{"id":2417,"date":"2012-12-03T22:15:24","date_gmt":"2012-12-03T21:15:24","guid":{"rendered":"https:\/\/stevepedwards.today\/quad\/wp\/?p=2417"},"modified":"2012-12-03T22:15:24","modified_gmt":"2012-12-03T21:15:24","slug":"marshall-mercury-2060-repair-week-2-3-circuit-theory","status":"publish","type":"post","link":"https:\/\/stevepedwards.today\/ElectronicsStuff\/marshall-mercury-2060-repair-week-2-3-circuit-theory\/","title":{"rendered":"Marshall Mercury 2060 Repair – Week 2-3, Circuit Theory"},"content":{"rendered":"

Marshall Mercury 2060 Repair - Week 2-3, Circuit Theory
\n----------------------------------------------------------------------------------------------------------------------------------
\nDISCLAIMER: The following is NOT to be taken as a definitive procedure for repairing ANY electronic device and the author takes NO responsibility for any damage or injury that results from anyone using this guide. It is intended for educational purposes ONLY.
\nIf you have ANY doubt about making modifications or repairs to your own equipment then seek advice from relevant qualified persons.
\nValve amplifiers use and can store high AC and\/or DC voltages that can KILL. You have been warned!
\n------------------------------------------------------------------------------------------------------------------------------------
\n\"\"
\nAs I'm still awaiting the arrival of the transistors I ordered last Sunday, I thought I'd research and document some other aspects of guitar related electronics.
\nThe key elements to most guitar circuits involve sections where capacitors and\/or inductors are in series or parallel with resistors, as they utilise the concept of resonant frequency.
\nFor details on these concepts, I found the site:
\nhttps:\/\/www.electronics-tutorials.ws<\/a>
\nextremely interesting and very in depth - if a bit maths heavy. It covers the basics on all aspects of this (and most other) audio circuit related topics such as amplifier biasing (the voltage divider \"main chain\" example I gave in the last Post), Class A, AB and C amplifier designs, transformer coupled outputs and loads more.
\nHere's an example of a Class AB \"push-pull\" circuit transistor power amp section, showing the npn and pnp type power transistors Q3+4:
\n\"\"
\nThe section on capacitors in parallel with inductors and resistors (RLC circuits) is important to try to understand, as this is the very circuit you have in your guitar - the pickup (inductor), the tone control (resistor and capacitor in series), and the volume pot (resistor in parallel with the pickup and tone control circuits). In its most basic form:
\n\"\"
\nYou can see that when the volume control pot is full up, the maximum voltage produced by the coil will be dropped across the volume resistor, preventing the signal from going to earth, and so passing it on to the tip of the cable jack. Similarly, the tone pot acts as a path for higher frequencies to preferentially pass to earth or not. The lower the frequency, the closer the signal is to a non-varying signal - i.e DC - which would be blocked completely from passing across the capacitor even if the tone pot is turned down (extreme non-real example), but the higher frequencies would still cross the capacitor preferentially, so not be passed on to the jack tip when the tone pot is turned down. The sound would be made up of only the bassier frequencies. Turning the pot resistor full up prevents the higher frequencies that would normally pass the capacitor to earth, to be sent to the jack tip, and so to the guitar cable and on to the amp.
\nFor reference, I just measured my guitar output with my multimeter on AC mV setting, strummed really hard and got nearly 200mV AC output from the guitar cable, on humbucker pickup, and 107mV from the single coil pickup.
\nThe best article I've read on pickup anatomy, by Helmuth Lemme,<\/span>
\n<\/span><\/em><\/strong><\/span>is here:
\nhttps:\/\/www.buildyourguitar.com\/resources\/lemme\/index.htm<\/a>
\nThis explains exactly how and why various pickups and associated circuits give the particular tone that they do, and is a must read if you are into understanding how the pickup, tone, volume circuits, and guitar lead itself alter the resonant frequencies of your guitar output - REALLY interesting! If you want to change the tone of your guitar, read this to see how putting larger\/smaller value resistance pots (e.g. 10M Ohm down to 100k Ohm) and\/or larger\/smaller value capacitors with your tone pot will alter the resonant frequency of the output signal, and hence, its \"brightness\" (apart from changing the actual pickups themselves of course). Why these values for volume\/tone pots?
