Hi Tek Group,

I have a Tek 2465 that has power supply issues. I have followed the
instructions from the Tek Service Manual, and found that the INVERTER DRIVE
portion of the inverter board is problematic. However, I don't know how to
identify the faulty components further. When power is applied there is no
output, and apparently the drive signals for Q1060 and Q1070 are missing.
Also I could smell smoke, and looks like R1071 (430 Ohms) is getting dark,
although its value still holds well.

Any suggestions?

Thanks,
Luca

Due to start-up circuit and loopbacks, difficult to debug

Use isolation transformers on mains, following PSU flowchart

Check the power transistors, rectifiers

Any blown up power semiconductor will often take the driver out as well .

See many previous threads here and tekscopes2, as well as eevblog

Bon chance

Jon

Luca,

Since R1071 is burning, C1071 is bad. This is one of those RIFAs that needs to be changed out with new. This type can be 630V type film. The two that need to be an X1 or X2 type are C1016 and C1018. The RIFAs are clear amber colour. There are several on the power supply board. Replace all of these! C1034 and C1066 are known to be open. I replaced these with R82 series 4,7mfd 50V film types. There are other condensers and resistors that should be replaced. If you want a list of parts to rebuild the supply and other parts, ask.

Mark

Thanks for the tips! I wonder why they design the start-up circuit this way rather than using a separate always-auxiliary supporting power supply. One reason may be to save one transformer. The other I could think of is to have some kind of protection so that if the regulator or inverter is malfunctioning, the driving circuit is cut off automatically. It would be helpful if some experienced designers can educate me. I can see a lot of work went into the engineering of it, and hats off to the documentation, although I have found several inconsistencies as well.

Thanks Mark. I just educated myself on the RIFA capacitors watching Dave's EEVblog video. I will replace mine. It would be helpful if you can provide a list of parts that need to be replaced. Thanks in advance. - Luca

Luca,

Going from the first in the schematics to last:
A1- 100mfd 25V (order 10) 647-ULD1E101MED1TD
(14) 47mfd 25V (order 41 or more) 647-ULD1E470MDD1TD (UHE series will work)
(2) 220mfd 16V 647-ULD2A221MED1TD
(2) 39meg order (4) 594-VR37000002005FA100 R710 and R910
A2- (2) 10mfd 100V 647-ULD2A100MED1TD
(5) 100mfd 25V
(3) 1mfd 50V film 80-R82DC4100DQ60J
(2) 47mfd 25V
A3- (2) 290mfd to 390mfd 200V 647-LGR2D391MELZ35
(3) 100mfd 25V
(3) 4,7mfd 50V film 80-R82CC4470AA30J
3,3mfd 350V 647-ULD2G3R3MPD6
(3) 250mfd 20V to 330mfd 25V 647-UHE1E331MPD
(2) 180mfd 40V to 330mfd 35V 647-UHE1V331MPD
(2) 10mfd 100V
(2) ,068mfd X1/X2 594-222233814683
,056mfd 600/630V 871-B32032B4563K000
(2) ,0047mfd 600/630V 871-B32620A6472J
(2) ,0022mfd 594-F339X122248MDA2B
,01mfd 80-F872BB103M480R
270,000ohms 2W R1020
(2) 390,000 ohms to 100,000 to 150,000 ohms 1W R1018 and R1019
A5- (5) 47mfd 25V
A9- 100mfd 25V
High Voltage board- 430,000 ohms 1W 279-H4P430KFZA
150,000 ohms 1W 279-H4P150KFCA
180,000 ohms 1W 279-H4P180KFDA
332,000 ohms 1W 603-FMP200FRF52-332K
442,000 ohms 1W (2) 71-CMF60221K00FHEK (put in series)
100 meg 588-SM102031006FE or 588-MOX-300001006FE
(2) 360 1W 279-H4P360RFZA
330 1W one of each of these two are near the input ICs. The originals are 1/2W carbon. The second 360 is near the horiz. output IC, also 1/2W carbon. 279-H4P330RFCA
For B version owners: the battery, if you can remove it, would be 667-TL5903P.
The part numbers are from Mouser. What they have in stock now is unknown. I went by my list when I did mine. Ones with part numbers missing will be listed above. Count the qty. needed. R1010 and R1019 can be removed if in the piece. I hope this list helps you and others.

