Calculation of Output ImpedanceFor a better estimation of the interaction between speakers and a tube power amplifier sometimes it might be useful to measure the tube amplifier output impedance. A very simple method requires two resistors, an AC voltmeter and some basic calculations. Figure 1 shows the experimental setup.Any power amplifier can be seen as a voltage source (UX) with a resistorRX in series. RX resembles the amplifiers‘s output impedance.Even in high power amplifiers with huge global feedback RX might be verysmall but rarely will be zero. RL is the load, U1 and U2 are the outputvoltages to be measured with varying load resistor values. Fig. 1: Basic setup to measure output impedance.EQ 1 and 2 describe the relation between resistor values and voltages.UX / U1 = (RX + RL) / RL(equation 1)UX = (RX + RL) * U1 / RL(equation 2)RL is known and U1 is measured. However, the equations contain the two unknown variables UX and RX. To solve this problem RL is multiplied by the factor a for a second measurement which givesUX / U2 = (RX + aRL) / aRL(equation 3)UX = (RX + aRL) * U2/ aRL(equation 4)UX disappears by substituting UX in EQ 2 by the right side of EQ 4 which gives(RX + aRL) * U2/ aRL = (RX + RL) * U1 / RL(equation 5)RL cancels out which leaves(RX + aRL) * U2/ a = (RX + RL) * U1(equation 6)Multiplication on both sides of EQ 6 reveralsRX * U2/ a + RL * U2 = RX * U1 + RL * U1(equation 7)Subtracting RX * U1and RL * U2 on both sides of EQ 7 reveralsRX * U2/ a - RX * U1= RL * U1- RL * U2(equation 8)RX * (U2/ a - U1) = RL * (U1- U2)(equation 9)Thus output impedance RX equalsRX = (RL * (U1- U2)) / (U2 / a - U1)(equation 10)RL is known and both U1 and U2 are measured. For measurement of U2 RL is multiplied by the factor a. EQ 10 can easily be programmed into an EXCEL table for measurement series. Let us briefly check whether EQ 10 is correct:If UX is 10V, RX is 9Ohm and RL is 1Ohm U1 must be 1V according toUX / U1 = (RX + RL) / RL10 / 1 = (9 + 1) / 110 = 10If RL is changed from 1Ohm to 3Ohm (a = 3) U2 must be 2.5V according toUX / U1 = (RX + RL) / RL10 / 2.5 = (9 + 3) / 34 = 4Determination of RX from EQ 10 with U1 = 1V, U2 = 2.5V and a = 3revealsRX = (RL * (U1- U2)) / (U2 / a - U1)RX = (1 * (1- 2.5)) / (2.5 / 3 - 1)RX = - 1.5 / -0.1666667RX = 1.5 * 6RX = 9which equals the preset value of RX.Output Impedance Nashville SE 20Output impedance was measured at 1W output power (2V@4Ohm, table 1a) and 9W output power (6V@4Ohm, table 1b) at 100Hz, 1000Hz and 10kHz according to the above equation 10. The following two tables show the results. Output impedance is around 1Ohm and almost constant with respect to frequency and output power.Table 1a and b: calculated output impedance at 1W and 9W output power into 4Ohm.Simplified the output impedance is given by three resistances in series: output impedance of the plain tetrode-BJT-MosFet circuit plus the primary coil resistance of the output transformer plus its secondary coil resistance. Secondary coil resistance is 0.2Ohm, reflected primary coil resistance is 0.24Ohm (15Ohm copper resistance divided by the transformer impedance ratio 250Ohm / 4Ohm). This leaves roughly 0.6Ohm reflected impedance for the circuit itself. Multiplied with the transformer impedance ratio the output impedance of the tetrode-BJT-Mosfet circuit is a surprisingly low 40Ohm.Back to top of the pageFrequency Response Nashville SE 20Frequency response was measured at 1W output power (2V@4Ohm) and 9W output power (6V@4Ohm) from 10Hz to 40kHz. Figure 1 shows the results. Below 20Hz the output voltage increases slightly probably due to the limited capacitance of the capacitor in the feedback loop. At 20KHz voltage drop is 0.4db (1W) and 0.75db (9W). At 40KHz voltage drop is 2.5db (1W and 9W) probably due to limits of the output transformer.Unfortunately the Lundahl data sheet on the LL1693 does not provide data on the inherent frequency response. However, for a bulky high power audio transformer for single ended application the frequency response of the LL1693 can be considered excellent.Output Power Nashville SE 20Output power was measured using a 4Ohm load resistor at 1kHz with a trusty old analog oscilloscope. Figures 1 to 3 show three screenshots from the oscilloscope just before clipping (fig. 1), with clippping (fig. 2) and the 1KHz square wave response (fig. 3). The scale on the screen is 5V/cm. Thus the sine wave is just above 30Vpeak to peak before clipping.From figure 1a maximum voltage swing of + / -15V was assumed. Thus the real maximum output power at 1kHz is U2 / R = (15 / square root 2)2 / 4 = 28W. Fig. 1 to 3: Maximum voltage swing of the amplifier output into 4Ohm. Square wave response (1kHz) at +/- 10V with slight overshoots. The red arrow indicates the scale of +/- 15V on the screen of the oscilloscope.