Marshall Mercury 2060 Repair – Week 2-3, Circuit Theory

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Marshall Mercury 2060 Repair – Week 2-3, Circuit Theory
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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 ANY doubt about making modifications or repairs to your own equipment then seek advice from relevant qualified persons.
Valve amplifiers use and can store high AC and/or DC voltages that can KILL. You have been warned!
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As 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.
The 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.
For details on these concepts, I found the site:
http://www.electronics-tutorials.ws
extremely 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.
Here’s an example of a Class AB “push-pull” circuit transistor power amp section, showing the npn and pnp type power transistors Q3+4:

The 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:

You 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.
For 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.
The best article I’ve read on pickup anatomy, by Helmuth Lemme,
is here:
http://www.buildyourguitar.com/resources/lemme/index.htm
This 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?
http://www.langcaster.com/Pickup-Anthology.html
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.”


Fig. 14. Response of a Fender Stratocaster pickup with 470 pF load capacitance and different Ohmic loads



Fig. 15. Response of a Fender Stratocaster pickup (1972) with 10 MOhms ohmic load and eight different capacitive loads

You 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).
Just 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.
Here are the tone and volume pots of my Ibanez 230 – quite simple:

5 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.
The capacitor is marked 223K 50R, and the explanation of capacitor values can be found here:
http://www.ecawa.asn.au/home/jfuller/electronics/capacitors.htm

“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”

[10 x 10, 000]

Milli, micro, nano, pico; 10 to the minus 3, 6, 9, 12 resp.
In 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.
This Ibanez has 2 single coils and a humbucker, so is innately trebly for a “rock” guitar – more like a Strat than a Les Paul.
Also 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:

The other end of this wire is soldered to the casing of one of the volume/tone pots.
The 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.
I 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.
Note 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).

The 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:
http://www.mds975.co.uk/Content/Impedance_Matching_ESP_Rod_Elliott.txt

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!
Some Guitar Circuit Elements
Just 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:
http://www.electronics-tutorials.ws/accircuits/parallel-resonance.html
Notice 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 LC tank circuit acts like an open circuit with the circuit current being determined by the resistor, R only. So the total impedance of a parallel resonance circuit at resonance becomes just the value of the resistance in the circuit and   Z = R as shown.



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.
So, 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.
Just 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:
http://www.ibanez-rg-review.com/dimarzio-guitar-pickups.php
Using 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…



Resonant Frequency, ƒr


Inductive Reactance at Resonance, XL


Quality factor, Q


Bandwidth, BW


The upper and lower -3dB frequency points, ƒH and ƒL


So, for my partly fictitious Ibanez:
Pickup inductance = 7H
Capacitor = 22, 000pF = 22nF = 0.000 000 022F
Resistance = 250k Ohm
Resonant Frequency = 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
Inductive Reactance at Resonance = 2 Pi f L = 2 x 3.142 x 405.6 x 7 = 17,839 Ohms
Quality factor Q = R / 2 Pi f L = 250,000/17,839 = 14
Bandwidth = f/Q = 405.6Hz /14 = 29Hz
The upper and lower -3dB frequency points = f LOW = f – ½ BW = 405.6 – ½29 = 377 – 14.5 = 362.5Hz
= f HI = f + ½ BW = 377 + 29 = 398 Hz
Passband (-3dB drop each side) = 362.5 to 398 Hz
Well, 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?

The 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?
Ah well, the maths did me good…Anyway…moving on…
The 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):
http://www.soundonsound.com/sos/jan03/articles/impedanceworkshop.asp
Everything you need to know about Impedance and much more in that article.
If the input has too low an impedance, the most noticeable effect will be a loss of high end — 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.”
Passive Band Pass Filter section:

The 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.
http://www.electronics-tutorials.ws/filter/filter_4.html
C3 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.


The 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.


