# What are these transistors doing and how are they working in this VU Meter?

I was thinking of making KITT's voice box from Knight Rider, which is essentially a VU meter. I did some researching and found an appropriate IC, LM3916,

http://www.mouser.com/ds/2/405/lm3915-443929.pdf

and some LED Bar Graphs. I actually ended up finding what seems to be an old schematic, in this link (it uses the LM3915). There's a group of PNP BJT's in it. Schematic It's a bit hard to follow, I made my own so I can follow the wiring a bit better, if you would like to look it's below(It's missing the resistors in the LED bar graphs, it should be 180ohms like the original schematic) . (My design uses 625ohms due to a mistake I made reading the datasheet. They should be 1k)

The author writes the BJT's are for to light multiple LED's (since each LED output on the chip is suppose to only supply 1 LED). So essentially he uses the IC to select between the transistors which will turn on the LED's.

I'm confused on how they are doing this. It's a PNP BJT. According to the LM3915 datasheet, the current this IC will be outputting at the LED outputs is 12.5mA, since 12.5V/1k. Then 12.5mA going through the 1k resistors at the outputs give 12.5V again.

So first, Vb = 12.5V. Ve = 12V from the supply. Veb should be .7V, but this gives -.5V. Vb would have to be 11.8V to have a Veb of .7V

Second it should be Ie = Ib+Ic, meaning the base current should be flowing out, but the current is coming in from the IC.

I tried making a simple overview of 1 LED, not sure if it's correct Third, let's say there was just 12V going to the LED's. The design the person has uses 180ohms. (12-Vf)/180 = If. Red LED's I read generally have around 2V Vf, so 12-2/180 is approximately 55mA. That seems extremely high, and it also produces really high wattage.

Thanks!

LED Bar Graph Data Sheet: https://www.sparkfun.com/datasheets/Components/LED/YSLB-102510R3-10.pdf

• your only question is referring to "these transistors" .... what transistors are you talking about? – jsotola Jul 3 '19 at 23:30
• The LED datasheet says the Absolute Maximum current is 20 mA. The LEDs will work (and be happier) with lower currents - you might try 10 mA. (The higher the current the brighter the LED, but I find most are bright enough at 10 mA or less.) – Peter Bennett Jul 3 '19 at 23:32
• ? Not sure what you mean, there's a group of transistors with a box around them in the schematic there that all serve the same purpose, Edit: @PeterBennett, ah, well even then, the designer would have 55mA running through them, unless he found some LED's that need around 10.75 forward voltage, and I couldn't find any LED's like that haha – TheCoolest2 Jul 3 '19 at 23:32
• I didn't check the datasheet, but 12.5mA is probably the maximum current before the chip malfunctions. It doesn't always output 12.5mA. – user253751 Jul 4 '19 at 0:12
• @immibis You are right! Thanks. – TheCoolest2 Jul 4 '19 at 1:25

1. Not sure where you get 12.5V from. If the supply is 12V at the emitter, then use that. The base would be 12V - 0.7V = 11.3V through the resistor 11.3V / 1000Ω = 11.3mA. And while the Collector would not be exactly 12V, we can assume no Vce drop for practical use here.

2. You are right. The current at the Emitter should be the current flowing through the base and the current through the collector. But the current is being SUNK by the LM3915 to turn the PNP transistor ON. Which means the IC is pulling the line low to turn on. And it's an open-collector internally, so it goes High-impedance to turn off. So Ie is indeed Ib+Ic. If the output SOURCED current, into a NPN's base, it would flow into the base as you expect.

3. You are correct, if it's using 12V. But we can't really tell. The Supply is listed as 12V AND the regulator is listed as 12V, you typically don't need a regulator to regulate a voltage to the same voltage. If we assume a 7805 regulator, so 5V out, (5V - 2V) / 180 = 16mA. Which is a reasonable current for a bargraph red led. Which we do not know if it actually is a red bargraph either. And we don't know what voltage the audio transformer works at.

Basically there's a lot of guess work required for that specific schematic which lacks part numbers or specs. But between it and the LM3915 data sheet and app notes for LM3915 circuits, you can figure out how to replicate it with parts you have.

Sidenote: Since the V+ voltage is likely 5V, the current through the base resistor would actually be 5V - 0.7V ) / 1000Ω = 4.3mA, which with a gain of 25~30 is plenty to put the transistors in saturation for the <100mA the 6 leds per output would take. The output current isn't too critical here as the transistors are just being used as an on/off switch.

