voltage regulator

Hi, I am a grad student at Stanford. I am trying to make a ECL to TTL converter board. So, I will use MC10H125. These chips need -5.2V. And I am trying to get this from NIM crate's power supply of -6V. I am wondering which voltage regulator to use. I was thinking of LM2990-5.2, but the dropout voltage could be as much as 1V in which case, it won't work. Any advice would be appreciated. Thank you.

Reply to
Kazumi.Ishii
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Thank you for your quick reply. I will need 8 of 10H125. Would you still use LM2990-5.2? I was also thinking of using -12V from the NIM power supply, but it would be wasteful... And, I will investigate switching regulator.

Reply to
Kazumi.Ishii

In both CAMAC and NIM, it was traditional to use a single diode to drop the +-6 to approximately 5. 1N5400 or something, good enough.

John

Reply to
John Larkin

Just use a normal rectifier diode (1N4005 or similar) in series with the -6 supply - this will give an 0.7 volt drop which will give you "close enough" to -5.2 volts.

We have lots of commercial CAMAC modules that just use a diode off the

+6 bus to provide power to TTL circuits.
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Reply to
Peter Bennett

Hello John,

One situation where that could be a problem is noise sensitive gear. We had an ECL memory bank driver modulate a wimpy -5.2V rail in an ultrasound machine. Since it was doing that at the video repetition rate there was no way to sufficiently filter this out other than design a stiff regulator. That got rid of the noise.

Regards, Joerg

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Reply to
Joerg

How many 10H125 are you going to use? The max 1V drop out voltage of the LM2990 is for a maximum current of 1A, i.e. about 25 10H125 if I remember their supply current requirements correctly. If you are only going to use a couple of them you should be fine. Besides that NIM crates also have higher output voltages rails, which would make it unnecessary to us an LDO regulator anyway, provided your total ECL chip count is low enough to keep the supply current low enough for a linear regulator. If not you may consider a switching regulator, e.g. one of National's Simple Switcher in inverting configuration. Using one of their adjustable Simple Switchers you can generate the -5.2 from e.g. the +6V rail.

HTH Klaus

Reply to
Klaus Bahner

Hello Peter,

Doesn't that exceed the tolerance limits a little? While most CMOS would be ok the old LS series isn't guaranteed to work properly above 5.25V.

Regards, Joerg

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Reply to
Joerg

You don't need any voltage regulation because the ECL inputs are referenced to GND, and the VEE deviation from nominal -5.2 is a common mode effect that does not influence the signal thresholds.

From the ON Semi family characteristics app note:"Power supply regulation which will achieve 10% regulation or better at the device level is recommended. The ?5.2 V power supply potential will result in best circuit speed. Other values for VEE may be used. A more negative voltage will increase noise margins at a cost of increased power dissipation. A less negative voltage will have just the opposite effect. (Noise margins and performance specifications of MECL 10H are unaffected by variations in VEE because of the internal voltage regulation.)"

Plan on 8x50=400mA Vee sink current, include a feel good 1uF power entry capacitor on the board, and 0.001u bypass Vee to GND every few chips or so depending on board type and layout. Do the same for Vcc. Add a Schottky 3A diode in series with -6V if -5.x makes you feel better.

Reply to
Fred Bloggs

The LS family is not guaranteed to work properly, period. Max Vcc is

7.0V, where did you get 5.25V from? -You could be thinking noise margin degradation for > |+/-5%| Vcc deviation of commercial grade, and that over 0-70C.
Reply to
Fred Bloggs

Hello Fred,

Page 6 is where I got it from:

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7V is the abs max but they never guarantee it to function that high, just that it doesn't instantly die. I guess it depends on the application but in mine (mostly med) you can never operate outside the min-max values as stated in the data sheet under "recommended operating conditions". Anything else would get flagged.

Regards, Joerg

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Reply to
Joerg

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