I have been working on an Arduino project for my dad's model train sets. I have all the code down and ready, i just need transistors to drive all of the electronic parts such as lights and engine motors. All of the lights work with our current power transformer so i can send the power from it through transistors switched by my Arduino. My problem lies in my lack of electrical knowledge and I am not sure about what kind of transistors to get based on the power output of the transformer.

This is a picture of the transformer showing it's input and output voltages.

What i would like to know is exactly how many amps and are put out in AC and DC current, as this transformer has power outputs for both. Thanks for reading this and if you help me, even more thanks.

The label states it has "7.5VA" total output. That "VA" refers to volts * amps, so using the 12V output alone it's roughly (7.5VA / 12) .625 amps

Your transistor selection doesn't really depend on the output of the transformer, however, but rather the current consumption of the light or motor that the individual transistor will be switching power to.

WRONG.
VA (Volt Amps) is the equivilent of WATTS for systems where the power factor is not equal to 1 . In the case of a train, the load is purely inductive so the power factor is NOT EQUAL 1, hence the use of VOLT AMPS, instead of WATTS.
REGARDLESS
POWER = I (A) * V (V) , which can be rearranged as I (A) = P (VA)/V(V)

CURRENT I (Amps) = 7.5 VA/12V = 0.625 A = 625 mA.

Your mistake was here:

output alone it's roughly (12V / 7.5VA)

Your dividing voltage by power (VA, WATTS, DOESN'T MATTER , SAME EQUATION)
P = I * V (divide both sides by V)
I = P/V (NOT V/P)

Thanks for both of your answers, i just tried hooking a multimeter in series with a light on a circuit running on the transformer and the multimeter read that the curcuit was drawing 20 amps, is this right, even though your math said no more than 1.6 amps? Is it possible that I may have wired the multimeter into the circuit wrong?

Also, raschemmel said:

In the case of a train, the load is purely inductive

I thought I pointed this out in my original message that I would not only be running a model train engine on the transformer, but also lights. Would this impact the inductive and power factor of the load on the transformer?

i just tried hooking a multimeter in series with a light on a circuit running on the transformer and the multimeter read that the curcuit was drawing 20 amps,

Something doesn't add up here.
Post a photo of the multimeter display. Most multimeters can only measure current up to 10A.

i just tried hooking a multimeter in series with a light on a circuit running on the transformer

Post a photo of the circuit running on the transformer that the light was connected to.

even though your math said no more than 1.6 amps?

I never said that . You said that. I said 0.625 A * 12V = 7.5 VA.

I thought I pointed this out in my original message that I would not only be running a model train engine on the transformer, but also lights. Would this impact the inductive and power factor of the load on the transformer?

Ok, this is the part where I ask you if the lights are incandescent or leds. (incandesent lights have filaments, which makes them inductive.) Leds are not inductive.

It's a model train transformer. It is possible is has both ac and dc. (I am not sure which is used for the motor in the engine).
If I knew more about the capabilities of the system I might venture a guess.

incandesent lights have filaments, which makes them inductive.

I disagree. Incandescents are practically entirely resistive. Power factor is a negligible consideration for small appliances under normal operating conditions. However, people have been known to do some illogical things. IF you are using it for its intended purpose, it should be ok. Given the 7.5va rating I would estimate it should be capable of .5 amps. Hard to say though since it's a dual voltage supply and only gives one power rating, so you can only assume it's a combined rating.

Also the current rating is an indication of how much current you can draw without causing excessive voltage drop or overheating the transformer. You can usually draw more current at the cost of losing voltage. So simply shorting the transformer out with an ammeter is not the best indicator of how much current it can reasonably produce. You really need to increase the load while monitoring current and voltage because open circuit voltage is significantly higher then short circuit voltage (or rather overcurrent).

Theory aside...The transistor selection depends on whether you want to control AC or DC currents. AC currents require a triac (BTA136 perhaps) or scr and usually an optocoupler driver (MOC4031) for isolating ac an dc. The common way for controlling dc is to use a logic level mosFET.