Mark

I've got a slightly toasted R1071 as well, and a supply that won't come up.
Initially, I could hear a ticking sound, U1030 would come alive, drive
signals were reaching the drivers, and the supply would come on for about
50mS and then stop. I went through the Inverter Troubleshooting Flowchart, and took
the PS503 power supply branch of the flowchart. Using that approach, I determined that
Q1021 and Q1022 weren't passing current, so I changed them out, and and it
started to work, albeit with levels at only about 25% of normal, due to PS503 limitations.
Now I've hooked back up to to normal 120VAC, and now it's not working, at all.

I've been pouring over the schematic, trying to understand how it develops
the 'start-up' voltages it needs to run, and the startup processes involved.
I suspect that the circuitry attached to T1020's secondary winding forms an
oscillator condition that pumps the C1025 via CR1022, until it reaches a level
high enough to turn on the TL475...
But there are 3 main difficulties to deal with:
1) The board layout is a nightmare. Component placement is scattered, and 90%
of the reference designators are missing in the silkscreen. So locating part positions
is so slow and tedious, that by the time I locat the part, I've forgotten why I was looking for it.

2) It's a complex board, conceptually. Easily one of the most challenging that I've dealt with.

3) I'm not as sharp as I once was. Not dull yet, but I'm no razor anymore either.

Swapping out all these caps may not be a shotgun approach, but it sure looks and feels like one
to me, and that might be the most sensible approach, but I'm not quite there yet. If I find something
useful, I'll let you know, and I'll watching here, in case someone has a good theory of operation, or
just a solution that doesn't involve 25 different components.

Thanks,
Lyle

On Tue, Apr 26, 2022 at 03:39 PM, <lylejlarson@...> wrote:

...
I've been pouring over the schematic, trying to understand how it develops
the 'start-up' voltages it needs to run, and the startup processes involved.
I suspect that the circuitry attached to T1020's secondary winding forms an
oscillator condition that pumps the C1025 via CR1022, until it reaches a level

C1025 initially charges from rectified line voltage through R1020. Once C1025 has enough voltage across it U1030 powers up. However, current coming through R1020 is not enough to supply U1030, running switcher needed to keep charging C1025 through CR1022. If the switcher is not starting U1030 will cycle on/off.

Initial pulses will flow through C1072, this current in T1050 charges up C1066. Once switcher starts current in T1050 continues to charge C1066.

If you look at voltage across C1066 wrt "REF", does it initially charge up or stay flat? If there is no voltage building up on C1066, C1072 could be bad, not giving the initial kick. A toasty R1071 could be the result of it too.

A minimum load is needed at +5VD as described in the SM if the PSU is not connected to the scope.

Ozan

Luca,

I offer Complete Overhaul Parts Kits for the entire 24x5 series, with a solid discount for Tekscopes members.

If you want to save yourself a whole bunch of work researching and sourcing all the correct parts, then write me off forum

yachadmATgmailD^Tcom

Menahem Yachad
www.condoraudio.com

Thank you so much Mark.

I have replaced a few capacitors that I can tell or measure to be bad. I have not gone systematically, because of the limited supply. I have order a bunch of more generic ones to replace some other capacitors as needed. I am also ordering a desoldering gun. Without it, it is very hard to put the new capacitor back on, because of the remaining solder.

Thanks again - Luca

On 2022-05-02 5:05 p.m., lpb612 wrote:

Thank you so much Mark.
I have replaced a few capacitors that I can tell or measure to be bad. I have not gone systematically, because of the limited supply. I have order a bunch of more generic ones to replace some other capacitors as needed. I am also ordering a desoldering gun. Without it, it is very hard to put the new capacitor back on, because of the remaining solder.

If your ordinary iron is hot enough to melt it, you can remove just about 100% of the existing solder with desoldering wick.

--Toby

Thanks again - Luca

How does current that initially charges C1025 flow? Is it like the following?

T1020 PIN4 -- R1020 -- C1025 -- CR1035 -- R1046 -- T1020 PIN6

This is not the only path, but seems the least resistant one. There is also a path through CR1072 and C1072, but that path will only allow C1025 to be charged to about 10V, because C1072 is about 30 times smaller than C1025.

Luca

On Tue, Apr 26, 2022 at 05:39 PM, <lylejlarson@...> wrote:

1) The board layout is a nightmare. Component placement is scattered, and 90%
of the reference designators are missing in the silkscreen. So locating part
positions
is so slow and tedious, that by the time I locat the part, I've forgotten why
I was looking for it.

