voltage drop

[George L.]

hello,

I am having a problem with a circuit I am building. I have attached the "problem" part of the circuit to this message.....

(circuit is simplified)

The problem is the relay will not turn on, I have been told that there is not enough voltage drop across the relay, but I have not been offered a solution.

The votage acroos both 10K resistors are almost exactly the same at 7.04 volts, and the current os extermely low (1.5 mA).

If anyone has a solution please help me.

Thank You,

George

 

[Klaus]

Its a strange way to turn on a relay .
take the 10K resistor on the right out altogether, its only wasting power.
Reduce the value of the resistor in series with the relay until the relay pulls in securely. If your battery is 7V (6V battery?) then it would need to be considerably smaller than 10K. If you provide the relay coil resistance it can be calculated easily.
Klaus

 

[evandude]

if it's a 5v relay you could probably run it at 7 volts anyway. Or if you want to be safe, just use a multimeter to measure the relay coil resistance, and add a resistor in series with it that is 40% of that value.

 

[lord loh.]

I was once trying to turn on a 5v / 100 ohm relay. My calculation revealed that I would have to put a 10 ohm resistor in series to get the desired voltage drop....so it's power dessipation was 2W !!! or so....
[Do not remember all the figures...but whatever I mentioned is correct]

My teacher told me to put the relay at the collector of a transistor...He said that magnetization of coil depends on current than on voltage. I still have not been able to figure this out... But may be it might help you.

 

[ljcox]

My teacher told me to put the relay at the collector of a transistor...He said that magnetization of coil depends on current than on voltage. I still have not been able to figure this out... But may be it might help you.

Relays have an electromagnet that is energised by current through the coil. This operates an armature that moves the springs to make or break the connection.

Ohm's law tells us that the current is dependent upon both the voltage and the resistance of the coil. So if you look at the relay data, it should give you the operate current and the release current.

You should ensure that the current is about 1.5 times the stated operate current to ensure reliable operation. Once the relay is operated, you can reduce the power consumption (if you wish) by switching in a series resistor to reduce the current to about 1.5 times the specified release current. This is not normally done as it adds extra complexity, but is useful if the power consumption is critical (eg. battery operated devices)

Len

 

[Miles Prower]

Here is a simple relay driver that will operate with greater efficiency.

To design, you need to know the relay coil current. This will become your collector current. The current through the bias resistors should be selected to be: I(bias)= [10 X I(C)]/h(FE). The voltage at the base is: 0.6 + 0.7= 1.3V. This will be the voltage across the lower bias resistor. The upper resistor simply absorbs the excess voltage from V(CC): V(CC) - V(base).

This circuit won't waste power as would using a droping resistor in series with the coil. Closing the switch will open the relay, and vice versa.

Hope this helps.

 

[ljcox]

The resistor between the base and gnd is not necessary in this case since the switch shorts the base to gnd when closed.

Len

 

[Nigel Goodwin]

The resistor between the base and gnd is not necessary in this case since the switch shorts the base to gnd when closed.

Len

The circuit also doesn't make any sense?, it's dropping the entire supply voltage (minus the Vce of the transistor and 0.7V drop across the emitter diode) across the relay. So effectively all it's doing is putting a diode in series with the relay coil and dropping 0.7V?.

I 'suppose' by carefully selecting the base feed resistor you could only turn the transistor partly ON, but this would dissipate the excessive heat inside the transistor, most probably causing it to fail?.

 

[audioguru]

This circuit won't waste power as would using a dropping resistor in series with the coil.

Sorry Miles. It will waste exactly the same power as a series dropping resistor if it dropped the correct amount of voltage, plus a little more for the transistor bias current.
Only a switching regulator wouldn't waste much power.

 

[George L.]

Ok,

Maybe the person that helped me didn’t lead me in the right direction…

I posted the entire circuit.

The problem originally was that I couldn’t turn on more than one relay using the circuit.

I would like to be able to turn on up to three relays wired in parallel or series at one time. Currently, I am having trouble getting one to turn on! It worked with one relay and no resistors, but the current is 37mA !!!

