Threaded Mode | Linear Mode

cruff · 04-06-2019, 06:24 PM

(04-06-2019 01:34 PM)albertofenini Wrote: I'm using for all the diodes some germanium diodes marked AA117, these have a forward voltage drop
of 0,328 Volts versus the original that have a forward voltage drop of 0,275 V
I have a few original from hp left, not enough to make a whole PSU, but I can substitute just CR5
Do you think that changing just CR5 would make the difference ?

Probably not much of a difference for those values. Not enough to explain the higher voltages you are seeing. One thought occurred to me is that those are the open circuit unloaded voltages. You could add some resistive loads on the outputs and see how much the voltages might drop. It may be that your circuit is functioning correctly now and it just needs loads to bring them into range.

albertofenini · (This post was last modified: 04-06-2019 09:22 PM by albertofenini.)

That is a good indication,
how much should be the value of the resistors that simulate the load ?
Should be connected between the Voltage point and ground, right ?

On the other hand, the guide that we use to repair the HP65 shows how to test PSU
off load, de-soldering cathode of CR2 and the anode of CR1 and
de-soldering the cathode of CR4
and the cathode of CR5 from the logic board, lifting them up, and solder them together

This means that good values should be available also off-load, right ?

cruff · 04-07-2019, 08:33 PM

(04-06-2019 09:22 PM)albertofenini Wrote: That is a good indication,
how much should be the value of the resistors that simulate the load ?
Should be connected between the Voltage point and ground, right ?

Yes, between the voltage and ground. The resistors would be sized to sink a current similar to that of the individual loads. For example, I believe that Vgg is used as a substrate bias voltage for the ICs, so the current flow is probably below 1 mA. If I remember correctly Vcc is the LED drive voltage and Vss is the voltage for the remainder of the calculator. Has anyone taken measurements? Perhaps several hundred mA for Vcc and 50 mA for Vss (which might represent idle and no card reading)?

Quote:On the other hand, the guide that we use to repair the HP65 shows how to test PSU
off load, de-soldering cathode of CR2 and the anode of CR1

What does that guide say you should see for the voltages?

Quote:and de-soldering the cathode of CR4 and the cathode of CR5 from the logic board, lifting them up, and solder them together

That would eliminate a bad circuit board trace between CR4 and CR5 as a possible problem, and removes the load the digital circuits placed on the +6V output. What does that guide say you should see for voltage?

Quote:This means that good values should be available also off-load, right ?

That assumes the power supply has sufficient capacity to source the load at the regulated voltage. It might not, and thus may be designed for a higher no-load voltage and at the design current will drop to the required voltage.

albertofenini · 04-08-2019, 05:52 AM

Hi
this is the guide we are using :
http://home.comcen.com.au/therobinsons/h...repair.pdf
if you look at page 7, after disconnecting the measurement points the values should be :
Vcc should be just around 8.2V.
Vgg should be about –12.4V
Vss should be about 6.28V
which are the same we encounter on a working HP65
our are a little higher, which to me is difficult to explain considering the similarity of all of the components used
The idea was to have a working PSU and use it for both testing unknown components from broken units, or to power units (classic only) under troubleshooting whose PSU condition is uncertain ...
but with these values we are afraid to use ...
Thanks for all the support and take care !

cruff · 04-08-2019, 01:08 PM

(04-08-2019 05:52 AM)albertofenini Wrote: Hi
this is the guide we are using :
http://home.comcen.com.au/therobinsons/h...repair.pdf

That guide is really nicely written. It should be very useful in diagnosing your test rig supply.

albertofenini · 04-08-2019, 01:14 PM

you are right, it is very well done, it's a shame that the author did'n complete the
card reader part.
However my breadboard prototype is exactly like the circuit shown,
and that's why I don't understand why my values are higher.
As you see, they suggest to measure Vcc, Vgg and Vss off load.
I'll check it again ... but if you have any idea ... let me know
thanks in advance

AndiGer · (This post was last modified: 04-08-2019 02:24 PM by AndiGer.)

Dear Alberto,

looking at the guide (page 6 in turn to page 7) I am not sure voltages are off load.

