How To Calculate Input Impedance - Page 1
Forum section: electronics / general

author: Gaussian
author_rank: Contributor
timestamp: December 29, 2018, 12:29:06 am
content: Can someone help me calculate the input impedance looking into the power stage of this amplifier circuit from CC2and CC3?Is the following correct looking in from CC2?Zin= R8|| (B4re4' + B5(re5' + RL)) || R9Where B = beta for the corresponding transistor and reis the emitter junction resistance (25mV/IE).Do I need to include the internal resistances of the diodes too? Not sure how to do this ...Thanks in advance for your help.

author: Gaussian
author_rank: Contributor
timestamp: December 29, 2018, 01:57:42 am
content: I *think* the AC model for the circuit (looking into the power stage from CC3is as shown in the attached schematic.Are the internal resistances of the diodes calculated the same way as the transistors (25mV/I) where I is the current through the diodes? If so, the internal resistance of each diode is ~ 29 ohms assuming a Vccof 20V (the current through the biasing circuit should be 860 uA).This puts R5|| rd1+ rd2+ rd3+ rd4+ R6|| R7Where I'm guessing R7= B3(re3+ B4(re4+ RL|| B6(re6+ B5re5)))Thus:Zin= R5|| rd1+ rd2+ rd3+ rd4+ R6|| B3(re3+ B4(re4+ RL|| B6(re6+ B5re5)))But I'm totally guessing ...Surprisingly, I can't find any examples of this calculation online.

author: boB
author_rank: Frequent Contributor
timestamp: December 29, 2018, 02:24:37 am
content: What is the positive rail voltage ?since the output stages are darlingtons and very high Hfe, I would disregard those at first.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 29, 2018, 02:31:33 am
content: Note that the speaker is getting ~half supply voltage; I assume a coupling capacitor was meant there.Your analysis is close, but omits:- Diode dynamic resistance (give or take, Rd(total) = 4 * Vth / If)- Impedance of the PNP outputs- The NPN and PNP outputs in parallel driving the speaker- You also need to account for the signal coupling capacitors as well, if this is a small signal AC analysis.You'll ultimately be more interested in the impedance seen by the driving stage, not just after Cc2, but assuming the capacitors aren't connected and looking into the given node, this should be correct.It's rather a lot of bother to do by hand, so a rough estimate, or a quick SPICE analysis, is adequate for practical purposes.Tim

author: Gaussian
author_rank: Contributor
timestamp: December 29, 2018, 02:40:39 am
content: Sorry, I forgot to include that on my schematic.Vcc= 20V.Using:Zin= R5|| rd1+ rd2+ rd3+ rd4+ R6|| B3(re3+ B4(re4+ RL|| B6(re6+ B5re5)))I calculated a Zin= 3.09 KOhmThis gives me an RACof 2.43 KOhmAnd a coupling capacitance = ((2*pi*(20Hz)(2.43K))^-1 = 3.27uFI have this circuit on a breadboard, I'll see how it sounds (and see what it looks like).

author: Gaussian
author_rank: Contributor
timestamp: December 29, 2018, 02:55:45 am
content: Thanks Tim. Yes, I neglected to draw in the coupling capacitor on the output - sorry about that.I'm attaching a corrected schematic.To make sure I understand:QuoteAre you saying the equation I provided is correct (Zin=3.09KOhm) but I should run it through Spice to verify as I neglected to take into account the transistor output impedances? I tried to take into account the diode resistances in the model. Thanks, just trying to understand.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 29, 2018, 04:57:15 am
content: It's still not quite right, Q7 is shown in the wrong place -- it's not in parallel with the C2 node, but with R9, so, in series with Rd's.But, again, you shouldn't need to compute that impedance, because nothing connects to just that node.The impedance that Q2 is driving, is both capacitors in parallel, and we expect/hope that the Rd's actually cancel out completely!BTW, this circuit has some other shortcomings, like the double-follower architecture losing a few volts in clipping (near the supply or ground rail), and also being limited by the saturation (clipping) of the first stage, which also produces a fair amount of distortion even within its "linear" range. The middle emitter follower stage doesn't seem to be helping much (indeed, R7 is larger than RC1 so the pull-down current available to the final follower is less!). These are some reasons why the conventional amplifier circuit is the way it is -- that is, with a current sink/source, and a "volt amp stage" after the input stage, which together is able to deliver a wide voltage swing for the outputs.Tim

