I'm designing a circuit with an LM3914 dot/bar display driver and want to shut off the entire bank of LEDs based on conditions in the circuit. I'm having some difficulty turning off the driver from the information in the datasheet.

The LM3914 doesn't have an "enable" pin, but the datasheet (PDF here) states the following on page 19:

enter image description here

Perhaps I'm misunderstanding the text, but I take this to mean that the internal current sources (e.g., for the LEDs) can be turned off by supplying 100 µA through the REF_OUT pin. I tried this on a breadboard by connecting pin 7 to +5V through a 47 kOhm resistor, but the LEDs remained illuminated (I also verified the current draw with a multimeter). Am I missing something?

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    \$\begingroup\$ Best way to kill the display is a tiny NPN or N-channel MOSFET used to short input to ground. A 2.2K series resistor (at the input) can prevent the signal source from being shorted to ground. \$\endgroup\$ – Sparky256 Mar 22 at 22:10
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    \$\begingroup\$ I haven't tried this, so I'll make a suggestion. If you try it and it works, then you can make an answer of it. Simply disconnect pin 7. If no current flows through pin 7, then the LED current will drop. The LED current is proportional to the current out of pin 7. No current out, no LED current. Put a transistor between pin 7 and the resistors used to set the LED current. \$\endgroup\$ – JRE Mar 22 at 22:26
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    \$\begingroup\$ Try pulling the REF-ADJ pin high (connecting it to V+) as they seem to suggest, the only mention the min current because needed to keep the circuit on, as far as I understand. This should pull the non-inverting inputs of all 10 current source op-amps high pulling the outputs high and turning the LEDs off (my understanding of the datasheet, I may be wrong) \$\endgroup\$ – axk Mar 22 at 22:31
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    \$\begingroup\$ The current through 47K from 5V to 1.25V is less than 100uA. \$\endgroup\$ – Spehro Pefhany Mar 23 at 0:45
  • \$\begingroup\$ @Sparky256 I like this approach and it's also recommended via the datasheet excerpt above. I've included this method in my answer below. Could you please comment if I implemented your suggestion correctly? \$\endgroup\$ – higrafey Mar 23 at 21:17

Thanks to everyone who commented. Based on the responses, I have two solutions for blanking the LM3914. I think the first method gets close to making sense of the datasheet in my original question. The second method illustrates a technique to "gate off" the input signal, also per the datasheet. I tested both of these solutions on the bench and they seem to work fine.

Note: The LM3914 provides 1.25 V via REF_OUT when REF_ADJ is connected to ground. In the following schematics, the internal voltage reference of 1.25 V is tied directly to the divider network, producing 125 mV between steps. I left off the transistor pull-down resistors for the sake of brevity.

Method 1: Shorting the Voltage Reference

The LM3914 display will turn off if no current flows through the voltage reference output REF_OUT. According to the datasheet, "pulling positive on the reference" will turn off the internal current sources, essentially shutting off the chip. In my testing with a 5 V supply voltage, I measured a standby current of only 150 µA in this state.

In the schematic below, R1 sets the LED current according to the datasheet formula \$I_{LED}\approx\frac{12.5}{\mathit{R1}}\$. When Q1 is on, current flows across R1 to ground and the display operates normally. When Q1 is off, REF_OUT is pulled to +5V via R1 and R2, shutting off the display (and just about everything else inside the chip).

Note that if R_HI is tied to REF_OUT, and R_LO is tied to ground (as shown below), leakage through the internal resistor network may keep the display dimly illuminated. Thus, R2 must be sufficiently low to offset the leakage and flow enough current from the +5V rail. I chose 22 kΩ empirically and measured about 180 µA through REF_OUT. If the internal resistor network is connected to a different voltage reference, I've been able to blank the display (and shut off the chip) with as little as 40 µA through REF_OUT. I think the 100 µA mentioned in the datasheet is more of a guideline and depends on the wiring arrangement.

enter image description here

Method 2: Shorting the Input Signal

The LM3914 display will turn off when the input signal is grounded. This can be accomplished with an NPN transistor as shown below. When Q1 is off, the LM3914 sees the signal voltage and operates normally. When Q1 is on, SIG is essentially grounded through the transistor. R2 was chosen to limit the Q1 "on" current to about 125 µA maximum; the input bias current at SIG is only 25-100 nA, so this should be fine. R3 was chosen to ensure Q1 is fully saturated when turned on; it passes about 44 µA and could probably be a lot higher than 100 kΩ. During testing, I found that the \$V_{CE}\$ of Q1 (and thus the voltage on SIG) was less than 30 mV, which is considerably lower than the minimum 125 mV required to illuminate the first LED. So, the display stays off.

Unlike Method 1, which draws hardly any current when blanked, all of the internal current sources are still operating when the input signal is grounded. In my testing with a supply voltage of 5 V, this amounted to about 6 mA. Not much, but still far greater than the 150 µA observed with Method 1.

This approach worked when I built the circuit. Is it an appropriate (i.e., "best practices") method to ground an analog signal?

enter image description here


Your solution is good, and a simple way to kill the display. Shorting REF_OUT and R_HI to ground is not a direct short, because resistors of specific value are used. The first option does not make sense, as in your lower option REF_OUT and R_HI are connected to ground via the same 1.2K resistor.

The lower schematic does kill the signal, which usually comes from a peak or average detector. R2 could be 10K, but only if source is an op-amp with low output impedance. The 100K driving a 2N3904 is ok, but it will not pull input down to zero volts.

Consider a small logic MOSFET which can pull the input down to zero volts. The LM3914 and LM3915 can have the lowest level LED come on with very little signal, so not sure a 2N3904 can shut off the most sensitive LED's. Your choice of 100K at the input may help with this issue.

EDIT: Due to better explanations from the OP: Ok, I see what is going on now. In the first schematic, the logic has to be reversed to cut off the LEDs, but it will work just fine, and more simple than trying to clamp the input to zero volts.

  • \$\begingroup\$ Thanks for the feedback. I'll try a MOSFET with Method 2 and see how it goes. Can you elaborate on your concerns with Method 1? The idea is that Q1 will connect R_HI and REF_OUT to "almost" ground (Vce ≈ 100 mV) when turned on and set the LED current via R1. When Q1 is off, R_HI and REF_OUT are connected to +5V through R1+R2; the reverse current flow shuts off the chip and seems to match the operation described in the datasheet. \$\endgroup\$ – higrafey Mar 23 at 23:57
  • \$\begingroup\$ Ok, I see what is going on now. In the first schematic, the logic has to be reversed to cut off the LEDs, but it will work just fine, and more simple than trying to clamp the input to zero volts. Good luck. \$\endgroup\$ – Sparky256 Mar 24 at 2:18
  • \$\begingroup\$ Right, I should have mentioned the logic was opposite. Thanks! \$\endgroup\$ – higrafey Mar 24 at 12:31

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