\nhttps:\/\/www.langcaster.com\/Pickup-Anthology.html<\/a>
\n\"Pickups vary in resistance and inductance but range between 2.5 and 10 Henrys. Inductance has a reactance (impedance) which rises with frequency. So at 4,000 Hz, a 10 Henry pickup with a resistance of 8 KOhms has a total reactance of (Z =2?fL) 251.3 KOhms for an impedance of 251.4 KOhms. This is calculated using the Pythagoras Law of vectors. For this reason we use 250K or 500K pots. Ever wondered why? At 4KHz, half our signal (-6dB) is lost! Any self-resonance is highly damped as well.\"<\/em><\/span>
\n\"\"
\n<\/span>
\n<\/span><\/span><\/p>\n

Fig. 14. Response of a Fender Stratocaster pickup with 470 pF load capacitance and different Ohmic loads
\n<\/strong><\/span><\/p>\n

\"\"
\n<\/span>
\nFig. 15. Response of a Fender Stratocaster pickup (1972) with 10 MOhms ohmic load and eight different capacitive loads<\/strong><\/span>
\n<\/span>
\nYou can see how smaller value capacitors shift the high frequency resonance up. As we know from my prior Posts, the human ear is most responsive to 1kHz-5kHz, so you may want to try 680 - 1000pF capacitors as in the blue\/black examples if your sound is a bit dull, or too toppy with the tone full up (assuming your amp tone settings are all set to mid - 5\/10 - values). Also try a higher value volume pot so all the higher frequencies are forced to the output, (if your sound is dull) rather than passing to earth (4.5kHz red line in top diagram).
\nJust be aware you will have less fine control over the output if you increase your volume and tone pots, which may be important if you use the volume to start overdriving valve amps subtly.
\nHere are the tone and volume pots of my Ibanez 230 - quite simple:
\n\"\"
\n5 position switch, two pots, a yellow ceramic capacitor at the tone pot and earth. I can't tell from the markings what resistor value they are (500k probably - I can make out a 50 on one), and it's not possible to read the value with a multimeter without de-soldering them from the other components first - obviously, you would be reading the whole parallel circuit resistance instead.
\nThe capacitor is marked 223K 50R, and the explanation of capacitor values can be found here:
\nhttps:\/\/www.ecawa.asn.au\/home\/jfuller\/electronics\/capacitors.htm<\/a>
\n<\/span>
\n\"Smaller capacitors, such as 'greencaps' use a numerical system where the first place represents the first digit, the second place; the second digit and the third place is the number of zeros. (the multiplier) The capacitance so indicated is in picofarads! 104 K = 100,000pF or 0.1uF\"
\n<\/em><\/span>
\n[10 x 10, 000]<\/span>
\n<\/span><\/em>
\nMilli, micro, nano, pico; 10 to the minus 3, 6, 9, 12 resp.
\nIn my case then, it is a 22,000pF or 22nF or 0.022uF, the K representing a 10% tolerance, 50V working limit, which is not represented in the response diagram above.
\nThis Ibanez has 2 single coils and a humbucker, so is innately trebly for a \"rock\" guitar - more like a Strat than a Les Paul.
\nAlso note that the guitar strings\/guitar lead act as a radio aerial and can pick up mains hum from fluorescent lights and computer screens etc. when you are close enough. Make sure the bridge or whammy bar system has an earth wire attached to it to minimise this, as on mine here:
\n\"\"
\nThe other end of this wire is soldered to the casing of one of the volume\/tone pots.
\nThe next thing to consider in your tone chain, is good quality guitar leads. These are quite expensive for screened (3 wire: hot, cold and braided earth) types, for what they are, but you should steer clear of cable with moulded plastic ends as these are probably cheap 2 wire, unscreened types, so noisy.
\nI invested in a 12m length of Klotz XLR microphone cable (3 wire: hot, cold and braided earth) from Ebay, at about 1 pound per metre - 20 quid with postage, as I had some decent Neutriks guitar jacks kicking round for the last 25 years(!), so soldered up my own 3m length cables. I can use this for 3 wire XLR microphone leads then also. For guitar, you just use the hot (e.g. red) wire for the jack tip, and as the guitar cold (e.g. black) is also the earth on a guitar jack, you join the cold and screen braid together and solder to the body (earth) of the Neutriks jack at each end.
\nNote that XLR mic connectors are usually balanced by an inductor when plugged into a mixing desk for example, and have 3 pins, so each wire and braid are soldered separately. This inductor then cancels out any interference that hits both hot and cold wires simultaneously, in phase, so that the two cancel each other out at one end of the cable when they cross the inductor\/transformer (i.e. balancing).