Let’s see if the numbers here for R1, C1 and R2, C2 make any more sense than the last lot!
For R1 (220k), C1 (2n2):
Freq 1 = ½ Pi RC = ½ Pi (220 000 x 0.000 000 002 2) = 1/0.00304 = 328.8Hz
For R2 (100k), C2 (2n2):
Freq 2 = ½ Pi RC = ½ Pi (100 000 x 0.000 000 002 2) = 1/0.001 38 = 723.4Hz
So input gain stage T1 band pass is 723-328 = 395Hz wide, centred on 525 Hz?
It 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!?
Gain stage 1 input:

R3, 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.
C4/C11/C23 Function:
http://www.talkingelectronics.com/projects/TheTransistorAmplifier/TheTransistorAmplifier-P2.html#NegativeFeedback

NEGATIVE FEEDBACK

“The circuit shows a capacitor between the base and collector. It provides NEGATIVE FEEDBACK.
If we remove the capacitor, when the base “moves down,” the collector “moves up.”  In other words the signal is inverted.

When 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.
When 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.

This is the opposite effect to the signal moving through the transistor.

This means the capacitor is working against the action of the transistor.

The capacitor will have more effect on high frequency signals while the low-frequency signals will be affected less.
Because the capacitor is working against the natural flow of signal through the circuit, it is called NEGATIVE ACTION or NEGATIVE FEEDBACK.”

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The performance characteristics of the BC184 transistor can be found here:
http://images.ihscontent.net/vipimages/VipMasterIC/IC/ONSM/ONSMS04085/ONSMS04085-1.pdf
Some 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:
SOLDERING PRECAUTIONS

The melting temperature of solder is higher than the rated

temperature of the device. When the entire device is heated

to a high temperature, failure to complete soldering within a

short time could result in device failure. Therefore, the

following items should always be observed in order to minimize

the thermal stress to which the devices are subjected.

??Always preheat the device.

??The delta temperature between the preheat and soldering

should be 100?C or less.*

??When preheating and soldering, the temperature of the

leads and the case must not exceed the maximum

temperature ratings as shown on the data sheet. When

using infrared heating with the reflow soldering method,

the difference should be a maximum of 10?C.

??The soldering temperature and time should not exceed

260?C for more than 10 seconds.

??When shifting from preheating to soldering, the maximum

temperature gradient shall be 5?C or less.

??After soldering has been completed, the device should be

allowed to cool naturally for at least three minutes.

Gradual cooling should be used since the use of forced

cooling will increase the temperature gradient and will

result in latent failure due to mechanical stress.

??Mechanical stress or shock should not be applied during

cooling.

* Soldering a device without preheating can cause excessive

thermal shock and stress which can result in damage to the

device.”

So, 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.
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Gain stage 1 Output:
VR1 = Volume, VR2 = Tone

This is a bit more complex and I can’t decipher it so well.
The 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.
When 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?
This 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).
As 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.
As 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.
From:

http://www.electronics-tutorials.ws/rc/rc_3.html

Sine Wave Input Signal

If we now change the input RC waveform of these RC circuits to that of a sinusoidal Sine Wave voltage signal the resultant output RC waveform will remain unchanged and only its amplitude will be affected. By changing the positions of the Resistor, R or the Capacitor, C a simple first order Low Pass or a High Pass filters can be made with the frequency response of these two circuits dependant upon the input frequency value.

Low-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, ƒc) is used to define the filters bandwidth and a loss of 3dB corresponds to a reduction in output voltage to 70.7 percent of the original value.

Cut-off Frequency


where RC is the time constant of the circuit previously defined and can be replaced by tau, T. This is another example of how the Time Domain and the Frequency Domain concepts are related.

If we treat the two RC parts separately as before we can try the maths and see what nonsense I get this time!