Please help

Thank you,

George

 

[audioguru]

1) The 10k resistor in series with the relay coil forms a voltage divider: 0.2V across the relay coil and 8.8V across the 10k resistor because the resistor's value is much too high. A 180 ohm resistor in series with the relay coil would would allow 5.23V to be across the relay coil and the current will be about 21mA. I figure that the relay coil is 250 ohms since it draws 37mA from 9V.
2) Each relay coil will need a 180 ohm resistor in series with it or you can parallel 3 of them and put the combination in series with a single 62 ohm resistor.
3) The 10k resistor across the relay coil and its series resistor won't do much to stop the hundreds of volts generated when the relay is turned off. The voltage spike can destroy the transistor. A reverse diode across the coil or paralleled coils instead of the resistor would arrest the voltage spike completely.
4) When you turn down the "timing" pot, it will be shorting the battery when the switch is turned-on and it will burn and kill the battery. :lol:

 

[eblc1388]

Hi George L.,

To help you in putting the circuit together, I drew up the following circuit which included ideas by Audioguru.

If you tell us the time you would like the relays to stay ON after you release the push button, Audioguru can quickly tell you what the value of C and R to use as he loves RC time constant.

 

[ljcox]

The resistor between the base and gnd is not necessary in this case since the switch shorts the base to gnd when closed.

Len

The circuit also doesn't make any sense?, it's dropping the entire supply voltage (minus the Vce of the transistor and 0.7V drop across the emitter diode) across the relay. So effectively all it's doing is putting a diode in series with the relay coil and dropping 0.7V?.

I 'suppose' by carefully selecting the base feed resistor you could only turn the transistor partly ON, but this would dissipate the excessive heat inside the transistor, most probably causing it to fail?.

Nigel,
Having designed many electronic/relay interfaces, I can assure you that the circuit does make sense. The relay is operated when the switch is open - provided there is enough base current to saturate the transistor.

Len

 

[audioguru]

I agree with Nigel.

 

[George L.]

WOW!

thanks everyone...

since I don't know if eblc1388's diagram is correct, I thought I would draw out Audioguru's suggestions. The diagram is attached. I am trying to keep the potentiometer because I may need to adjust the amount of time the relay stays activated.

Will the attached circuit work, and also will it work with up to 5 relays.

Thanks,

George

 

[audioguru]

Sorry George,
I could tell that you are a noobie because you don't know anything about voltage, current, resistors and relays.
I didn't realise until now and nobody else did either that the capacitor, diode and battery are all backwards!

I think I said before that each relay needs a series 180 ohm resistor, or all 3 relays can be paralleled then the combination needs a single 62 ohm series resistor.
The 2N3906 transistor is rated for 200mA so can drive up to 10 of your 20mA relays.

 

[Klaus]

George, you picked the wrong type transistor for your latest circuit. You need a NPN type for it to work. Also, you need to make sure the combined current from the 5 relays can be handled by the transistor without overheating it.
If you are unsure of the transistor specs just type the number into 'Google' and you'll most likely find the specs amongst the hits. Ditto for the relay.
Klaus

 

[George L.]

Audioguru, DO NOT talk to me like that! :evil:

obviosly I am a newbie, why would would I be asking these questions if I knew the answers.

I have this circuit on my breadboard and it is currently working with one relay. I think I just didn't draw the circuit correctly.

Will my circuit work with with the 2N3906 PNP transistor or do I need to change it.

Thanks,

George

 

[audioguru]

Hi Klaus,
A PNP transistor will work exactly the same as an NPN transistor in his circuit, except of course the polaritiy of the cap, diode and battery must be correct.

The transistor saturates so it doesn't dissipate much heat. With the rather high value for its base resistor and more added from the pot, if the transistor barely saturates with a 1V drop, it will dissipate only 93mW when it is driving five 20mA relays. :lol:

 

[audioguru]

Well George, I am commenting on the schematics that you posted, aren't I?
I suppose it is for a school assignment. Didn't your teacher teach you anything?
Before I said I was sorry and I guess I still am. :lol: