transistor uses more power?

[littlebill]

noobie alert,

im trying to learn transistor currents now.

i have a relay that use 65ma to run its coil, when i run this through a transistor specifically a bc500, that current goes up to 91ma, i have a voltage drop from 4.85 to 4.61, seems like a significant increase for a small voltage drop, or does the transistor use power to run as well and thats what i am seeing?

[c4757p]

You can't increase current by adding something in series, unless that something is an energy source or a magical negative resistor.

I'd suspect the way you're driving it. Can you draw up a diagram of precisely what you're doing?

[littlebill]

does this help im doing the most basic thing, now before we get mad, i am just learning, i realize the 4.7k is the wrong resistor, but the coil does engage and the relay works which is very confusing because going off the base i get .8ma which means i should be only able to move about 20ma through the transistor, but it looks like over 90 ma is pushing through it.

if i use a 100 ohm resistor the base goes to 34ma, but i only push about 61 ma through the transistor..

[xrunner]

If you turn on the transistor , it won't make the coil of the relay draw more current than it would without the transistor (assuming you are still using 5V). I think you are measuring the current wrong, but I don't know how yet because I'm not in the room with you.

[Richard Crowley]

How exactly did you measure "relay that use 65ma to run its coil"?
Assuming you have the negative side of the power source (battery?) connected to circuit ground.
How much current is flowing through the base resistor (and into the transistor)?  (should be ~ 1 mA)
How much collector current is flowing between the relay coil and the collector?  (should be 65 mA)
As c4757p said, it should be impossible for your circuit to be drawing 91mA unless you are doing something wrong.

Since your transistor (BC550) has a minimum gain (h_FE) of 100, and you are injecting 1mA of base current, it should be within the range that will allow the transistor to pass ~100 mA.  Although, to be safe, it would be preferable to drive the transistor into saturation using perhaps 3~4x more base current (change the resistor to something like 1 ~ 2 K ohms)

[littlebill]

i break the connection at the emitter and run it through my meter, i use 2 different meters, both agree,

after saying that, i broke the connection at the plus side and now its reading right? around 62 ma....

ok if break the connection at the collector, the ma is fine, why is this only a issue when breaking the emitter side?

[xrunner]

You say the relay draws 56 mA with 5V supplied. If the gain the the transistor you chose is 100 (it's prolly more) then you only need the tranny to allow 56 mA, but in this situation you just need it to turn on fully, so you can just say you want more than 56 mA for design purposes - say 200 mA.

So 200 mA / 100 (hFE) = .2 mA or 200 uA base currrent. To get that you would want a base I calculated from:

(5 - 0.6) / .2 mA = 22k resistor. That would turn on the transistor enough and the relay would simply draw about the same current as it was without the transistor.

Perhaps others can chime in ...

[littlebill]

How exactly did you measure "relay that use 65ma to run its coil"?
Assuming you have the negative side of the power source (battery?) connected to circuit ground.
How much current is flowing through the base resistor (and into the transistor)?  (should be ~ 1 mA)
How much collector current is flowing between the relay coil and the collector?  (should be 65 mA)
As c4757p said, it should be impossible for your circuit to be drawing 91mA unless you are doing something wrong.

Since your transistor (BC550) has a minimum gain (h_FE) of 100, and you are injecting 1mA of base current, it should be within the range that will allow the transistor to pass ~100 mA.

i broke one side of the connection put my meter in line

please explain how you figure 1ma works, according to my data sheet

"IC= 100 mA, IB= 5 mA " doesnt this mean it needs 5ma to get it to pass 100ma through the collector?

[Richard Crowley]

i break the connection at the emitter and run it through my meter, i use 2 different meters, both agree,

after saying that, i broke the connection at the plus side and now its reading right? around 62 ma....

We don't know what you mean by "broke the connection at the plus side"?
If you break the connection to a current (amp) meter, then you open the circuit and should measure ZERO current.
And the circuit will stop working, also.  So your statement "broke the connection at the plus side" isn't making any sense here.

[littlebill]

sorry, i put my meter in between the emitter and ground. or the plus side and the relay coil, or between collector and the other side of the coil.

i hope that makes more sense

[Richard Crowley]

please explain how you figure 1ma works,

Ohm's Law.  5V across 4.7K is 1.06 mA

according to my data sheet "IC= 100 mA, IB= 5 mA " doesnt this mean it needs 5ma to get it to pass 100ma through the collector?

No. Those are probably maximum rated values.

If your transistor has a current gain ("h_FE") of 100, then puttig 1mA into the base should cause the transistor to pass 100 mA of current.  The collector current should be the base current multiplied by the transistor gain ("h_FE").

[littlebill]

this is the sheet i am using

https://www.fairchildsemi.com/datasheets/BC/BC547.pdf

so my hFE is 110?

if i can pass 100ma and i only need 65 ma, what is gained by pushing into saturation? how do i determine what saturation would be?

[Richard Crowley]

sorry, i put my meter in between the emitter and ground.

Yeah, not exactly the right place, but.....
If the current flows from the + supply rail down through whatever components to ground, then.....
The emitter current (which you are measuring) should be the collector current (which we asked you to measure but you have not reported) PLUS the base current (which we also asked you to measure).
There is no way within the laws of physics as we know them on this planet for you to have MORE current flowing out of the emitter than is flowing into the base added to the current flowing into the collector.
If you are measuring MORE current, then you are measuring something wrong somewhere or you meter is broken.

[Richard Crowley]

this is the sheet i am using

https://www.fairchildsemi.com/datasheets/BC/BC547.pdf

so my hFE is 110?