\n\"\"
\nThe length of a guitar cable should not be an issue with audio frequency losses between guitar and amp regarding voltage impedance matching (the combination of cable capacitance and resistance losses), as explained here:
\nhttps:\/\/www.mds975.co.uk\/Content\/Impedance_Matching_ESP_Rod_Elliott.txt<\/a><\/p>\n

\"In the hi-fi audio world, this is not an issue, since this is 50 times the highest frequency we can hear, and few instruments create appreciable harmonics above 20kHz anyway. In theory, we could send an audio signal 12km without having to worry about impedance matching, although at extreme line lengths, matching can reduce high frequency signal losses. To understand the reasons is beyond this article, as it involves transmission line theory - not one of the easiest concepts to grasp.\"\n<\/span><\/em><\/span><\/code><\/pre>\n
This does not mean a longer cable won't be more susceptible to picking up longer radio wavelength interference from radio stations, walkie-talkies etc. that you may hear through your amp, or be a bigger trip hazard when your band is running about on stage!
\nSome Guitar Circuit Elements
\nJust to have a play with some numbers, probably not completely correct, but to get an idea, I'm going to assume the output of my Ibanez is the measured voltage, 200mV, and assume the tone and volume pots are 500k each. From:
\nhttps:\/\/www.electronics-tutorials.ws\/accircuits\/parallel-resonance.html<\/a>
\nNotice that at resonance the parallel circuit produces the same equation as for the series resonance circuit. Therefore, it makes no difference if the inductor or capacitor are connected in parallel or series. Also at resonance the parallel-\u00a0<\/span>LC<\/span>-\u00a0tank circuit acts like an open circuit with the circuit current being determined by the resistor,-\u00a0<\/span>R<\/span>-\u00a0only. So the total impedance of a parallel resonance circuit at resonance becomes just the value of the resistance in the circuit and-\u00a0-\u00a0-\u00a0<\/span>Z-\u00a0=-\u00a0R<\/span>-\u00a0as shown.
\n<\/span><\/span><\/p>\n

\n\"\"
\n<\/span><\/p>\n
As the AC source is the guitar pickup itself, it could be viewed as being in series with the capacitor, and tone knob, with the volume in parallel, so the combined (parallel) resistors values in series. This is why the resonance between the capacitor and an inductor in series or parallel yields the same equation as stated above - the energy stored by both in turn flows from one to the other either way, 180 degrees out of phase.
\nSo, from the above, the impedance of the guitar, as the pickup and capacitor phases cancel out at resonance, the total impedance of the guitar = Z = R, which is just the two resistors in parallel, so 1\/R = 1\/500,000 + 1\/500,000 = 0.00004, so R = 1\/0.00004 = 250k Ohms.
\nJust for arguments sake and fun, let's assume that the inductance of my humbucker pickup is 7 Henries, from the possible values for an Ibanez RG with EMG pickups for example:
\nhttps:\/\/www.ibanez-rg-review.com\/dimarzio-guitar-pickups.php<\/a>
\nUsing the formulae examples below for an RLC circuit, let's see what numbers get spat out for my semi-fictitious guitar and see if they would make sense...<\/p>\n
\"\"\"\"
\n\"\"
\n<\/span><\/p>\n
Resonant Frequency,-\u00a0<\/span>\u00c6\u2019r<\/sub><\/span>
\n<\/span><\/span><\/p>\n
\"\"
\n<\/span><\/p>\n
Inductive Reactance at Resonance,-\u00a0<\/span>XL<\/sub><\/span>
\n<\/span><\/span><\/p>\n
\"\"
\n<\/span><\/p>\n
Quality factor,-\u00a0<\/span>Q<\/span>
\n<\/span><\/span><\/p>\n
\"\"
\n<\/span><\/p>\n
Bandwidth,-\u00a0<\/span>BW<\/span>
\n<\/span><\/span><\/p>\n
\"\"
\n<\/span><\/p>\n
The upper and lower -3dB frequency points,-\u00a0<\/span>\u00c6\u2019H<\/sub><\/span>-\u00a0and-\u00a0<\/span>\u00c6\u2019L<\/sub><\/span>
\n<\/span><\/span><\/p>\n
\"\"
\n<\/span><\/p>\n
So, for my partly fictitious Ibanez:
\nPickup inductance = 7H
\nCapacitor = 22, 000pF = 22nF = 0.000 000 022F
\nResistance = 250k Ohm
\nResonant Frequency<\/span> = 1 \/ 2 x Pi (LC)^0.5 = 1\/ 2 x 3.142 x (7 x 0.000 000 022)^0.5 = 1 \/ 0.000 000 2466 = 405.6Hz
\nInductive Reactance at Resonance <\/span>= 2 Pi f L = 2 x 3.142 x 405.6 x 7 = 17,839 Ohms
\nQuality factor Q <\/span>= R \/ 2 Pi f L = 250,000\/17,839 = 14
\nBandwidth<\/span> = f\/Q = 405.6Hz \/14 = 29Hz
\nThe upper and lower -3dB frequency points<\/span> = f LOW = f - -\u00bd BW = 405.6 - -\u00bd29 = 377 - 14.5 = 362.5Hz
\n= f HI = f + -\u00bd BW = 377 + 29 = 398 Hz
\nPassband (-3dB drop each side) = 362.5 to 398 Hz
\nWell, that could be right? My capacitor is really large by Lemme's highest value (2,200pF - by 1 order of magnitude - 10 x), as mine is 22 000pF, and judging by the way the larger value capacitors give lower resonant frequency values as they get bigger, I guess a 10x lower peak may be around 200-400Hz?