• Thanks for the response For 1), if you look at the datasheet for the IC, page 2, there's an equation for the LED current. ILed=12.5/R1. (That's if you do not use Pin 8). So in that schematic, 12.5/1k = 12.5mA. More on page 8. 2) Thanks for the explanation! It's a bit late so I will try to understand more tomorrow but I got an idea of it now. For 3)Thanks. A lot of people I asked seemed to ignore the 55mA, I figured there has to be a mistake. Yeah a lot of guessing on this schematic! I did this quick as I have to sleep, but do my calculations look right? imgur.com/a/Vb74lM0 – TheCoolest2 Jul 4 '19 at 3:49
• That formula is specifically for Figure 1, a 0 to 10V volt meter. The schematic you provide is different, but you get a similar number. The 1.2V out of the reference voltage, pulled down by a 1K you get 1.2mA as the reference current which is multipled by 10 to get the 12mA of the led output current. There's no current/voltage regulation happening at the outputs because the 1K base resistor essentials caps the current being pulled. You can't pull 1 gallon from a 16 oz bottle type situation. In this schematic, 12.5V is not relevant at all. – Passerby Jul 4 '19 at 4:30
• @TheCoolest2 the calculations look okay. You have to keep in mind that leds work on a range of currents, and 20mA are the nominal safe max current for small leds. You can run them at a lower voltage/current and they will still work just less bright. With a 180Ω resistor, it will draw sightly less current but due to the logarithmic visual response it will seem almost as bright as 20mA. Plus real life isn't as fine grain as the math we use. The transistor may not drop exactly 0.2V, the led may draw 2.1V at 20mA, the resistor is 10% tolerance, etc. So all these small changes affect the real thing – Passerby Jul 4 '19 at 4:35
• Thanks! Also, I figured 20mA was the min, not the max. After some reading I learned yeah, 20mA is usually the max :). As for the not at .2V, yeah, this is ideal. I would have to factor in stuff like tolerances and non-ideal scenarios. Thanks! :) – TheCoolest2 Jul 7 '19 at 2:39

I built a project with LM3915s almost 40 years ago. The part has a programmable current sink so you can connect LEDs directly to it. Depending on the color, you can put 4 or so LEDs in series at 12V. Four red LEDs for sure, since that is what I used. More than this, you can use a PNP, but you should still put some in series so you don't waste a lot of power in the resistor. For your design you seem to want 6 LEDs in each circuit, I would put 3 in series, each path needs a resistor. The transistor will saturate, so you need to limit the current.

Rough calculation for the series resistor: Vce = ~0, each red LED drop = ~2V

R = (12 - 2 * 3) / 10 mA = 600 ohms simulate this circuit – Schematic created using CircuitLab

Edit: Another idea: The LM3915 is not expensive, use two, then you can use 12V without transistors. Send your audio input to both. Divide the 6 LEDs into two groups, 2 & 4 or 3 & 3.

Note that the original manufacture doesn't make these anymore, you are most likely buying a Chinese clone. Therefore, they may not match the datasheet exactly.

• Thanks! I was wondering if you can put multiple LED's in series with this. I'll get back to you :) – TheCoolest2 Jul 4 '19 at 3:51
• As this whole thing is self-contained, the easiest scheme is to put all corresponding LEDs in series and drive them directly from the LM3914. The 3914 is rated at 25V on its outputs, so 6 bargraphs could be driven from a 15V supply, and 8 from a 18V supply. This gets rid of all PNP transistors and a load of resistors. (As the power supply looks strange - 12V in and out? - this is probably the easiest area to change.) Another advantage is that the bargraphs can be wired side-to-side or fixed to stripboard. – henros Jul 4 '19 at 9:25
• @henros - I assumed that the OP was in a car. If yes, the regulator should be replaced with a filter. Or, use a boost regulator to get to 15V and put all 6 LEDs in series. – Mattman944 Jul 4 '19 at 11:38
• @TheCoolest2 - Your audio input is very crude, the LM3915 input expects DC, the AC will never allow the LEDs to be at full brightness. I used a peak detector with decay. Since this is a completely different topic, I recommend you ask a new question when you get to this. – Mattman944 Jul 4 '19 at 11:46
1. The output level from a microphone is much too low to feed an LM3915 or LM3915. You need a preamp.

2. The 12V regulator will not work unless its input is +14V or more.

3. The base resistors are not needed because the IC has a programmable output current which is 10 times the current from pin 7 to ground.

4. If each group of 6 paralleled LEDs use 60mA then the base current of the transistors should be 6mA then the pin 7 current should be 0.6mA then the resistor from pin 7 to ground should be 1.28V/0.6mA= 2.1k ohms. Use 2.2k.