Quote:This is all you can do with the logic board removed, so now
install the logic board, and connect power to the battery contacts.
Measure Vbsw again to make sure it hasn’t dropped too far below the
original value. Also, measure the voltage at pin 2 of T1 to check if it
is the same as Vbsw. Now measure Vcc, Vss and Vgg.
Vcc should be just around 8.2V.
Vgg should be about –12.4V
Vss should be about 6.28V

Andi

EDIT: In the working unit I think you measured anything installed as well?

albertofenini · 04-08-2019, 02:29 PM

Dear Andi,
in the working unit I measured everything under load,
but in the guide it says that if the value are those listed you can re-connect the diodes,
which makes me think that the values should be pretty much the same.
To tell you more, we recently repaired a HP25 PSU, with all the chips removed (except anode and cathode drivers)
and the values were the same that we encountered with later with load.
Don't know what to say ....

AndiGer · (This post was last modified: 04-08-2019 03:28 PM by AndiGer.)

Dear Alberto,

I think you are right! Sorry I confused you. I only read that part of the guide but not the part that begins after figure 6 but is crucial.
If you followed the guide from scratch you removed all 3 voltages from the rest of the calculator circuitry (CR1, CR2 and CR4/CR5).
Installing the board at this part of the guide only is for getting connection to Vbsw I guess (what could be done also otherwise).

Andi

teenix · 04-08-2019, 09:21 PM

(04-07-2019 08:33 PM)cruff Wrote: If I remember correctly Vcc is the LED drive voltage and Vss is the voltage for the remainder of the calculator. Has anyone taken measurements? Perhaps several hundred mA for Vcc and 50 mA for Vss (which might represent idle and no card reading)?

The LEDs and card reader motor circuitry is controlled by the CPU but powered directly from the battery. it is a complicated setup but works well and keeps the power supply design simpler.

cheers

Tony

cruff · 04-09-2019, 12:49 AM

If the transistor you selected for Q1 is out of tolerance for its operating parameters, as compared to the original type, that might be enough to keep it from performing the regulation function in the way the circuit was designed. If that is the case, then selecting a lower value zener diode would suffice to adjust the voltages down. You could check to see what the waveform on the base of Q1 is doing, perhaps there is not quite enough drive current to switch Q1 on more fully to reduce the Vss voltage, and indirectly the other two voltages. Also verify that R1 is close to the listed value on the schematic.

albertofenini · 04-09-2019, 10:12 AM

(03-28-2019 11:06 PM)teenix Wrote:
(03-28-2019 04:06 PM)albertofenini Wrote: The 1954-0094 however can be replaced with PN3646 which is very hard to get

You can replace the PN3646 with a 2N3703, MPS3638 or PN3638

PN3638 similar to PN200

PN3904 similar to PN100

cheers

Tony

Ciao Tony,
unless I order it on Mouser which takes order from Italy but ships globally from US and you can imagine the cost,
there is noway I can find a pn3646, nor a 2N3703
do you have any other idea on how to replace the 0094 ?
thx Alberto !!

albertofenini · 04-09-2019, 10:15 AM

(04-09-2019 12:49 AM)cruff Wrote: If the transistor you selected for Q1 is out of tolerance for its operating parameters, as compared to the original type, that might be enough to keep it from performing the regulation function in the way the circuit was designed. If that is the case, then selecting a lower value zener diode would suffice to adjust the voltages down. You could check to see what the waveform on the base of Q1 is doing, perhaps there is not quite enough drive current to switch Q1 on more fully to reduce the Vss voltage, and indirectly the other two voltages. Also verify that R1 is close to the listed value on the schematic.

Thanks for the idea, Q1 is actually an old original 0550,
I'll check the HFE and I'll check the R1 value but I'm pretty sure is close to what it should be, and this R1 is also from an old logic board
I'll also check the waveform on the base of Q1
thx !!