author: soldar
author_rank: Super Contributor
timestamp: December 29, 2018, 11:04:33 pm
content: Quote from: T3sl4co1l on December 29, 2018, 04:57:15 amI am confused by this. Can you explain?Quote from: T3sl4co1l on December 29, 2018, 04:57:15 amI agree that it can be improved but I do not want to hijack the thread if it is just an exercise for calculating the input impedance. If the OP would like to discuss ways of improving it then I will be happy to give my opinion. I dealt with this kind of design when I was starting, shortly after the electron was invented.

author: spec
author_rank: Frequent Contributor
timestamp: January 01, 2019, 07:25:51 pm
content: Quote from: Gaussian on December 29, 2018, 12:29:06 amHi Gaussian,Your circuit is a class B (no quiescent current) or class AB audio power amplifier, depending on the relationship between the combined VF of the four diodes and combined VBE of the four output transistors, which is indeterminate, especially as the diodes are 1N4148 types. The four diodes have no practical effect on the input Z (in some similar amplifier designs there is a high value electrolytic capacitor across the four diodes, which is effectively the case in this amplifier).Assuming that the capacitors have a capacitance value so large that their impedances can be taken as zero Ohms, for practical purposes, then:The input Z for the positive excursion of the output wave form will be RL(Q4hfe * Q5hfe) where RL is the loudspeaker ZThe input Z for the negative excursion of the output waveform will be RL(Q7hfe * Q8hfe) where RL is the speaker ZSo, assuming that Q4hfe and Q7hfe are 30 and Q5hfe and Q8hfe are 20, the input Z for each half of the output waveform would be 8(30 * 20) = 4k8R.If the amplifier were operating in class B there would be a high, but indeterminate input impedance, at the 0V output point.If the amplifier were operating in class A/B the input impedance would be roughly the same as for the positive and negative excursions of the output waveform at the 0V output point.But BJT hfe varies greatly between samples of the same transistor and, for a given transistor, hfe varies greatly with:vceicjunction temperaturefrequencySo the bottom line is that the input impedance of the output quad transistors will vary greatly.Not only that, but the loudspeaker Z will vary greatly with frequency to further add to the variation of of the input Z of the output stage.In practice the input Z of the output stage would be of little use. Instead you would make sure that the stage driving the output stage had sufficient voltage and current drive to generate the maximum worst case current that the load could demand (afraid to say that this is not the case for the amplifier schematic posted). And in high-end audio power amplifiers you would ensure that the output stage could supply 5x to 10x the calculated worst case current that the load could demand.NotesR8 and R9 are both effectively in parallel with the input impedance of the quad output transistors, but the effect of R8 and R9 have been omitted for clarity.The effect of transistor re has also been omitted for clarity but, in view of the other large variations in parameters mentioned above, the effect of re will not be significant, especially for power transistors (Q4re/Q7re = 0R, Q5re/Q8re = 330mR average).HFE is the DC current gain, IC/IB, of a transistor while hfe is the AC signal gain ic/ib of a transistor.https://www.onsemi.com/pub/Collateral/TIP41A-D.PDF

author: Zero999
author_rank: Super Contributor
timestamp: January 02, 2019, 10:01:08 am
content: Q1 and its associated components determine the input impedance. The buffer and output stage make very little difference.See the tutorial linked below:https://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.htmlAs mentioned above, it's indeterminate whether it's class B or AB. If the circuit does become biased in class AB, then there's a high risk of thermal runaway, resulting in destruction of the output stage. The base-emitter voltage of BJTs has a negative temperature coefficient, which means that if all the transistors in the output stage conduct simultaneously, they'll heat up a bit, resulting in their base-emitter voltages dropping, causing increased base current, more collector current and increased power dissipation. This is a classic positive feedback loop which will cause the output transistors to overheat.To overcome thermal runaway, there should be emitter resistors to limit the maximum current and D1 to D4 should be on the same heat-sink as the power transistors, so their forward voltage also goes down, as they're heated up by the power transistors.