This is a picture of the circut i wired up for reading the amperage drawn. The left power terminal is hooked up to one lead on the light, the other lead on the light is connected to the ground of the multimeter, and the positive of the multimeter is connected to the right power terminal.

I also noticed as i was taking this picture the "u" next to the A in the multimeter display, some quick research tells me that this stands for microampere, which seems way too low for what the light should be drawing. I posted the picture for reference anyway.

I think we can safely rule out the "20 A" measurement as bogus.
I agree 18.5 uA is very small amount of current.
I might be wrong about the lightz being inductive. Let's disgard that assumption for the moment and focus on Ohms Law that ststes
that I (A) × V (V) = P (W or VA)

W and VA are not the same thing, VA = sqr(W^2 + VAR^2)
If you have no caps or inductors VAR = 0, then VA = W

You know in math class when they make you do sin/cos/tan calculations? Congratulations you just found a real world application for it. Welcome to the world of phase angles and power factor calculations.

.0.625 A × 12 V = 7.5 VA, close enough :P. As with most applications you would never design a system to actually pull 0.625A and given the unknown induction of the system I'd think a value like 0.5A or lower is what you'd aim for.

W and VA are not the same thing, VA = sqr(W^2 + VAR^2)
If you have no caps or inductors VAR = 0, then VA = W

Agreed. We can assume the train motor is inductive but until the OP posts the inductive reactance of the motor I don't think we can use your equations.

BTW
I work at a company that builds power factor correction equipment for utility companies all over the world.
If the OP can post the inductive reactance we can calculate how much capacitance it would take to correct the PF to 1.

Hi, the train controller/transformer has how many outputs.
If only 2, one DC and one AC, the DC will be controlled by the control on the transformer to control the train speed, so you it will not be available as a supply.
Unless you are using another transformer for the motor control, also the controlled DC is unsmoothed DC, that is it is 100Hz rectified AC.
The 17V AC is the supply that will not be controlled, it is used for point motor and lights and accessories.
A picture of the output terminals please.

Don't get bogged down in high falooting theory, the loco motor will have a cap across it to help suppress noise, it will be a simple DC brushed motor.
Forget about power factor.

If you want to use the AC of the transformer as a DC power supply, you will need to rectify it, smooth or filter it, and regulate it to say 12V.

You need to know how much the max current will be when you are controlling the train, then subtract that from the 0.625A, the result will be how much you can draw from the other output.

Tom......

My dad, my brother and myself had 20ft x 12ft HO layout, 250ft track, 12 - 15V Triang Hornby modelled on GWR, God's Wonderful Railway.

I don't know about other countries but in USA , an HO train is required equipment for any ( male) child. You got extra peep points for smoke and sound.

raschemmel:
WRONG.
VA (Volt Amps) is the equivilent of WATTS for systems where the power factor is not equal to 1 . In the case of a train, the load is purely inductive so the power factor is NOT EQUAL 1, hence the use of VOLT AMPS, instead of WATTS.

The output is DC, whether the load is inductive or not isn't very relevant!

Also a DC motor's impedance is complicated and tied to the mechanical load's
nature.

The output is DC, whether the load is inductive or not isn't very relevant!

In most hobby or house hold applications this is true, the induction isn't large enough to make a noticeable difference at any given point in time.

What changes is how much power you pull.
ex.
Plug in a 5W resistor you would pull 5VA
Plug in a 5W DC motor, you now need to pull 5.2VA.

That's a random number but as you can see not taking induction and capacitance into account can cause you to over source your power supplies. As long as you stay a modest amount below the power supplies rating it won't make a lick of difference.

Because transformers work off AC and va is the units for ac power. It is convention... basically. In this situation, it is safe to assume that VA roughly equals watts in rms. Power factor is really only an important consideration in industry and large scale power distribution where efficient power transfer is crucial. i have never needed to calculate power factor for any of the small projects I have built.