The h_FE is typically a RANGE of tollerance because transistor fabrication has variations.  The data shows that your BC550 should have a MINIMUM DC current gain of 110x.  It could be as high as 800x.  But, in any case, plenty to control the current of your relay.

if i can pass 100ma and i only need 65 ma, what is gained by pushing into saturation?

You ensure that the transistor is at is minimum "resistance" so the transistor itself isn't dissipating any significant amounts of power (which would heat up the transistor and is quite undesirable).
Saturation ensures that the transistor is always turned FULLY ON even through variations of supply voltage, load current, base drive current (output current, resistance variations, etc.).  It is VERY desirable to drive the transistor well into saturation just to remove that unknown from the already-complex circuit variables.

how do i determine what saturation would be?

Collector current = base current x transistor gain (h_FE)
You drive it into saturation by reducing the base resistor to cause plenty more than enough base current to be injected, even under worst-case conditions.

[littlebill]

Richard,

i learned more in the last 15 minutes then 2 weeks reading , thank you.

can the base voltage be different, say i wanted to use 3.3v how would i calculate that?

also i retested the emitter current it still shows .93, maybe i will make a video when my GF gets home so some one explain what i am doing wrong, i have 2 meters both showing this

[xrunner]

What Richard said -  anything above 56 mA is ON as far as you are concerned. But on is on, the harder you drive the transistor on, it won't cause the relay to pull more current. You just want it ON with an assurance margin, but the margin isn't going to cause any problem.

[littlebill]

can the base voltage be different, say i wanted to use 3.3v how would i calculate that?

[IanB]

You could make your circuit easier to understand if you drew the full circuit on the schematic, like this:



I know it is common for people to use a "ground" symbol, but personally that is something I would avoid. It is something that may reduce clutter on complex circuits, but it really doesn't improve the clarity on simpler circuits.

As you can see, the currents around the transistor must add up. If you take measurements in each place you should be able to confirm this (approximately--the meter resistance will upset the measurements a bit).

[Richard Crowley]

i find it easiest to work "backwards".
You know you want to control 5V @ 56mA.
So you need a transistor with at least V_CEO of 5V.  The BC550 is good up to 45V, so that is fine.
And the transistor needs to be able to control at least 56mA. BC550 is rated ("I_C") for 100mA. You probably wouldn't want to use a transistor with any less than 100mA rating.

Your BC550 has a "worst case" (minimum) current gain of 110x.
And we would like to drive the transistor into AT LEAST 2x "saturation"
So, doing the calculation in my math-deficient brain, if I drive a 100x transistor with 2mA of base drive current, it will cause it to have a maximum of 200mA of collector current.  Of course we know it will never actually reach this current because we know what the load voltage and current is from our previous design steps.

So, Ohm's Law says that if we have a 3.3V drive voltage (from an Arduino, perhaps?) and we want 2mA of current into the base of the transistor, we need around a 1.6K ohm resistor.  I use my Ohm's Law online calcuator here:  http://www.rcrowley.com/eirp.htm).  Actually I would use a much lower resistance value more like 1K or even lower just to remove any shadow of doubt that we are driving the transistor into saturation.

[wraper]

IMO "4.7k resistor" which OP have used actually is not 4.7k at all. Something more like 100 ohms.

[littlebill]

So, Ohm's Law says that if we have a 3.3V drive voltage (from an Arduino, perhaps?) and we want 2mA of current into the base of the transistor, we need around a 1.6K ohm resistor.  I use my Ohm's Law online calcuator here:  http://www.rcrowley.com/eirp.htm).  Actually I would use a much lower resistance value more like 1K or even lower just to remove any shadow of doubt that we are driving the transistor into saturation.

guess my math sucks as well.. how do you calculate out to 1.6k, volts(3.3)/ amps(.02), isn't that 165 ohm? sorry, im close


my math was wrong, i failed it in school, amps should be .002...

[wraper]

also i retested the emitter current it still shows .93

0.93 what? 0.93 cannot be 93mA. It can be either 0.93mA or 0.93A. If the relay is not connected, 0.93mA would be the current to expect with 4.7k base resistor.

[littlebill]

IMO "4.7k resistor" which OP have used actually is not 4.7k at all. Something more like 100 ohms.

i maybe new, but i can measure resistance, they are 4.7k considering i measured them and i bought them.

[Richard Crowley]

how do you calculate out to 1.6k, volts(3.3)/ amps(.02), isn't that 165 ohm?

If I go to my online Ohm's Law calculator:  http://www.rcrowley.com/eirp.htm
I put in 3.3 for the Voltage
I put in 0.002 Amps ( 2 milliAmps)  (NOTE THE VALUE IS IN AMPS!)
And I click the [SOLVE] button.
The page says: 1649.99...  Ohms.

It also says 0.0066 Watts (6.6 mW) which means that even a tiny 1/8 watt (125 mW) resistor is plenty big enough.

[littlebill]

how do you calculate out to 1.6k, volts(3.3)/ amps(.02), isn't that 165 ohm?

If I go to my online Ohm's Law calculator:  http://www.rcrowley.com/eirp.htm
I put in 3.3 for the Voltage
I put in 0.002 Amps ( 2 milliAmps)  (NOTE THE VALUE IS IN AMPS!)
And I click the [SOLVE] button.
The page says: 1649.99...  Ohms.

It also says 0.0066 Watts (6.6 mW) which means that even a tiny 1/8 watt (125 mW) resistor is plenty big enough.

yea i figured it out after a bit, dude, really appreciate it, have a stronger understanding now