope.

With regard to the output transformer, we can say that its primary must have an DC resistance of 500 ohms and 10000 ohms as impedance plate-to-plate. Its windings ratio should be given by the formula: n=SQRT(Rl/Ro) where n is the ratio, Rl is the required load impedance as per the tube specs and Ro is the output (speaker) impedance. This means that the original choke should have n1=SQRT(10000/500)=4.5  and n2=SQRT(10000/500)= 1.4  for the 500 and 5000 ohms outputs. If you want (or need) to substitute it with a 8 ohm output, the formula becomes: n=SQRT(10000/8)= 35 (it can be challenging to build such a choke in the same space as the original one… some  people report Hammond has made such a transformer for their guitar amplifiers…).

Here is an excerpt from the user manual (as you can find @ bama) describing the LF-amplifier and the power supply;



Waiting for the net to clarify:

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Last update: 15/09/01 22.14.00