[GreenKeys] Model 33 Q1 transistor replacement

Tue Jul 22 11:01:35 EDT 2025

Oh, gosh.  That schematic reminds me of the good old days at AT&T.  I used
to have mild arguments with drafting as to how a schematic should be shown
on paper.  Their approach was often that the schematic was just a graphical
netlist and the goal was to use as few pages as possible.  My position was
that a schematic told a story, in graphic form, a story about how a circuit
worked, so that the reader could easily understand its operation.

Anyway, here's a redrawn version of the schematic with some commentary
below that.  Please let me know if you spot any transcription bugs in the
schematic.

[image: image.png]

A couple of things to note here.  First, I don't see any reference to earth
ground in the original schematic.  To me, that means that this circuit
"floats" with respect to earth ground, and the relationship with ground is
determined by the applied 20mA (or 60mA) current loop (one side of which is
likely grounded directly or through some finite resistance).  When I call
out a voltage on the schematic or text, that voltage is with respect to the
Zero volt bus.

Second, the selector current is BIG, about 0.5 A.  And when Q2 is
conducting,  Q2 is operating in its linear range, so it will get HOT.

Let's look at what happens in the MARKing condition.  In the marking
condition the 20 mA (or 60 mA) of positive current is shoved toward the
base of Q1.  In this condition the loop current flows through CR5 and the
R3 (0.8 ohms) as well as through R1 into the diode network.   Also in this
condition the voltage at the base of Q1 is clamped to about +1.1V, and the
BE junction of Q1 is cut off, so Q1 is out of the picture in the MARKing
state.  The base of Q2 is held at -5.4 volts by the diode network and the
voltage the the emitter of Q2 is held at 4.7 volts (due the diode voltage
at the base, less the diode drop across the Q2 BE junction).  So with a
constant voltage at the emitter of Q2, and a constant resistance between
the emitter and the 0V bus, Q2 forms a constant current driver.  If Q2 has
a decent gain value, then Ic = Ie = 4.7V / (0.8 +1.5 + 8) = 456 mA.  So
that's the marking condition.

In the case of the spacing condition, the 20 mA loop current is disabled
and Q1 is biased on via R1 and the diode network.  This causes Q1 to turn
on and current is sent from the collector of Q1 toward R8, which tends to
rob current from the diode network, sending the voltage at the base of Q2
toward zero volts.  It won't get to zero but it will be close (maybe -1
volts).  With lower voltage at the base of Q2 (and 0.7 to 0.5V less than at
Q2 emitter), that will reduce the selector magnet current significantly
(but I don't think it will go to zero).  In a working circuit the actual
MARKing and SPACEing selector current can be determined by measuring the
voltage across R7 and dividing by 14.

Paul, ad7i

On Mon, Jul 21, 2025 at 5:04 PM RODNEY HOGG <radio_man at wbsnet.org> wrote:

> Paul,
> Do you have a type number or p/n for the transistor.  I have 1000s of
> transistors, civilian and military types, be glad to help you out.
> Rod
> K0EQH
>
> ------------------------------
> *From: *"Erik Bruchez" <erik at bruchez.org>
> *To: *"Paul Newland, ad7i" <ad7i at ad7i.net>
> *Cc: *"Greenkeys" <greenkeys at mailman.qth.net>
> *Sent: *Monday, July 21, 2025 1:58:36 PM
> *Subject: *Re: [GreenKeys] Model 33 Q1 transistor replacement
>
> Paul,
>
>> It's probably not all that difficult to find a working replacement, but
>> in a different form factor.   Can you show a larger portion of the
>> schematic?  I'd like to see if this transistor is operating as an on/off
>> switch, or if it's trying to work some magic to regulate the voltage on the
>> magnet for fast pull in and then low current idle.
>>
>
> I attach the whole thing. There are explanations in the schematics.
>
> I measure about 0 V on the base of Q1, but also about 0 V on the base of
> Q2. My understanding (which could be wrong) is that when one is on, the
> other one should be off, and vice versa, which means that 0 V on Q1 should
> lead to about 4.7 V on the base of Q2. In any case, I am not getting any
> voltage to the selector magnet through Q2 (the large transistor on the heat
> sink, separate from the board), which is the original problem I am having.
>
>
>> Some basic DC measurements might give you some insight into the health of
>> the transistor without removing it from the circuit board.
>>
>
> This is what is leading me to Q1: it seems that there is continuity
> between emitter and collector, which I don't think there should be! I have
> another instance of the board where this is not the case. (However, that
> second board also doesn't seem to work, but there might be something else
> wrong with it.) I could desolder the transistor on that one and transplant
> it to see if that gets my first board working, but I was trying to avoid
> doing that immediately.
>
>
>> I'd be curious to know the voltage from base to emitter when the
>> transistor is "ON" (or supposed to be on).   Also (can't see that part of
>> the schematic) but if there's a resistor in series with the base I'd be
>> curious to know the voltage across the resistor (and the resistance of the
>> resistor), and then use ohms law to determine the base current.  Same thing
>> for the selector magnet current, if that's possible.
>>
>
> -Erik
>
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