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I have a DC power source which can provide up to 5V, 50mA (LabJack maximum output current, voltage bus-powered) and I want to power my relay (SDT-S-105LMR2,000) with 5V, 100mA. I have an option to use NPN transistor with a DC current gain of approximately 2, but the 0.7V voltage drop is not that nice.

First question - Is the circuit logic correct (more attention on the added current booster)?

Second question - Because I am building 16 channel relay board, is there more convenient option for boosting the current for each of the relay (preferably less components)?

Third question - Will the LabJack supplied power will be a good enough source for 16 relays (working at different times)?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Why not use a more sensitive relay or an opto-isolator to pick the relay using Vin and its common? \$\endgroup\$ – Transistor Oct 4 at 16:15
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You can't in the way you try it. Your source is 250mW, your relay requires 500mW. You have not got enough energy.

A transistor can amplify current but not out of thin air. It needs to come from somewhere.

What you can try is to see what the relay hold current is. It is often much lower then the 'attack'* or "pick" current. In that cause you might store energy in a capacitor and use that to get the relay to 'start' en then your 50mA may be enough to 'hold' it.

*I can't remember if that is the right term...

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  • \$\begingroup\$ Unfortunately, the minimum current required is already 100mA, as it says in the datasheet. Thank you, I now know that I need to use different power supply for this relay. \$\endgroup\$ – Gabriel Oct 4 at 16:33
  • \$\begingroup\$ @Gabriel - You've missed the point. If you put a large capacitor on the 5 volts, it will be able to provide a much greater current than 50 mA - but only for a short time. With a big enough cap, you can provide 100 mA long enough to activate the relay. Once this has happened the current can fall to the power supply level (50 mA), and if the holding current is equal or less than 50 mA, the relay will continue to hold closed. \$\endgroup\$ – WhatRoughBeast Oct 4 at 16:39
  • \$\begingroup\$ Is "pick-up current" what you are thinking of Oldfart? :) \$\endgroup\$ – rdtsc Oct 4 at 18:26
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A slight modification to your design.

A mosfet can have a lower voltage drop than a BJT due to small Rds-on (10mΩ is common). It's not about transistor current gain; you need to operate the transistor in saturation mode. This is much easier to do with a mosfet.

No need for the upper transistor Q2.

I added C1 as a current reservoir, like @Oldfart alludes to, to get the pick current. If you don't have a 1000uF cap laying around you can parallel up a few smaller values to get you close.

All solenoid/relays have a pick current and a hold current. Although the manufacture may not tell you both. The hold current can easily be as low as 10% to 20% of the pick current. The pick current is the amount of current needed to get the relay to move its armature.

R1 can be anything from 10Ω to a few kΩ.

schematic

simulate this circuit – Schematic created using CircuitLab

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Q2 is not needed and R1 could be larger, say 10k. I suggest you check out the ULNxxxx series of relay drivers. There you have say 8 drivers in one DIL package. Anyway, you will need an extra power source.

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Thank you for the answers, it really made me think and I find solution of @Aaron and @Oldfart well fitting for one relay. Only, what I find disturbing is, when I have 16 charging capacitors by only 5V, 50mA power source, which means that the current will be divided. My conclusion is to use extra power source with suitable power rating as @AndersG said (I did not want to use ULNxxxx series of relay drivers, because I need I2C or SPI).

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