\n\"\"
\nThe passband (where the -3dB drop is either side of the resonant frequency) may be about right also, assuming the curve shape doesn't change much?
\nAh well, the maths did me good...Anyway...moving on...
\nThe next stage - the guitar plugging into the HIGH input - below, means that the guitar output impedance (250K) is being put in parallel with R1 and R2 (in series = 220k + 100k = 320k). Note this is a relatively low input impedance by today's standards (470k min):
\nhttps:\/\/www.soundonsound.com\/sos\/jan03\/articles\/impedanceworkshop.asp<\/a>
\nEverything you need to know about Impedance and much more in that article.
\n\"If the input has too low an impedance, the most noticeable effect will be a loss of high end \u00e2\u20ac\u201d in fact, even using guitar cables with too high a capacitance can audibly reduce high frequencies (see 'Impedance & Frequency Response' box for details of this effect). The sustain is also affected, giving a 'dead' sound.\"<\/span>
\nPassive Band Pass Filter section:
\n\"\"
\nThe input stage has a band pass filter comprised of C1, R1, R2 and C2 (these are in parallel with the guitar pickup now). This only allows a certain frequency range to pass to the input of the first gain stage.
\nhttps:\/\/www.electronics-tutorials.ws\/filter\/filter_4.html<\/a>
\nC3 is a DC blocking capacitor, so the bias voltage for T1 is isolated from the guitar input section. C3 then only allows the guitar and bandpass section AC signal to pass to the input of T1.
\n\"\"
\n\"\"
\nThe upper and lower cut-off frequency points for a band pass filter can be found using the same formula as that for both the low and high pass filters, For example.
\n<\/span><\/p>\n
\"\"
\n<\/span><\/p>\n
Let's see if the numbers here for R1, C1 and R2, C2 make any more sense than the last lot!
\nFor R1 (220k), C1 (2n2):
\nFreq 1 = -\u00bd Pi RC = -\u00bd Pi (220 000 x 0.000 000 002 2) = 1\/0.00304 = 328.8Hz
\nFor R2 (100k), C2 (2n2):
\nFreq 2 = -\u00bd Pi RC = -\u00bd Pi (100 000 x 0.000 000 002 2) = 1\/0.001 38 = 723.4Hz
\nSo input gain stage T1 band pass is 723-328 = 395Hz wide, centred on 525 Hz?
\nIt will be interesting to check all this with my DSO Nano oscilloscope when it arrives, and as the new transistors are put in also, to see if any of this correlates at all!?
\nGain stage 1 input:
\n\"\"
\nR3, 4, and 5 are part of the voltage divider chain that holds the bias voltage constant between the collector (26V) and the emitter (8V) so the only variation at the base is the AC signal generated by the guitar, which is amplified and output between C6 and VR1.
\nC4\/C11\/C23 Function:
\nhttps:\/\/www.talkingelectronics.com\/projects\/TheTransistorAmplifier\/TheTransistorAmplifier-P2.html#NegativeFeedback<\/a><\/p>\n
\n\n\n\n<\/colgroup>\n\n\n
<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
NEGATIVE FEEDBACK
\n\"\"<\/strong><\/span>
\n\"The circuit shows a capacitor between the base and collector. It provides-\u00a0NEGATIVE FEEDBACK.<\/strong>
\nIf we remove the capacitor, when the base \"moves down,\"-\u00a0the collector \"moves up.\"-\u00a0 In other words the signal is inverted.
