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Consider this simple constant current source. The voltage is 27 V, yet the indicator LED is rated to a much lower reverse breakdown. Is the LED safe?

Will the LED draw current away from the output load? Output would be less than 1.5 mA.

Simulations in LTSPICE proved LED won't lower output current below 1.5 mA but I don't know for sure.

Circuit

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    \$\begingroup\$ Welcome to EE.SE. Your circuit is good and LED current should be 4.33 mA at most. Since most small LED's have a maximum of 20 mA, this is fine. The LED is not exposed to any reverse voltage. \$\endgroup\$ – Sparky256 Apr 20 '18 at 20:59
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    \$\begingroup\$ It is the transistor that has any excess voltage across it. A bigger issue is R6 -- 1 Mohm only allows 27 uA at most into the base of the transistor. This requires a transistor with a gain of 200 or so to achieve the regulated current. I would reduce it to 220k or thereabouts. \$\endgroup\$ – Dave Tweed Apr 20 '18 at 21:14
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    \$\begingroup\$ Which "output load"? Put it into your schematics explicitly. \$\endgroup\$ – Ale..chenski Apr 20 '18 at 21:20
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    \$\begingroup\$ Can you please clarify what "Output" is? If the two outputs are connected together, the circuit is fine. But if that is the case, why not draw it that way? I agree with Dave that R6 is probably too big. Not sure how they came up with 1.5mA. I think Sparky's calculation is closer to correct. \$\endgroup\$ – mkeith Apr 20 '18 at 21:49
  • \$\begingroup\$ For some reason I assumed LED was exposed to reverse voltage. \$\endgroup\$ – user33915 Apr 21 '18 at 6:49
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The LED is not exposed to any reverse voltage, so that is not an issue. The 2 diodes with Q1 and R7 form a constant current sink. As long as the LED source voltage is 5 volts or more (up to the 40 volt rating of Q1) and R7 is 150 ohms the current will be 4.33 mA, even if output is tied to +27 volts. You could put 5 LED's in series and still the current would be fixed at 4.33 mA.

Typical round 5mm LED's usually have a maximum current of 20mA, so R7 should not be less than 43 ohms, which gives you 15 mA for the LED (Very bright!). A typical 47 ohm resistor for R7 would give you 13 mA of current.

The math is simple. The 2 diodes provide a fixed 1.3 volt reference to the base of Q1. The base-emitter junction of Q1 drops that by .65 volts, leaving you with a fixed 0.65 volts at the emitter and across R7.

So if all things stay the same then adjusting R7 adjust the constant current at the Q1 collector. Even if you bypass the LED Q1 will only sink 4.33 mA.

This simple math means that if R7 is 650 ohms, then the current is fixed at 1.00 mA. At 65 ohms it becomes 10.0 mA, etc. At 6.5 ohms the current would be 100 mA and likely destroy the LED, and Q1 is rated for only 200 mA.

R7 has a limited range to light the LED from very dim to very bright. To protect the LED R7 should be no lower than 43 ohms (15 mA), and above 10 K ohms the current (6.5 uA) would not be enough to light the LED and may reach the cut-off current for Q1. At 1K ohms the LED current is 650 uA, which would make it very dim even in a dark room. So your 'safe' range (R7) for the LED current is ~43 ohms to ~1K ohm.

By the way, if the LED has no connection to power then the circuit has no current flow. If the LED output were shorted to ground (with no power attached) then there is no harm. The collector of Q1 cannot source current, or less than 1 uA if it leaked from Q1 base which has a 1 M resistor for a current source. Worst case would be .027 uA, which is not enough to do any harm.

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