The cross reference table listed on 4 pages before the Schematic 9 are for this purpose.
It is pretty easy to locate components on the board and on the schematic this way.
I found the scanning quality for Tek 2467 service manual better for this purpose.
See https://w140.com/tekwiki/images/4/41/070-6019-01.pdf

2) It's a complex board, conceptually. Easily one of the most challenging that
I've dealt with.

No disagreement here! I am trying to understand it and I hope I am getting getting closer.

Luca

My current understanding of the start up process is as follows:
1. C1025 is charged using the following paths

T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- C1072 -- T1020 PIN6
T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- R1069 -- C1066 -- T1020 PIN6
T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- R1069 -- R1061 -- VR1062 -- T1020 PIN6

Initially the first 2 paths are used. After C1072 reach about +40V and C1066 reaches 13V in about 250ms, the only path that is active is through the Zener Diode VR1062 because it is 12V limited. It will take about 2 seconds for C1025 to reach +21V.

2. After C1025 reach about +21V relative to REF, Q1021 and Q1022 saturates, and TL494 (U1030) starts to work. The initial cycles are almost 100% duty cycle. And Q1050 conducts. The current will flow through CR1072 to further charge C1072 towards regulated line voltage (325V).

The manual states that the initial current pulses creates enough secondary voltage to charge C1066 to its working voltage, and INTERTER DRIVER starts to work. However, it seems that C1066 is already charged in the initial stage to about +13V before C1025 is charged to +21V. So I have some doubts here as to whether C1072 is really needed to act as a initial load. I think the circuit should start even if it is removed. It does provide a path for the current, in case the INVERTER DRIVER is not working yet.

3. Once INTERTER DRIVER starts to work, the secondary windings of T1050 will generate voltages to power TL494 the PWM controller and the INTERTER DRIVER itself.

The following is a simulation program for the charging circuit. It should run using NGSPICE.

* Charging circuit for C1025

Vin 1 0 PULSE(0, 325V, 10u)
R1020 1 2 270K
C1025 2 3 100u IC=0V
D1034 3 4 Diode
R1046 4 0 240K

D1072 3 5 Diode
C1072 5 0 3.3u IC=0V

R1069 5 6 36k
C1066 6 0 4.7u IC=0V

R1061 6 7 2k
D1062 0 7 Dzener

.MODEL Diode D
.MODEL Dzener D BV=12V

.control
tran 1m 5s
plot V(2)-V(3) V(5) V(6)
.endc

.END

On Tue, May 3, 2022 at 10:30 AM, lpb612 wrote:

My current understanding of the start up process is as follows:
1. C1025 is charged using the following paths

T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- C1072 -- T1020 PIN6
T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- R1069 -- C1066 -- T1020 PIN6
T1020 PIN4 -- R1020 -- C1025 -- CR1072 -- R1069 -- R1061 -- VR1062 -- T1020
PIN6

Initially the first 2 paths are used. After C1072 reach about +40V and C1066
reaches 13V in about 250ms, the only path that is active is through the Zener
Diode VR1062 because it is 12V limited. It will take about 2 seconds for C1025
to reach +21V.

2. After C1025 reach about +21V relative to REF, Q1021 and Q1022 saturates,
and TL494 (U1030) starts to work. The initial cycles are almost 100% duty
cycle. And Q1050 conducts. The current will flow through CR1072 to further
charge C1072 towards regulated line voltage (325V).

The manual states that the initial current pulses creates enough secondary
voltage to charge C1066 to its working voltage, and INTERTER DRIVER starts to
work. However, it seems that C1066 is already charged in the initial stage to
about +13V before C1025 is charged to +21V. So I have some doubts here as to
whether C1072 is really needed to act as a initial load. I think the circuit
should start even if it is removed. It does provide a path for the current, in
case the INVERTER DRIVER is not working yet.

Good analysis and great idea to simulate the start up section. Your spice netlist is missing Q1062 and the load at its emitter. For inverter driver to function Q1062 should be able to support at least the current through R1066 and R1068 (in alternate phases) in addition to other components. R1069 is too large to supply such current. With a single 2k from Q1062 emitter to (REF) node there is not enough voltage across C1066 and at the emitter of Q1062 for proper driver operation without CR1062-CR1065 working.

Ozan

On Tue, May 3, 2022 at 03:05 PM, Ozan wrote:

Good analysis and great idea to simulate the start up section. Your spice
netlist is missing Q1062 and the load at its emitter. For inverter driver to
function Q1062 should be able to support at least the current through R1066
and R1068 (in alternate phases) in addition to other components. R1069 is too
large to supply such current. With a single 2k from Q1062 emitter to (REF)
node there is not enough voltage across C1066 and at the emitter of Q1062 for
proper driver operation without CR1062-CR1065 working.

