Please explain me some of this analog wizardry found in a tape preamp - Page 1
Forum section: electronics / beginners

author: fabiodl
author_rank: Frequent Contributor
timestamp: March 21, 2025, 08:01:36 am
content: I am reverse engineering a tape player board. I attach the schematic. I traced it from pictures of the board. I triple checked it, but there may be some errors. Especially in the resistor values,which are difficult to judge color from the images.Anyway, I have several questions.1) The tape head comes through a coax. Is there any standard? (For instance 50/75ohm impedance).2) Why are there several "rails" built with resistors and caps? I am referring to R8+C22, R27+C1. Is the main reason providing a cleaner power supply? (The power supply is either an external AC or batteries). Is this a common approach?3) What is the role of R3?4) What is the role of the filter R6, C5,R7. Why is it placed between the output and the emitter of Q1?5) In general, is this configuration (around Q1 and Q2) somewhat standard? Does it have a name (besides Darlington if we consider only the two transistors)?6)The rightmost part is a crude "VU meter". What is a first order approximation of what the LED is displaying? The average of the signal positive part over 66ms? (5*tau, with tau=(R28+17)*C15?7) Do you see any errors on the "VU meter" part? Simulating it shows that the catode of D1 swings between -0.63 and 1.3V, but Q7 never turns on. If I disconnect the base of Q7, C16 never charges.

author: Haenk
author_rank: Super Contributor
timestamp: March 21, 2025, 12:27:44 pm
content: Why don't you state manufacturer/device? Most devices have their schematics on the interweb, or someone is able to supply one...

author: fabiodl
author_rank: Frequent Contributor
timestamp: March 21, 2025, 02:39:30 pm
content: Quote from: Haenk on March 21, 2025, 12:27:44 pmTiger 2-XL (a 1990s toy)

author: iMo
author_rank: Super Contributor
timestamp: March 22, 2025, 08:14:54 am
content: Quote from: fabiodl on March 21, 2025, 08:01:36 am1) no2) yes, yes3) to set the operation point of Q1 (and Q2 as well)4) to improve freq characteristics somehow and limit the output signal as well5) yes, no special name but a rather standard amplifier wiring6) and 7) imho it will show modulation peaks only, it is just a crude peak meter, its setting depends on many factors.

author: magic
author_rank: Super Contributor
timestamp: March 22, 2025, 08:56:56 am
content: 3) Set the operating points and also supply Q1 base current - it has to be coming from somewhere. R2 pulls Q2 up until R3 voltage rises enough that Q1 turns on and sinks all R2 current minus Q2 base current.5) It's a common old school audio topology. It could be classified as a two stage current feedback operational amplifier. Q1 base is the noninverting input and emitter is the inverting input, actively driven to follow the noninverting input (with -0.6V offset). The difference between actual emitter voltage and open circuit output voltage of the feedback network causes the feedback network to source/sink additional current into the emitter, which is passed to the collector and produces some AC voltage, which in turn drives the inverting second stage (Q2) with final output taken from its collector.4) This is the feedback network of the opamp. 1548x gain at DC, falls 6dB/octave from 1Hz to 30Hz, flattening out at 48x. Not entirely sure about the purpose of that.

author: PGPG
author_rank: Frequent Contributor
timestamp: March 22, 2025, 09:48:51 am
content: 5. It has nothing to do with Darlington.Powering Q1 base from voltage at R5 gives a very strong negative DC feedback perfectly stabilizing DC working point.AC shorting R5 with C4 blocks this feedback for AC what allows to get high AC gain.Long time ago, being young (12..14) having such circuit with pnp germanium transistors, having no computer with Spice or even a calculator (such times) I discovered my method of calculating for this circuit the working point.Starting from assumed voltage Q1c you can calculate voltages in circle in one of two directions One is: Q1c->Q2e->Q1b (from Q1c you have Q1Ic and R3I=Q1Ic/beta) ->Q1e -> new Q1Ic ->Q1c. Second is calculating this in opposite way. Calculating in one direction gives results gradually converging to the true operating point, while counting in the reverse direction gives divergent results. I don't remember which direction is this one good - you can check it.4. With signal from tape head is like with signal from dynamic microphone. From electret microphone you have signal proportional to membrane deflection so for the same vibration amplitude at 100Hz and 1kHz you get the same signal amplitude - you need not to correct anything. But with dynamic microphone for the same vibration amplitude you will get 10 times smaller signal for 100Hz than for 1kHz as signal depends on membrane speed and not deflection. To get signal proportional to amplitude you need to make correction (the lower frequency the higher gain is needed).The same happens in tape head. Signal you get is proportional to speed magnetic field changes and not to its amplitude value (for stable field you will get signal 0).

author: magic
author_rank: Super Contributor
timestamp: March 22, 2025, 10:15:24 am
content: The operating point should be quick to find.C5 conducts nothing and R7 almost nothing, so ignore.R3 voltage drop probably is low, less than 1V.Hence Q2 base and Q1 collector are maybe 1~2V above Q1 base.Say, there is 2V between Q1 emitter and Q1 collector.Subtract 2V from C1 voltage, divide by 10.1kΩ to get Q1 bias.Then you can also know Q1 base current and correct the initial guess of R3 voltage.Then you can know Q2 current and R27+R8 voltage drop, which gives exact C1 voltage.

author: fabiodl
author_rank: Frequent Contributor
timestamp: March 22, 2025, 12:50:17 pm
content: Thank you for all your comments, it's much clearer now.I also discovered this bookhttps://www.amazon.com/Transistor-circuits-recording-photofact-publication/dp/B0007EG594which describes a configuration very close to the one around Q1 and Q2.I have one open problem though. I tried simulating the circuit, and even with a 20mV amplitute head signal, the peak detector does not seem to work. Specifically, Q7 does not turn on. I attached its collector to VCC (6V) by a 2K resistor and here are the simulation results