[Bri4n]

I have a G540 with a 48 volt power supply, I did not think things through and bought 2 32 volt relays. I am guessing hooking up the positive side of the relay would spell death for the relay since it is only rated for 3-32 volts. Other than buying 2 new relays what are my options for making these work without cooking the G540? I have another power supply laying around it is only 24 volts so I don't want to use it with the 540. Any suggestions?

[Al_The_Man]

Sounds like these are SSR type, you could use a simple resistor in series with a 5~24v zener across the SSR to clamp the voltage.
Al.

[Bri4n]

What size resistor would I need? Or do you have a formula I could use to figure it out? Which diode would you reccomend? I have some of both that I scavanged but I don't know how to figure out what size of each to use.

Thanks

[Al_The_Man]

3-32 volt SSR's are typically 1k input impedance (you would have to confirm the one you have). If you go with a 24v zener across the input to clamp it to 24vdc, you would need around a 1k resistor to drop 48-24=24v 1/4w.
Al.

[Bri4n]

Thanks for the help, what are the chances of me burning my power supply or 540 up if I get this wrong?

[Al_The_Man]

As long as you put the resistor in series with the 3-32v input, there is not much chance of doing any damage.
Al.

[Bri4n]

Thanks alot it is appreciated.

[john-100]

Hi Bri4n

what is the part number of your solid state relays ?

if its the SSR-25DA that is a popular result in an ebay search
its input impedance is 1500 or 1600 ohms depending on which data sheet you look at


this one - SIPIN TECHNOLOGY CO., LTD.
gives the input current as 7.5 mA at 12v
this works out as 12V/7.5 mA = 1600 ohms

on this second sheet
Ssr 25da datasheet and application note, data sheet, circuit, pdf, cross reference | Datasheet Archive
gives the input impedance as 1500 ohms

if the input impedance of the relay is 1600 ohms
using a 1000 ohm resistor to connect terminal 3 to +48V
would give you a voltage of 1600/2600 x 48 = 29.5V between terminals 3 & 4
when the G540 output switches on
and connects the relay terminal 4 to the 48V power supply negative terminal

I'd increase the resistor to 1500 or 2200 ohms to avoid being so close to the 32V maximum input
( BTW a 22000 ohm resistor would reduce the relay input to 1600/23600 x 48 = 3.25V )

John

[Al_The_Man]

Surely the 24v zener would prevent anything higher appearing at the 3-35v input?
Al.

[john-100]

Hi Al

yes ,
the addition of a 24V zenner across terminals 3 & 4 would limit the relay input voltage to 24V
my thought is by selecting the resistor value to limit the voltage to a safe value without a zenner
the circuit would fail safe if a connection to the zenner was broken
belt and braces if you like

to drop 24V (with the 24V zenner connected) ,
I think a 1200 ohm is the minimum for a 1/2 W resistor
24/1200 x 24 = 0.48 W
which so happens to limit the relay input to 27.5V without the zenner
assuming the relay input impedance is 1600 ohms



john

[Bri4n]

Yes it is the SSR-DA. Can you all show me a simple drawing of how to make the connections I thought I knew but I am now wondering. The way I understood it was to solder the resistor to the diode and then connect them to pos pin on the relay. It sounds now like put the resistor inline and bridge the 2 terminals with the diode???

[Al_The_Man]

Yes you wire the resistor in series with the SSR and the Zener across the input terminals.
IOW, +48v source - resistor - SSR + in, and zener across SSR + and SSR - with cathode of zener (stripe) at the SSR + terminal.
I can draw it up if you wish?
Al.

[john-100]

is this OK ?

Attachment 191500

[Bri4n]

Thanks I got it now.