5. Connect 3 LEDs in series as a string then connect two strings in series for each transistor then the current limiting resistors can heat much less. Then the resistor from pin 7 to ground can be 6.2k.

• Yeah, I will not be using a microphone, that was just a placeholder. I think a 3.5mm audio jack should suffice for testing. 2) Why will the regulator not work? 3) Yeah, that's what I was thinking. What's confusing me is some people are saying that iLed equation is only for specific applications. I figured this would be one of them, no? 5) Alright will do, thanks! – TheCoolest2 Jul 11 '19 at 23:01

I'm confused on how they are doing this. It's a PNP BJT. According to the LM3915 datasheet, the current this IC will be outputting at the LED outputs is 12.5mA, since 12.5V/1k. Then 12.5mA going through the 1k resistors at the outputs give 12.5V again.

Yes, but you've fundamentally misunderstood the output stage of the LM3915. It does not source current, it sinks it. That is, the output pins look like an NPN transistor to ground. It is intended to work by effectively shorting the output pin to ground in order to draw current from a higher voltage. (Actually, it works as a current sink rather than a short, but this is a detail.) To get an idea of this, the first thing you should have done was looked up the data sheet. Notice how the LEDs are connected in the Typical Applications (page 2).

With this said, this is actually a pretty crappy circuit. Each output pin, when active, will attempt to pull something like 15 mA from the bases of the PNP transistors. Since the LED current will be something like 55 mA (12 volts minus about 0.2 volts (PNP Vce) minus about 2 volts for the LEDs, divided by 180 ohms), a base drive of about 5 mA would be quite adequate. A good rule of thumb for driving BJTs as switches is to use a gain of 10. If the data sheet were followed with a base current of 5 mA calculated, each output would cause about 60 mW of power dissipation in the 3915 (12 volts times .005 amps), and the worst-case power dissipation in the chip would be about 600 mW, which would only occur when all segments are lit. For a normally-operating bar graph, this simply does not happen for any period of time.

So first, Vb = 12.5V. Ve = 12V from the supply. Veb should be .7V, but this gives -.5V. Vb would have to be 11.8V to have a Veb of .7V

Nope, 12.5 is used in the data sheet as a maximum value. If you are actually using 12 volts, Vbe will be 0.7 (more or less) and Vb will be 11.3 volts. Keep in mind that the output stage of the 3915 will not produce a "higher" voltage - it always drops voltage.

Second it should be Ie = Ib+Ic, meaning the base current should be flowing out, but the current is coming in from the IC.

Yes, Ie = Ib + Ic, but base current flows out from the base of a PNP. Base current will be a bit less than 12 mA, and LED (collector) current will in fact be about 55 mA, for a total emitter current of about 67 mA per lighted segment.

I tried making a simple overview of 1 LED, not sure if it's correct

Almost correct. The current source is pointing the wrong way. Furthermore, the 3915 output stage is not a perfect current source. so the base current will be less than 12.5 mA. In practice, the left-hand side of the base resistor will appear to be at a voltage of about a volt or so.

Third, let's say there was just 12V going to the LED's. The design the person has uses 180ohms. (12-Vf)/180 = If. Red LED's I read generally have around 2V Vf, so 12-2/180 is approximately 55mA. That seems extremely high, and it also produces really high wattage.

Well-spotted. The 180 ohm resistor is a bit underpowered, as it is a 500 mW unit being asked to dissipate 550 mW. However, this is probably not an issue, since a 10% overpower condition is not usually fatal. Manufacturers do build making conservative assumptions about how the resistors will be used. If you put this in a small, closed box with little airflow, there may well be long-term problems, since the resistors will get hot. However, as I commented earlier, audio bar graphs spend most of their time with only one or two segments lit, so the heat will spread out some.

But, like I say, this is a pretty crappy version of the circuit, and should be a lesson to you about blindly grabbing circuits from the internet.

• Got it. Thinking it sources the current at the output was really confusing me with a lot of things. For the current values, isn't the collector current going to be 55mA, not the emitter? The LED current is coming from the collector. Also indeed, unfortunately it seems many people I've talked to about this circuit just seemed to copy it without understanding, which is why I posted to ask! :) – TheCoolest2 Jul 7 '19 at 2:45
• @TheCoolest2 - Again, well-spotted. I did mix up emitter and collector. I've corrected. Thanks. – WhatRoughBeast Jul 8 '19 at 20:59
• Thanks for the help @WhatRoughBeast! I will hopefully have the cash soon to try and put this together :) – TheCoolest2 Jul 11 '19 at 22:56