diode replacment
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slimecity
Super Member
Looks like you can use a 3.9V zener to replace this - if its small, go for 500mW.
Not totally sure if you can actually get a 3.9V zener, if not, try to look for a 4V one.
Translate this page from Italian - the datasheet for your diode is on there.
https://translate.google.co.nz/tran...c/82057-echivalent-dioda-tlr-112/&prev=search
Not totally sure if you can actually get a 3.9V zener, if not, try to look for a 4V one.
Translate this page from Italian - the datasheet for your diode is on there.
https://translate.google.co.nz/tran...c/82057-echivalent-dioda-tlr-112/&prev=search
Is there any chance that new D507 3.9V zener instead of tlr112 led couse transistors Q581 and Q583 run hotter?
schematics: https://drive.google.com/open?id=0B8KUNmX2TYSUOU1VbkdGRUs2S1U
schematics: https://drive.google.com/open?id=0B8KUNmX2TYSUOU1VbkdGRUs2S1U
It seems a TLR112 is a GaP red LED by the looks of it - not actually a zener. - I get this from the data sheet.
http://www.datasheet-pdf.com/PDF/TLR112-Datasheet-Toshiba-539928
Looking at the schematic, and then at the datasheet (thanks Hyperion ! ), these type of diodes serve to set fixed voltage drop levels, and as per datasheet around 1.5~1.7 Volt (as a function of the current drawn, which can result into below 1.5V and above 1.7V).
This is confirmed in the schematic, when looking at D505:
+51.8V on the positive side (left of R519)
+50.0V on the negative side
i.e. dV = 1.8V
For D507:
-48.9V (-49.4V + 0.5V) on the positive side
-50.7V on the negative side
i.e. dV = 1.8V
LED technology is used to create a single PN junction providing the set voltage drop.
This is easiest replaced by 3x 1N4148 in series
Hence, using a 3.9V zener (which are noisy by the way) is more than doubling the required voltage drop.
D507 is in fact "enforcing" the current through R581/583, by imposing a fixed voltage drop across them.
Taking a fixed voltage drop of 0.5V across the B-E junction of the transistor, the voltage across the resistors is forced to 1.3V, and hence the current to ~4mA (I=U/R)
As a consequence, the current through the transistor is forced to 4mA.
By using a 3.9V zener, the current through the transistors is increased proportionally !
And thus their heat dissipation (P=dVxI)
It is not the right replacement.....
So, I would replace the diode by 3x 1N4148 in series
This is confirmed in the schematic, when looking at D505:
+51.8V on the positive side (left of R519)
+50.0V on the negative side
i.e. dV = 1.8V
For D507:
-48.9V (-49.4V + 0.5V) on the positive side
-50.7V on the negative side
i.e. dV = 1.8V
LED technology is used to create a single PN junction providing the set voltage drop.
This is easiest replaced by 3x 1N4148 in series
Hence, using a 3.9V zener (which are noisy by the way) is more than doubling the required voltage drop.
D507 is in fact "enforcing" the current through R581/583, by imposing a fixed voltage drop across them.
Taking a fixed voltage drop of 0.5V across the B-E junction of the transistor, the voltage across the resistors is forced to 1.3V, and hence the current to ~4mA (I=U/R)
As a consequence, the current through the transistor is forced to 4mA.
By using a 3.9V zener, the current through the transistors is increased proportionally !
And thus their heat dissipation (P=dVxI)
It is not the right replacement.....
So, I would replace the diode by 3x 1N4148 in series
So, I would replace the diode by 3x 1N4148 in series
Did that and heat is gone
Thanks Oilmaster for detailed explanation!AudioNutz18
Active Member
I have a pioneer sx650 with a MZ322 diode only have a 1W 18 or 20v will one of these work for a replacement? Thanks
MZ322 is a 500mw 22v zener diode tolerance 20.9v --- 24.1v
The 20v zener you have may work.
Last edited:
AudioNutz18
Active Member
Ok thank you. Is 1W ok to replace the 500mw? Thanks
Yes..