AndiGer · 04-09-2019, 05:17 PM

If you can wait a little longer you can order PN3646 on TAS from China.
Offer no. 401500431782
10 pieces, US$ 1,99, free shipping ...
Andi

Zaphod · 04-09-2019, 10:34 PM

(04-06-2019 01:34 PM)albertofenini Wrote: Hi
I'm using for all the diodes some germanium diodes marked AA117, these have a forward voltage drop
of 0,328 Volts versus the original that have a forward voltage drop of 0,275 V
I have a few original from hp left, not enough to make a whole PSU, but I can substitute just CR5
Do you think that changing just CR5 would make the difference ?
Thanks for help!

Don't fear running out of germanium , just use modern day schottky's, they have a similarly low forward volt drop

albertofenini · 04-09-2019, 10:53 PM

(04-09-2019 05:17 PM)AndiGer Wrote: If you can wait a little longer you can order PN3646 on TAS from China.
Offer no. 401500431782
10 pieces, US$ 1,99, free shipping ...
Andi

Hi Andi, these are avaialble on TAS even from UK, it's just me that I'd like to get things by yesterday if you know what I mean
Our usual online shop is RS Components and they just have the 2N3904, will order the others on TAS and will wait, Mouser asks 20 € for shipping regardless of what you order ...

albertofenini · 04-09-2019, 11:05 PM

(04-09-2019 10:15 AM)albertofenini Wrote:
(04-09-2019 12:49 AM)cruff Wrote: If the transistor you selected for Q1 is out of tolerance for its operating parameters, as compared to the original type, that might be enough to keep it from performing the regulation function in the way the circuit was designed. If that is the case, then selecting a lower value zener diode would suffice to adjust the voltages down. You could check to see what the waveform on the base of Q1 is doing, perhaps there is not quite enough drive current to switch Q1 on more fully to reduce the Vss voltage, and indirectly the other two voltages. Also verify that R1 is close to the listed value on the schematic.

Thanks for the idea, Q1 is actually an old original 0550,
I'll check the HFE and I'll check the R1 value but I'm pretty sure is close to what it should be, and this R1 is also from an old logic board
I'll also check the waveform on the base of Q1
thx !!

We did few tests, and we re-designed the layout, now the position of the components is more close to the one on the electric circuit and this helps in troubleshooting.
We have now good Vcc and Vss, very close to the desired values, but still Vgg about -15,6 V
In the previous message I have erroneously indicated Q1 as a 0550.
This is wrong, Q2 is a 0550, Q1 is a 0094
We have three different Q1 0094, one have an HFE of 1, another one have an HFE of 2, the third one have an HFE of 16.
I think the first two are almost dead, the third one could be good.
We have tested the three in combination with both Q2 0550 HFE 16, and Q2 2n3904 with HFE > 100
The result is the some, Vgg equal to -16
No matter which Q1 or Q2 we use
R1 is 1470 ohm, which is a good approximation of the 1500 ohm requested.
Attached is the waveform from Base of Q1 (0094 with HFE=16)
What do you think ?

cruff · 04-09-2019, 11:41 PM

(04-09-2019 11:05 PM)albertofenini Wrote: We have now good Vcc and Vss, very close to the desired values, but still Vgg about -15,6 V

Add a 15K ohm resistor from Vgg to ground and measure it again.

Is C7 in the circuit? Looks like Q1 is getting enough current to perform regulation now, but I'm surprised a bit about the ringing on that downward part of the waveform.

albertofenini · 04-10-2019, 12:16 AM

Hi there,
yes C7 is in the circuit (lower left of the breadboard picture) and when measured is ok,
we will add (tomorrow, local time now is 2.13AM wednesday april 10th) the resistor and
will test again, but the aim was to reply an HP65 as in the scheme with no additional pieces
and have it working.
What do you think of the fact that Q1 with HFE equal to 1 or equal to 2 gives the same
behavior of a Q1 with HFE 16 ? couldn't this an HFE this low the reason ?
thx again !

cruff · 04-10-2019, 02:41 AM

(04-10-2019 12:16 AM)albertofenini Wrote: What do you think of the fact that Q1 with HFE equal to 1 or equal to 2 gives the same
behavior of a Q1 with HFE 16 ? couldn't this an HFE this low the reason ?
thx again !

A low hFE could certainly cause problems. It is possible there is sufficient base drive current available to make the low hFE transistors work well enough to regulate the circuit.