The SRPP Pre-Amplifier: Quick Down and Dirty, how this works for the purpose of troubleshooting

R1 =1M? = Input Load

R2 = 300?

R3 = 1M? = Output Load

C1= Output Capacitor

This is a Series (Shunt) Regulated Push-Pull  (SRPP) amplifier, many articles have been written about this amplifier because it is so simple it becomes complicated to explain because there are many configurations this amplifier can be set for and still be an SRPP amplifier. The configuration above is a basic bare-bones configuration used for explaining how the circuit functions without the complications of grid plate biasing circuitry that can be installed in the Plate, Grid, Cathode matrix formed by the plate of the bottom tube, the grid of the upper tube, and the cathode of the upper tube fed from the plate of the bottom tube via (Resistor anode to cathode (Rak)). Note the plate is also known as the anode of a tube. So, therefore, the resistor from the Anode of the bottom tube to the Cathode of the upper tube is (Rak) and the resistor from the cathode of the bottom tube to ground is (Rk).

Here we go, when the circuit is first turned on the upper and lower tubes will conduct for a moment and set the bias voltages, the grid of the lower tube will attempt to reverse bias because of the presence of R1 and R2. This process is caused when electrons flow from the cathode to the grid putting a negative-going or negative trending voltage on the grid (since this is a class A-B amplifier it is biased slightly above the cutoff point, so the voltage is not actually negative but it is low). This causes the bottom tube to conduct less until the quiescence point is met. This in effect causes the voltage on the plate (anode) of the bottom tube to go more positive, which puts a positive voltage on the grid of the upper tube causing it to effectively conduct more until the quiescence point is met. When both tubes have met their quiescence points, bias will be established on the tubes. If the B+ voltage is 400v each tube at quiescence should have a voltage drop of about 189.12 volts from anode to cathode with the remainder dropped across Rak and Rk which totals 21.754 or 10.88 per resistor. The current through the SRPP would thus be Vak/Rak or 10.88/1k = 10.88ma the same goes for Vk/Rk or 10.88/1k =10.88ma Knowing the current through Rk and Rak we can determine the effective resistance at the quiescence point which is 400v/10.88ma = 36.77K ohms from the plate of the top tube to ground (note: this is a calculation, not a measurement).

With these voltage readings, it is possible to tell if the components are functioning properly as far as Direct Current readings (Static Voltages).

Figure 2

Multisim 14 rendition of the SRPP Circuit

Knowing these values allow you to troubleshoot the DC voltage levels to test if the circuit is functioning properly as far as the bias voltages are concerned which are as follows:

Tube 2 Bottom Tube (ground reference)

Anode (Plate): 199.998v

Grid: 10.879uV or .000010879v

Cathode: 10.88v

Heater: 6vdc across the heater element

Tube 1 Top Tube (ground reference)

Anode (Plate): 400v

Grid: 199.998v

Cathode: 210.88v

Heater: 6vdc across the heater element

Understanding that the bias voltages are important because they set the tubes operational parameters for the dynamic voltage changes that occur when a signal is applied to the amplifier’s input.

Dynamic Voltage tests:

Since the gain is not known at this point because we did not use a load line to determine the RL = Resistance load, we can measure the input voltage and the output voltage to determine the dynamic gain of the circuit.

From the dynamic readings we start off with: (2vp-p input, Channel B, bottom trace on the oscilloscope) (7.45vp-p output, Channel A, top trace on the oscilloscope)

Vac input: 1vpk or (.707vac RMS, this is the voltage that produces the same power as its dc voltage counterpart)

Vac output: 3.72vpk or (2.634vac RMS)

Gain is (Vout/Vin = 3.72/1 = 3.72) so the output signal will be 3.72 times as large as the input signal.

Gain is 3.72 ≈ 4 this is good for a preamp.

Therefore we have our static and dynamic readings which show this circuit functions very well as a preamp circuit.

The output of the preamp is usually fed into a power amplifier tube which amplifies the ac signal giving you the final output.

The question is why did I do this?

To demonstrate there is nothing special about this circuit and the complications that I have seen in the explanation of this circuit are all based on the plate to grid connection and how the designer seeks to bias the tubes around the interconnecting matrix of added resistors. The circuit remains basically the same. The only thing that justifies the high cost of these amplifiers is the fact that you can put $700 dollar capacitors in this unit followed by a $600.00 audio transformer and a $500.00 power transformer, not to mention tubes which can vary in cost. Listen, I know people love this hobby and money is meant to be spent, but my knowledge of electronics tells me that I can build an amp from a kit that may not rival the cost of the more expensive amps but I can give them a run for their money as far as sound. Two of the most basic circuits in electronics should not cost that much money.