\n<\/span>
\nWhen the capacitor is fitted, we have to start with the collector because it has more \"power\" and it is the lead that is driving the action of the capacitor and then go to the base.
\nWhen the collector voltage \"moves down\" the right plate of the capacitor moves down and it charges and tries to pull the left plate down too.<\/span>
\nThis is the opposite effect to the signal moving through the transistor.
\n<\/span>
\nThis means the capacitor is working against the action of the transistor.
\n<\/span>
\nThe capacitor will have more effect on high frequency signals while the low-frequency signals will be affected less.
\nBecause the capacitor is working against the natural flow of signal through the circuit, it is called NEGATIVE ACTION or-\u00a0NEGATIVE FEEDBACK<\/strong>.\"<\/span>
\n----------------------------------------------------------
\nThe performance characteristics of the BC184 transistor can be found here:
\nhttps:\/\/stevepedwards.today\/images.ihscontent.net\/vipimages\/VipMasterIC\/IC\/ONSM\/ONSMS04085\/ONSMS04085-1.pdf<\/a>
\nSome general soldering advice - if you don't solder the legs of transistors\/diodes quickly, and use pliers as a heat sink, the heat can destroy the NPN or PNP junction, usually resulting in a shorted component:
\n\"SOLDERING PRECAUTIONS
\n<\/strong><\/span>
\nThe melting temperature of solder is higher than the rated
\n<\/span>
\ntemperature of the device. When the entire device is heated
\n<\/span>
\nto a high temperature, failure to complete soldering within a
\n<\/span>
\nshort time could result in device failure. Therefore, the
\n<\/span>
\nfollowing items should always be observed in order to minimize
\n<\/span>
\nthe thermal stress to which the devices are subjected.
\n<\/span>
\n??<\/span>Always preheat the device.
\n<\/span><\/span>
\n??<\/span>The delta temperature between the preheat and soldering
\n<\/span><\/span>
\nshould be 100<\/span>?<\/span>C or less.*
\n<\/span><\/span>
\n??<\/span>When preheating and soldering, the temperature of the
\n<\/span><\/span>
\nleads and the case must not exceed the maximum
\n<\/span>
\ntemperature ratings as shown on the data sheet. When
\n<\/span>
\nusing infrared heating with the reflow soldering method,
\n<\/span>
\nthe difference should be a maximum of 10<\/span>?<\/span>C.
\n<\/span><\/span>
\n??<\/span>The soldering temperature and time should not exceed
\n<\/span><\/span>
\n260<\/span>?<\/span>C for more than 10 seconds.
\n<\/span><\/span>
\n??<\/span>When shifting from preheating to soldering, the maximum
\n<\/span><\/span>
\ntemperature gradient shall be 5<\/span>?<\/span>C or less.
\n<\/span><\/span>
\n??<\/span>After soldering has been completed, the device should be
\n<\/span><\/span>
\nallowed to cool naturally for at least three minutes.
\n<\/span>
\nGradual cooling should be used since the use of forced
\n<\/span>
\ncooling will increase the temperature gradient and will
\n<\/span>
\nresult in latent failure due to mechanical stress.
\n<\/span>
\n??<\/span>Mechanical stress or shock should not be applied during
\n<\/span><\/span>
\ncooling.
\n<\/span>
\n* Soldering a device without preheating can cause excessive
\n<\/span>
\nthermal shock and stress which can result in damage to the
\n<\/span>
\ndevice.\"
\n<\/span>
\nSo, it seems it would be good to hold the legs of a tranny\/diode with pliers once it is in place on the circuit board, then preheat the lot (local area) using a heat gun (dont melt the plastic TO-92 casing of course!), or at least a hair drier, to raise the temp of the transistor so the heat from the solder tip transfers to the body of the transistor as slowly as possible. The spec sheet for a BC184 states an operating and storage junction temp of -55 to +150 C.<\/span>
\n----------------------------------------------------------------------------------------------------------------------------
\nGain stage 1 Output:
\nVR1 = Volume, VR2 = Tone
\n\"\"
\nThis is a bit more complex and I can't decipher it so well.
\nThe simple bit first - when VR1 arrow is at the bottom, the T1 output can go direct to earth via C6 and C7, with no resistance = no signal sent on to gain stage 2 = no volume. Fine.
\nWhen VR1 arrow is at the top, C7 is short circuited against itself, to the top of VR1 anyway, effectively out of the circuit, leaving only VR1, (C8+VR2), and R7, all in parallel with each other. When VR2 arrow is at the bottom and VR1 at the top, VR1 and R7 are in parallel so become a 50k resistance in parallel with C8+VR2. This is full volume with minimum tone. C9 seems to part of the bandpass?
