Dear All

I have a small circuit using an LM3914 to drive a 10-segment LED bar graph. The supply voltage to the circuit is 5 VDC.

If I measure the voltage across any LED when lit I get about 2.3 volts.

My question is, can I use that 2.3 volts to drive a simple transistor switch that will then turn on a 5 volt relay coil without effecting the LED brightness? I hope I've explained that adequately.

I've looked at sample circuits using a transistor as a switch but haven't succeeded in making it work.

Thanks and regards

Trevor

 

Most probably yes. See circuit below.

Attached Images

lm3914_relay.png (3.8 KB, 362 views)

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L.Chung

 

Hi LC

Many thanks for this. It's really helpful. I'll breadboard it up and see what happens.

Trevor

 

If you use a ULN2003 or 2803, you can have 7/8 switches @500ma each!

 

Hi LC

Thanks again for the circuit diagram. I can't get it to work, probably because I'm doing something stupid. I'm running the whole circuit from 12 V dc now but the voltage across the LEDs is only 2 volts. Can I still use this 2 volts to switch the transistor on and would I need to replace the relay with a 12volt one?

Sorry to be asking such dumb questions.

Trevor

 

2V across the LED is sufficient to switch a PNP transistor ON. Of course you'll need to use a 12V relay instead or places a resistor in series with the coil of the 5V one to make it operable using 12V.

Please posts the LM 3914 LED circuit you are using as there are so many variations. Also can you measure the coil resistance of the 5V relay so I can recommend the value of the series resistor.

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L.Chung

 

Hi LC

This is very kind of you. The coil resistance is 171.1 ohms. Do the other resistor values need changing for a 12v supply also?

Is it theoretically correct that any voltage could be used to say, drive a relay or LED directly as long as there's adequate current limiting?

The whole circuit can be found on this board under "Voltage Dividing Networks" which was my first post. The thread is 2 pages long. It's an attempt to build a fuel gauge for a motorcycle with a 10-segment bar graph display. The circuit I'm using is the one right at the end "Gas Gauge - Final" which was drafted for me by Audioguru. The system is pretty much complete but now I just want to add an additional "low fuel" alarm hence the transistor switch and relay.

I should say I've had an enormous amount of help from members on this board as you will see. As I didn't want to keep pestering the same people, I posted a new query about the LED voltage.

Your help is very much appreciated. This is a huge learning experience for me. I'm too old to think of sitting in a classroom to learn electronics. I'd much rather be learning while trying to make something I actually need (I have much more expertise with metal machining). Of course, there's enormous gaps in my knowledge and I have to ask for help.

Thanks again

Trevor

 

The calculation as follow: the relay is 5V and has a resitance of 171 ohms. Therefore the coil current is I=V/R = 5/171 = 29.2mA. When you connect the coil with a series resistor across +12V, the same current should flow. There is 5V across the relay means the rest is across the resistor, which is 7V. Thus the resistance is R=V/I = 7/29.2mA = 240 ohms.

Yes, as long as the voltage is higher than the LED minimum voltage(2V for red, 3~4V for white) or the relay operating voltage.

This is the bit that is worrying. You said "I'm now running it from 12V..." but the circuit design by audioguru is for working at +5V. There is a voltage level shifting required in this case and the previous relay circuit I posted will not work.

I still need you to describe how your circuit is powered, especially what is the voltage of the LED common rail(red arrow) as shown in the image. Is it +5V or +12V?

Think of it this way. If people don't want to answer your question, they just do nothing. If some forum users is clocking up thousands of postings, they are prepared to "see you through" with your problem, so don't worry about it.

Nobody is too old to learn new things. I always learn something myself by answering posts. Sometimes I have to go back to check with a textbook to avoid giving wrong advice so it is also good for me too.

Attached Images

gas_guage-final_x.png (8.8 KB, 259 views)

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L.Chung

 

LC, thank you again.

The LED common rail is 12 volts as I running off a 12 v supply. I've put a 1 watt resistor in series with the rail otherwise the chip gets too hot.

As you correctly point out, Audioguru designed the circuit for 5 volts because that's what I originally asked for. My idea was to use a 5v regulator to stabilize the supply and cut down the huge spikes from the charging circuit. I then thought, if the thing will run off the 12v battery, I can cut down the number of components. The fault is mine for confusing the design. That said, it seems to work ok with 12 or 5 with no apparent change in the points at which the LEDs turn on. If you look at the final photo, you'll see the reed switch assembly I made to fit in the tank. The vertical heights of the switches represent 10 steps of 1.5 litres of fuel. The height of the float for each LED doesn't appear to have changed by increasing the supply to 12 volts.

At this stage, I can easily put the regulator back in and power it from 5volts as originally designed if that's better.

Thanks once again.

Trevor

 

If you need that 1 watt resistor in place to prevent the LM3914 from overheating, then the simplest solution to your requirement is to get a solid-state relay.

A solid-state relay is a black box with 4 connections, 2 on each end. There is no electrical connections between the two ends. Inside the black box at one end is a LED which lights up and the "light" is used to turn on the load switch at the other end.

You simply connect this LED in series with one of the LEDs of the LM3914 so they light up together and bingo the load is turn ON.

Attached Images

ssr.jpg (22.2 KB, 246 views)

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L.Chung

 

Hi LC

That's interesting. I knew about solid state relays but didn't know they had an LED inside.


If I can get one with NC contacts then I'm in business. The relay is powered so it's NC contacts are open. As soons as LED No3 goes out (leaving only No1 and No2 lit) , the NC contacts close and the alarm goes on.

I'll shop around for one. Meanwhile, many thanks indeed for all your help. If it proves impossible to find one, I may have to ask again for help with the transistor driver for the conventional relay.

Regards

Trevor

 

Now I know exactly what you want, I can make a better suggestion for you.

You have 1% chance of find a normally close solid-state relay.

In the new circuit, an opto-coupler is used to replace the solid-state relay. If the LED #3 is on, then the MOSFET transistor will be OFF.

If your aim is to turn ON a buzzer or a warning indicator using the relay, then you might not need to use the relay and the diode afterall. A N-Channel MOSFET can switch a buzzer/indicator without problem. Just put them in place of the relay & diode.

You can choose any n-ch MOSFET with current of 2A and voltage of 100V or higher. (Ignor the 50V rating in schematic, I now consider 100V or hgiher is better)

Attached Images

tank_lvl_alarm.png (4.9 KB, 226 views)

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L.Chung

 

LC

Once again many thanks for doing this for me. I'm going to carry on playing around with the circuit and now, thanks to you, have a much better idea of how to do things (not to mention all the background info).

As I said to Audioguru - it's sometimes hard to know when one is writing too long an explanation or too short. In some cases I realise I didn't give enough clear info about what I was trying to do.

At the risk of being annoying, I may decide to switch back to a 5 volt supply as originally allowed for in Audioguru's schematic. Then I could still use your first transistor circuit. The reasons would be

1. I have all the components in front of me and don't need to shop for more

2. I'd prefer a relay since I don't know what load I might put on it yet. It might be just a red LED or I might want to run a couple of 12 volts lamps.

3. I think a regulated supply will smooth out spikes and dips from the charging circuit, turn signals and brakes light etc.

4. Your first circuit looks simpler.

However, you have given me a range of options. Thank you very much for all this.

Trevor

 

LC

Just to let you know that I got the original 5 volt circuit working. It was necessary to change the 22k resistor to 10k before there was enough current at the base to switch it on. Anyway, it's all working ok now. Many thanks once again for all the help.

Trevor