Author Topic: No resistor before clipping diodes in Bosstone?  (Read 236 times)

mark2

No resistor before clipping diodes in Bosstone?
« on: February 05, 2021, 03:58:10 PM »
I typically see 1k to 10k resistors from signal to clipping diodes, i.e. in series with C4 in the below schematic.

Can anyone shed light on why this may not have it?


from http://fuzzypedals.blogspot.com/2015/04/jordan-bosstone-fuzz.html

PRR

Re: No resistor before clipping diodes in Bosstone?
« Reply #1 on: February 05, 2021, 04:07:58 PM »
Q2 Emitter here is a high-ish impedance source. '4148 diodes can swamp that.

The 22n cap is even more impedance from bass to mids.
« Last Edit: February 05, 2021, 04:09:31 PM by PRR »
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antonis

Re: No resistor before clipping diodes in Bosstone?
« Reply #2 on: February 05, 2021, 04:19:54 PM »
In case of Q2 absence, Q1 output impedance (18k) should experience issues with directly driving diode pair..
(very low resistance to 600mV above ground - an almost short circuit..)

1k - 10k in that case should act as current limiting resistor..
 
edit: Paul already answered more concisely.. :icon_wink:
« Last Edit: February 05, 2021, 04:32:41 PM by antonis »
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mark2

Re: No resistor before clipping diodes in Bosstone?
« Reply #3 on: February 05, 2021, 04:49:59 PM »
Thanks to you both. I don't fully understand but it gives me enough to start googling.

One thing I'm struggling to find by searching: How do you know the output impedance of a transistor? Is it just a general rule based on its type and the configuration it's set up with?

iainpunk

Re: No resistor before clipping diodes in Bosstone?
« Reply #4 on: February 05, 2021, 05:08:20 PM »
an almost over simplified method is :
1) measure the collector voltage
2) calculate the current through the resistor with that voltage (remember that the voltage over the resistor is Vpower - Vc)
3) just apply Ohm's law to the transistor
4) then calculate parallel resistance with the collector resistor.

if the Collector voltage is 4.5v, the output impedance would be 9k
if the Collector voltage is 3v, its 6k
if the Collector voltage is 6v, its 12k


cheers, Iain

EDIT:
i just now see the 2nd transistor is PNP buffer, the impedance is complex here, when the integration of the output signal is positive (rising flank), the resistance is 18k, when the integration is negative (falling flank), its a few ohms.
this stage is common collector, the above method is still a handy thing to know for common emitter transistor stages tho,
« Last Edit: February 05, 2021, 05:13:50 PM by iainpunk »
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antonis

Re: No resistor before clipping diodes in Bosstone?
« Reply #5 on: February 05, 2021, 05:53:11 PM »
Although Iain points right it might confuse rather than enlighten you..  :icon_wink:

Under some acceptances (like BJTs are linear devices, infinite Collector internal resistance, infinite Early voltage, consistent transcondance, etc..), Common Emitter amp (Q1) output impedance is considered its Collector resistor (load) where Common Collector amp (voltage follower) (Q2) is considered its Emitter intrinsic resistor (0.025/ICollector)

For further clarifications, we have to refer to some kind of small signal model analysis..
(diode model, T-model, hybrid-π model..)
« Last Edit: February 05, 2021, 05:55:33 PM by antonis »
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..