\nThis would still allow AC bass tone signals to pass via the C10 DC block input to gain stage 2 when the tone control is at minimum (VR2 arrow at bottom).
\nAs VR2 is the tone control (I checked the PCB), this section is about frequency filtering I'd guess, with the varying VR2 value creating different resonance frequencies with the combination of C9\/R7, C8\/VR2 creating a changeable high and low bandpass filter depending on the position of VR2.
\nAs we've seen, C7 is not part of any band passing, as the closer VR1 comes to the bottom, the less volume output there is anyway as the signal grounds out; and the closer to the top, the more C7 is shorted out across itself. This leaves only C8\/VR2 and R7\/C9 to act as variable filters depending on the position of VR2.
\nFrom:<\/p>\n
https:\/\/www.electronics-tutorials.ws\/rc\/rc_3.html
\n<\/span><\/h3>\n
Sine Wave Input Signal
\n<\/span><\/h3>\n
If we now change the input RC waveform of these-\u00a0<\/span>RC<\/span>-\u00a0circuits to that of a sinusoidal-\u00a0Sine Wave<\/strong>-\u00a0voltage signal the resultant output RC waveform will remain unchanged and only its amplitude will be affected. By changing the positions of the Resistor,-\u00a0<\/span>R<\/span>-\u00a0or the Capacitor,-\u00a0<\/span>C<\/span>-\u00a0a simple first order-\u00a0<\/span>Low Pass<\/strong><\/span>-\u00a0or a-\u00a0<\/span>High Pass<\/strong><\/span>-\u00a0filters can be made with the frequency response of these two circuits dependant upon the input frequency value.
\n<\/span><\/span>
\nLow-frequency signals are passed from the input to the output with little or no attenuation, while high-frequency signals are attenuated significantly to almost zero. The opposite is also true for a High Pass filter circuit. Normally, the point at which the response has fallen 3dB (cut-off frequency,-\u00a0<\/span>\u00c6\u2019c<\/span>) is used to define the filters bandwidth and a loss of 3dB corresponds to a reduction in output voltage to-\u00a0<\/span>70.7<\/span>-\u00a0percent of the original value.
\n<\/span><\/span><\/p>\n
Cut-off Frequency
\n<\/span><\/h3>\n
\"\"
\n<\/span><\/p>\n
where-\u00a0<\/span>RC<\/span>-\u00a0is the time constant of the circuit previously defined and can be replaced by tau,-\u00a0<\/span>T<\/span>. This is another example of how the-\u00a0Time Domain<\/em>-\u00a0and the-\u00a0Frequency Domain<\/em>-\u00a0concepts are related.
\n<\/span><\/span>
\nIf we treat the two RC parts separately as before we can try the maths and see what nonsense I get this time!
\n\"\"
\nWhen VR2 is minimum, only R7 and C9 are active it seems?
\nFreq 1: C9 = 22n, R7 = 100k
\nf = -\u00bd Pi RC = -\u00bd 3.142 x 0.000 000 022 x 100 000 = 723.4Hz
\nFreq 2: C8 = 220p, VR2 = 100k
\nf = -\u00bd Pi RC = -\u00bd 3.142 x 0.000 000 000 22 x 100 000 = 7,234 Hz
\nThis seems a possible pass band overall for a guitar amp - but this is not the Bandwidth remember - measured at -3dB drops from 0dB at maximum and minimum frequencies, so seems possible at this point. You may not hear these low frequencies because they may be very low volume, and they have to get through T2 and the Valve gain sections yet.
\nA typical guitar amp bandwidth is around 5-10kHz according to here:
\nhttps:\/\/www.aikenamps.com\/ResistorNoise.htm<\/a>
\nGain stage T2 input:
\n\"\"
\nThe way to think of DC decoupling capacitors is they allow the AC ripple to be superimposed onto the next fixed DC voltage biased section of the amp, causing those DC bias voltages to ripple more, then less, positive and negative at an amount dependent on the voltage gain of the previous stage output. This way, just the AC signal we are interested in can traverse the circuit from end to end.
\nSo, what do we know about this circuit from the data given, and all the research I've done, and can I find out more from the values given?
\nFor straight testing for faults on the components most likely to fail - the transistors and the valve probably, this diagram gives all the key voltages needed, which is why I have ordered some replacement BC184 transistors.
\n\"\"
\nNote: T3 typo - T3 Emitter is at ground = 0V
\nThe collector voltage for all three should be 26V
\nThe emitter voltage for all three should be 8V (T3 typo - must be wrong as it's at 0V)
\nThe anode of the valve should be 250V
\nThe cathode of the valve should be 6V
\nFrom this info some figures can be generated for the quiescent bias currents etc - so what else do I know?
\nThe-\u00a0<\/span>common-emitter<\/em><\/span>-\u00a0current gain<\/em>-\u00a0is represented by ?F<\/sub>-\u00a0or hFE<\/sub>; it is approximately the ratio of the DC collector current to the DC base current in forward-active region.
\n<\/span>
\nThe current gain (Beta) of a common emitter circuit is Collector I \/ Emitter I; or the ratio of Collector R\/ Emitter R = R5\/R8 for T1 and R10\/R11 for T2 = 200k\/10k = 22<\/span>
\nThe collector voltage for all three transistors should be 26V ( the BC184 data sheet states a max of 30V for the CE voltage, max 45V for CB, and max 6V for BE
\nT1\/T2 Emitter voltage = 8V
\nEmitter current in R6\/R11 = 10k so 8V\/10 000 Ohms = 80uA
\nEmitter I = Base I + Collector I
\nBase bias must be at a voltage value approximately half that of the Collector - Emitter voltage so that it can amplify both positive and negative sides of the AC signal cleanly (no clipping), so:
\nThe CE Voltage = 26 - 8 = 18V, so Base voltage should be around 9V?
\nThe base bias voltage is set depending on the circuit requirements, and varies for what base input current ranges you want to amplify. The base saturation voltage changes depending on the input base current and the CE voltage, so you need to look at the spec. sheet
\nFor this exercise, I will take the BE voltage from the above base voltage:
\nBase Emitter voltage = 9 - 8 = 1V
\nR4\/R9 current = 9V\/6M8 Ohm = 1.3uA
\nTherefore, same current flows via R6\/R10 and R3\/R8, plus the base current amount that keeps the transistors turned on. The base current depends on the base-emitter resistance of the transistor.
\n\"\"
\nFrom above, the Emitter current is 22 times bigger than the Collector current, so Collector I = 80uA\/22 = 3.6uA
\nTherefore, the current through R10 = collector current + current through 6M8 = 1.3uA + 3.6uA = 4.9uA
\nSo, the voltage across R10 = V = IR = 4.9uA x 220 000 Ohm = 10.9V
\nThis means the voltage at Point A = Collector V + R10 V = 26V + 10.9V = 36.9V
\n\"\"
\nThe difference between points X and A then is 275V - 37V = 238V
\nThe key to visualising this next part is to see the valve itself and the speaker transformer as the primary voltage divider for the T1, 2, and 3 stages. The reason there is such large voltage drops over R16 and R17 is because the bulk of the combined currents that flow through the valve come via the speaker transformer, and all of the current that feeds the combined transistor voltage divider chains for T1, 2 and 3 comes through R16 and R17, so they drop the bulk of the 275V voltage available at point X, as the total currents for voltage divider sections R5 (T1 section), R10 (T2 section) and R18 (T3 section) pass through R16.
\n\"\"
\nThe 238V difference goes across R16 and R17 or 238V \/ 28K Ohms - about 8.5mA depending how much Valve gate current is also taken at R14.
\nSo, if the V drop over R16 = 8.5mA x 27k = 230V
\nThe V drop over R17 = 8.5mA x 1k = 8.5V
\nVoltage at bottom of R17 = 275 - 8.5 = 267.5V
\nThe V drop across the speaker transformer primary = 275 - 250 = 25V
\nWithout more info on the speaker transformer, its not possible to say what V or I is going through it or the speaker windings at this point without checking specs for the valve and the transformer.
\n\"\"
\nThe EL84 specs are:<\/p>\n
\n\n\n\n<\/colgroup>\n\n\n\n\n\n\n\n\n\n
Heater voltage<\/span><\/td>\n6.3V<\/span><\/td>\n<\/tr>\n
Heater current<\/span><\/td>\n760mA<\/span><\/td>\n<\/tr>\n
Max anode voltage (Ia=0)<\/span><\/td>\n550V<\/span><\/td>\n<\/tr>\n
Max anode voltage<\/span><\/td>\n300V<\/span><\/td>\n<\/tr>\n
Max anode dissipation<\/span><\/td>\n12W<\/span><\/td>\n<\/tr>\n
Max screen voltage (Ig2=0)<\/span><\/td>\n550V<\/span><\/td>\n<\/tr>\n
Max screen voltage<\/span><\/td>\n300V<\/span><\/td>\n<\/tr>\n
Max screen dissipation<\/span><\/td>\n2W<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
\n\n\n\n<\/colgroup>\n\n\n\n\n\n\n\n\n\n\n\n
Pin<\/span><\/td>\nFunction<\/span><\/td>\n<\/tr>\n
1<\/span><\/td>\ni\/c<\/span><\/td>\n<\/tr>\n
2<\/span><\/td>\nControl grid<\/span><\/td>\n<\/tr>\n
3<\/span><\/td>\nCathode + suppressor grid<\/span><\/td>\n<\/tr>\n
4<\/span><\/td>\nHeater<\/span><\/td>\n<\/tr>\n
5<\/span><\/td>\nHeater<\/span><\/td>\n<\/tr>\n
6<\/span><\/td>\ni\/c<\/span><\/td>\n<\/tr>\n
7<\/span><\/td>\nAnode<\/span><\/td>\n<\/tr>\n
8<\/span><\/td>\ni\/c<\/span><\/td>\n<\/tr>\n
9<\/span><\/td>\nScreen grid<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
The equation for power P = VI so taking the max values for the anode drawing current:
\n12W\/300V = .04 Amps or 40mA<\/p>\n
This max current would also pass through the speaker primary windings.<\/p>\n
This correlates well with the JJ EL84 datasheet for typical usage also - Anode current - Ia = 48mA:<\/p>\n
Class A1 amplifier:
\n<\/span>
\nUa = 250 V
\n<\/span>
\nUg2 = 250 V
\n<\/span>
\nRk = 135 <\/span>?
\n<\/span><\/span>
\nIa = 48 mA
\n<\/strong><\/span>
\nIg2 = 5,5 mA
\n<\/span>
\nRa = 5,2 k<\/span>?
\n<\/span><\/span>
\nUg1eff (50mW) = 0,3 V
\n<\/span>
\nUg1eff(N) = 4,3 V
\n<\/span>
\nN (10%)1) = 5,7 W
\n<\/span>
\nN 2) = 6 W
\n<\/span>
\n1) Ug1 fest fixed grid bias
\n<\/span><\/p>\n
2) Ig1 +0,3 <\/span>?<\/span>A<\/span><\/span>
\n<\/span><\/p>\n
The speaker transformer has a label with DE Ltd, C442, but Google found nothing for me.
\nThe only way now is desolder and measure the primary and secondary resistances with a Voltmeter. Another day.
\nMoving on...
\nI just found some electronic circuit simulator software today, so will have a go at simulating this circuit and see if these calculated values pan out...
\nValve pinouts:
\n\"\"
\nThe reason for Marshall's exaggerated 25W claim for this amp?:
\nValve specifications:
\nhttps:\/\/stevepedwards.today\/lenardaudio.com\/education\/14_valve_amps_2.html<\/a>
\n6BQ5-\u00a0EL84 (and 6GW8 EL86) is a small noval base power pentode. Many cheap stereo systems used one 6BQ5 per channel (Class A) approx 4 Watts. The majority of domestic high fidelity stereo systems used 2 x 6BQ5s in Class AB push pull, and are capable of 16 Watts. Many stereo systems that used these small power valves were marketed as 30 Watts total
\n<\/span><\/span>
\nIt looks like the transistors got lost in the post so are being re-sent...ho-hum
\nHopefully the next Post will be the last on this epic story.<\/p>\n","protected":false},"excerpt":{"rendered":"
Marshall Mercury 2060 Repair - Week 2-3, Circuit Theory ---------------------------------------------------------------------------------------------------------------------------------- DISCLAIMER: The following is NOT to be taken as a definitive procedure for repairing ANY electronic device and the author takes NO responsibility for any damage or injury that results from anyone using this guide. It is intended for educational purposes ONLY. If you have ...  Marshall Mercury 2060 Repair – Week 2-3, Circuit Theory<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-2417","post","type-post","status-publish","format-standard","hentry","category-tech-studies"],"_links":{"self":[{"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/posts\/2417","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/comments?post=2417"}],"version-history":[{"count":0,"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/posts\/2417\/revisions"}],"wp:attachment":[{"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/media?parent=2417"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/categories?post=2417"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/stevepedwards.today\/ElectronicsStuff\/wp-json\/wp\/v2\/tags?post=2417"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}