This is one of those Aha moments for me. Thanks for the correction. It makes perfect sense to me now.
With just the R1066 and R1068 as load, C1066 only charges to a bit less than 3V. More load (U1062-U1064) will lead to even lower voltage. So the initial push using C1072 to get C1066 charged is necessary.

It also removes another doubt for me. When Q1062 is not considered, C1066 would charge to about 40V at maximum. However, this exceeds its 35V rating. With Q1062 considered, that will not happen.

Can you help me with the following two questions? I have had difficulty figuring out an answer by myself.

1) The PWM control signal is taken from the REF line and 5V REF of TL494 through R1045 and R1046. However the REF line relative to PIN 7 of TL 494 (namely REF2 in later versions) is not a DC voltage. It is a switching voltage following the switching pattern of Q1050. Does C1035 serve as a filtering capacitor, together with the Thévenin equivalent resistance R1045||R1046?

2) There is no DC feedback in the error amplifier TL494/U2. Just R1033 in series with C1033. The same happens with U4 - no feedback at all. Therefore the output voltage V2 and V4 should be either LOW or HIGH, but not in the middle, as Figure 3-11 seem to suggest. It can work like this in a bang-bang way (100% or 0% duty cycle). But is this how the circuit is supposed to work?

Thanks so much.

Luca

On Tue, May 3, 2022 at 01:58 PM, lpb612 wrote:

Can you help me with the following two questions? I have had difficulty
figuring out an answer by myself.

1) The PWM control signal is taken from the REF line and 5V REF of TL494
through R1045 and R1046. However the REF line relative to PIN 7 of TL 494
(namely REF2 in later versions) is not a DC voltage. It is a switching voltage
following the switching pattern of Q1050. Does C1035 serve as a filtering
capacitor, together with the Thévenin equivalent resistance R1045||R1046?

Yes, C1035 acts as a filter capacitor as far as I can tell.

2) There is no DC feedback in the error amplifier TL494/U2. Just R1033 in
series with C1033. The same happens with U4 - no feedback at all. Therefore
the output voltage V2 and V4 should be either LOW or HIGH, but not in the
middle, as Figure 3-11 seem to suggest. It can work like this in a bang-bang
way (100% or 0% duty cycle). But is this how the circuit is supposed to work?

U4 is just for over voltage protection, it is inactive in normal operation. U2 controls the switcher operation.

C1033 is a large cap, you can think of U2 as an integrator rather than a comparator. DC level at input is set by switcher control loop (let me know if this needs more explanation). Optocoupler U1040 and feedback signal from sheet <10> regulates +5VD.

Ozan


There are two control loops R

On Tue, May 3, 2022 at 03:58 PM, lpb612 wrote:

There is no DC feedback in the error amplifier TL494/U2. Just R1033 in series
with C1033. The same happens with U4 - no feedback at all. Therefore the
output voltage V2 and V4 should be either LOW or HIGH, but not in the middle,
as Figure 3-11 seem to suggest. It can work like this in a bang-bang way (100%
or 0% duty cycle). But is this how the circuit is supposed to work?

Correction of typo: V2 and V4 should be U2 and U4, the error amplifiers.

On Tue, May 3, 2022 at 03:59 PM, lpb612 wrote:

On Tue, May 3, 2022 at 03:58 PM, lpb612 wrote:

There is no DC feedback in the error amplifier TL494/U2. Just R1033 in

series

with C1033. The same happens with U4 - no feedback at all. Therefore the
output voltage V2 and V4 should be either LOW or HIGH, but not in the

middle,

as Figure 3-11 seem to suggest. It can work like this in a bang-bang way

(100%

or 0% duty cycle). But is this how the circuit is supposed to work?

Correction of typo: V2 and V4 should be U2 and U4, the error amplifiers.

---
That is what I guessed in my reply. U4 is wired for over voltage protection so it is OK it be be on/off. When over voltage at the primary happens it needs to turn off the switcher.

U2 is the one that controls the loop. As I wrote in the previous message it is wired more like an integrator, it also has a zero coming from R1033 for compensating the loop, so in normal operation output of U2 is a control voltage and not on/off only. There is no local DC feedback between pins 2 and 3 but global feedback (around the full switcher control loop) still controls DC voltage at the input.

Ozan