Efficient relay driver design

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sam781

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I measured relay coil resistance using DMM.
5v relay coil resistance=102ohm
9v relay coil resistance=220ohm

I checked the relay with a NPN transistor and a base resistance like this https://i.stack.imgur.com/Oarz3.gif

I used base resistance as 1k ohm with supply voltage 5v. 5v relay works fine. When the relay is on, it the driver circuit consumes power. So, I need to design a efficient driver which will consume less power without sacrificing switching strength.
 

your 5v relay needs 5/100 ~ 50mA to operate. It will probably operate with lesser volts/ current too.
You need to ensure your NPN goes into saturation for mininmal power loss there, so your base current should be > Ic/hFE

typical Vsat would be ~ 0.2v, so your NPN will dissipate 0.2 x 0.05 ~ 10mW

Lets say you chose an NPN of hFE >= 100
then Ib > 0.05/ 100

Assume your drive voltage (from uC ?) is worst case 4v. Then your base resistor should be < (4 - 0.8) / Ib
so for this example you can choose base resistor to be ~ 6.4Kohm or less.

for a higher hFE transistor, you can increase the base resistor.
 



Use Ohm law I=U/R to see coil current. Always buy relays with higher coil resistance, this will ensure lower current.

But this is interesting to see, methods how to reduce relay coil current :

Simple Circuit Reduces Relay Coil Power
**broken link removed**

**broken link removed**

Reducing Relay Power Consumption
**broken link removed**
 

Thanks a lot both of you.

From post#2, it is clear that a relay can be driven without any series resistance. Earlier, I thought it needs a series resistance.

BJT power dissipation calculation is not clear? Is there no involvement of Ic?
For a particular amount of Ic we can reduce power dissipation by choosing higher value of hFE where we need low Ib. Can you please mention some commonly used BJT number having higher hFE which can be used in driving 5v relay?
 

Varouis options for efficient relay driving (ordered by efficiency)

1) Use latching relays --> only need power when changing its state.
2) Using a relay driver with high initial current "during the switching process" and lower current afterwards (the circuit posted by tpetar is a simple option to achieve this initial current peak)
3) Use high resistance relay (low power relays)
4) Drive the relay at its minimum voltage

I would not recommend simply limiting relay current too much as you may face switching reliability issues during production or with relay aging...
 

Regarding reliability - a variation on the latching relay mentioned above is to have a periodic pulse (e.g. every sec) as well. That way, if the relay drops out (e.g. due to it being knocked or vibration for example), then it will switch back on again eventually.
 

sam781 said:
So, I need to design a efficient driver which will consume less power without sacrificing switching strength.
Click to expand...

don't listen to these other ideas... they are all tricky ways which will fail if you don't know what you are doing.

Just put a 6.8K resistor, or better yet, a 4.7K resistor in the base circuit and you'll be fine. Use any ordinary NPN transistor like 2n2222 or 2n3904 or bc847B. All will work fine and give good enough efficiency.
 

The latching relay with periodic pulse is a well established, extremely low power method. Certain battery powered scenarios
that absolutely require relays do implement it.
Your method of reducing the base resistor, whilst it works, saves just a few mA.
 
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    tpetar

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If transistor drive current is a concern (although this may reduce only a few mA), use a n channel mossfet instead of NPN.

In the other hand, this "high efficency" should be clarified... how many mW do you want to consume? If you say 150 mW, there are high efficiency relays available. If you say 50 mW you will need other solutions such as:
A) Latching relay according to sky_123 explanations
B) Non-latching relay with nominal current during switching and reduced current in standby
C) Using solid state relays?
 

High efficiency with low power consumption like 50mW. Latching relay is not widely available here.
 

A well specified relay will have a "pull-in voltage" (or current) and "holding voltage" making a low power solution as suggested by sky_123 and tpetar perfectly possible.

Keith
 

If a relay coil has a given resistance, and requires a certain coil current for pull in, then it will inherently require some minimal amount of power to operate (I^2 * R). There is nothing your driver circuit can do to decrease that power requirement. To operate with less power, you would have to redesign the relay itself, which probably isn't very feasible.

For dissipation in the driver circuit itself, that depends on the voltages available to you and the voltage required by the relay. If you have a 5V rated relay coil and a 5V supply, then just drive it with a low Rds FET, and that will get great efficiency. If you have 12V available instead of 5V, then getting low losses is more complicated, and will likely require a pulse width modulation scheme somewhere.
 

Integrated energy efficient current controlled PWM relay driver allow the use of 5V or 9V on the same 12V supply. Like here described: **broken link removed** .
 


After pull-in it can reduce the voltage/current to the "holding" current, thereby reducing power.

Keith
 
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    mtwieg

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sky_123 said:
Your method of reducing the base resistor, whilst it works, saves just a few mA.
Click to expand...

the objective was not to save mA, but to ensure the NPN goes into saturation for minimal Vce drop & hence minimal power dissipation in the bjt.
The OP requires a basic answer here, not complicated pwm sophistry.

- - - Updated - - -

sam781 said:
High efficiency with low power consumption like 50mW. Latching relay is not widely available here.
Click to expand...

do you have any other parameters for your relay except for voltage & resistance ? This voltage would normally be the 'pull-in' voltage. With no other parameter it is impossible to design any other reliable method for your relay.

- - - Updated - - -

sam781 said:
Thanks a lot both of you.

From post#2, it is clear that a relay can be driven without any series resistance. Earlier, I thought it needs a series resistance.
Click to expand...
adding a series resistor will cause a lesser voltage across the relay, reduce its pull-in strength, and possibly cause it to not 'switch' at all.


power loss in the bjt is ~ Vce x Ic. If the bjt is in saturation, then power loss in bjt is 0.2 x 0.05 ~ 10mW
power used in relay is ( 5- 0.2)^2 / 100 ~ 230mW

so your driver is only using ~4%. That sounds fairly efficient
 

Thanks a lot kripacharya to clarify each and every important points. I've got all the necessary information that I had desired.
 
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