When VR2 is minimum, only R7 and C9 are active it seems?
Freq 1: C9 = 22n, R7 = 100k
f = ½ Pi RC = ½ 3.142 x 0.000 000 022 x 100 000 = 723.4Hz
Freq 2: C8 = 220p, VR2 = 100k
f = ½ Pi RC = ½ 3.142 x 0.000 000 000 22 x 100 000 = 7,234 Hz
This 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.
A typical guitar amp bandwidth is around 5-10kHz according to here:
http://www.aikenamps.com/ResistorNoise.htm
Gain stage T2 input:

The 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.
So, 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?
For 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.

Note: T3 typo – T3 Emitter is at ground = 0V
The collector voltage for all three should be 26V
The emitter voltage for all three should be 8V (T3 typo – must be wrong as it’s at 0V)
The anode of the valve should be 250V
The cathode of the valve should be 6V
From this info some figures can be generated for the quiescent bias currents etc – so what else do I know?
The common-emitter current gain is represented by ?F or hFE; it is approximately the ratio of the DC collector current to the DC base current in forward-active region.

The 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
The 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
T1/T2 Emitter voltage = 8V
Emitter current in R6/R11 = 10k so 8V/10 000 Ohms = 80uA
Emitter I = Base I + Collector I
Base 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:
The CE Voltage = 26 – 8 = 18V, so Base voltage should be around 9V?
The 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
For this exercise, I will take the BE voltage from the above base voltage:
Base Emitter voltage = 9 – 8 = 1V
R4/R9 current = 9V/6M8 Ohm = 1.3uA
Therefore, 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.

From above, the Emitter current is 22 times bigger than the Collector current, so Collector I = 80uA/22 = 3.6uA
Therefore, the current through R10 = collector current + current through 6M8 = 1.3uA + 3.6uA = 4.9uA
So, the voltage across R10 = V = IR = 4.9uA x 220 000 Ohm = 10.9V
This means the voltage at Point A = Collector V + R10 V = 26V + 10.9V = 36.9V

The difference between points X and A then is 275V – 37V = 238V
The 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.

The 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.
So, if the V drop over R16 = 8.5mA x 27k = 230V
The V drop over R17 = 8.5mA x 1k = 8.5V
Voltage at bottom of R17 = 275 – 8.5 = 267.5V
The V drop across the speaker transformer primary = 275 – 250 = 25V
Without 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.

The EL84 specs are:

Heater voltage 6.3V
Heater current 760mA
Max anode voltage (Ia=0) 550V
Max anode voltage 300V
Max anode dissipation 12W
Max screen voltage (Ig2=0) 550V
Max screen voltage 300V
Max screen dissipation 2W
Pin Function
1 i/c
2 Control grid
3 Cathode + suppressor grid
4 Heater
5 Heater
6 i/c
7 Anode
8 i/c
9 Screen grid

The equation for power P = VI so taking the max values for the anode drawing current:
12W/300V = .04 Amps or 40mA

This max current would also pass through the speaker primary windings.

This correlates well with the JJ EL84 datasheet for typical usage also – Anode current – Ia = 48mA:

Class A1 amplifier:

Ua = 250 V

Ug2 = 250 V

Rk = 135 ?

Ia = 48 mA

Ig2 = 5,5 mA

Ra = 5,2 k?

Ug1eff (50mW) = 0,3 V

Ug1eff(N) = 4,3 V

N (10%)1) = 5,7 W

N 2) = 6 W

1) Ug1 fest fixed grid bias

2) Ig1 +0,3 ?A

The speaker transformer has a label with DE Ltd, C442, but Google found nothing for me.
The only way now is desolder and measure the primary and secondary resistances with a Voltmeter. Another day.
Moving on…
I 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…
Valve pinouts:

The reason for Marshall’s exaggerated 25W claim for this amp?:
Valve specifications:
http://lenardaudio.com/education/14_valve_amps_2.html
6BQ5 EL84 (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

It looks like the transistors got lost in the post so are being re-sent…ho-hum
Hopefully the next Post will be the last on this epic story.