Switching transistor question

If you're driving the resistor to ground (i.e., one end of the resistor is connected to the transistor and the other is connected to ground and the transistor switches to 12V) you need to use a pnp transistor. Connect the base through a resistor to +12, then pull the base to ground to turn the transistor on.

If you're driving the resistor from +12V (i.e. one end of the resistor is connected to +12v and the other is connected to the transistor, which switches to ground), use an npn transistor. Apply 12V to the base though a resistor (1K is fine) to turn it on.

The problem with using an npn to drive a load to ground (resistor connected from the emitter to ground) is that it's the base to emitter voltage that turns the transistor on. Once current starts to flow through the resistor the emitter voltage rises. It will eventually rise to the point where V_BE will be less than the ~.7V required to turn it on. That's why the pnp solution is better.

But, the pnp solution requires a drive that's opposite from the npn. +12V turns on the npn, but turns off the pnp. So, if you want to keep the same control polarity, you'll need to invert the control signal through an npn transistor (ground the emitter, pull up the collector to +12v and use the collector voltage to drive the base of the pnp).

Hope this helps.

I'm almost understanding this. So essentially two transistors are needed to switch on a circuit to provide a 12v 47k ohm output? (The output is merely a signal for another circuit) Is the emitter follower switch setup the wrong application because there's not enough to saturate the transistor? I just thought that the second resistor of the right value would be enough. If not, would a led in its place do?

I'd do the two transistory setup if that's the only way to do it.

Sean

Your terminology of an output 12V 47k ohm is a little confusing.

Is the load you want to drive 47k ohm? If so that is nothing in amps, I=12/47,000.

Is that 47k ohm load refrenced to +12 v or to ground?

Normally interfacing logic level signals only requires 1 transistor. Only occassionally do you need active pull up and active pull down as in a totem pole driver.

How close to zero volts and 12 volts does the output have to be?

Relay outputs are often used for isolation between circuits. Do you need to maintain that isolation?

A picture is worth a 1000 words. Attach a sketch and then we can really help.

The purpose of the second transistor is just to give you the same polarity on your control input voltage, if you're using a pnp transistor, since the input voltage to turn on a pnp is opposite that to turn on an npn.

Look at the emitter follower circuit. With the transistor in saturation, let's say V_CESAT = .2V, then the voltage at the emitter is 12- 0.2 = 11.8 volts.

The base emitter junction is just a diode - it's needs about .7 volts to turn on - you can get precise numbers from your data sheet - but with V_E at 11.8 V, the most V_BE can be is .2 volts. See the problem?

Now, in actual practice this will probably work, the transistor just won't go fully into saturation, and you'll end up with V_BE = V_CE = 0.7V, and the voltage across your resistor will be 12 - .7 = 11.3 volts. With such a small current (I_CE = 11.3/47K = 0.24mA) this is not an issue for the transistor. With larger currents you would really want the transistor in saturation to minimize the power it has to dissipate (P = I_CE * V_CE).

An N channel mosfet is even worse - if connected in a source follower configuration - since it wants about 10 volts from gate to source to turn on - you would need to apply 22V to the gate. A p-channel mosfet would be just fine. Like the pnp transistor, you turn a p channel mosfet on by pulling the gate low (to ground) relative to the source.

So, your circuit will work, but it's not best engineering practice. Best engineering practice would be a pnp, or an npn transistor in grounded emitter configuration, but that may not be possible with your application.

Someone asked about isolation - and that's a valid concern. If the power and ground of your load are the same as the power and ground of your relay, you're OK.

The best thing to do is just build your circuit and see if it works.

2N3055 is rated at 15 amps 60 volts 115 watts. 2N2222a is 0.5 amps 40 volts maybe half a watt? Not sure about 2N3094. So it depends on what you want to do.

What is the relay driving?

You are probably better off with a MOSFET.

There are some very good 'FETs about which will work fine off 12v (or even 5v), a pull down to 0v on the gate is a good idea, say 47k and you can stick a series resistor between signal and gate if you like but it isn't essential. Some have a very low Rds ON... (sort of equivalent to Vce SAT in your switching transistor when used correctly)...e.g. not as an emitter follower!

Emitter followers aren't best for switching as they don't fully saturate.

If you want to use a transistor, use NPN, emitter to 0v, signal into the base (through a suitable resistor) Connect your load between +v and the collector...that will switch hard ON, (so you must make sure the transistor can handle the power/current)

Hmm the more I re-read the Q the more confused I am?

a) Is the output a 47K load ?

If yes....then any old small signal transistor in just about any configuration will be fine, depending on what the output load is driving into!

b) If the relay contacts are the output then my previous posted answer is fine. But it would be nice to know what the relay contacts are driving...a motor? This makes a big difference as any inductive load (motor, solenoid etc) will mean you need some diodes to help protect your switching device from back emf.

Post a bit of the circuit diagram or a sketch!

Yes, the output is a 47k load, very small but it was required neverless and it does work without a problem with relays. It is required to trigger a seperate circuit in a vehicle and isn't a variable option. I just input +12vdc on a 47k 1/4w resistor. The additional benefit of the relay is that there's a complete circuit isolation when off. I used a relay because at that time, I was not completely familiar with transistors.

Originally it was just a 47k resistor on a 12v signal input but the circuit was false triggering by residual loads even with a diode in place so a relay was used. Now I'd like to duplicate the exact function with transistors as switches and do away with the relays since the amperage is so tiny.

Can it be done with one transistor or will 2 be required?

Any small signal transistor will do. Choose a beta >100.

If you have a npn, use a 1k pull up with the emitter tied to ground. Use a 10k base resistor, output logic True (hi) will be when there is no base drive, False (lo) will be when you provide base drive.

In an automotive circuit you may actually drive your 47k resistor with the output of any common OP amp. Use a 1k resistor in series with the output for short circuit protection, (I also use diodes from the output to the v+ and ground for transient protection, probably overkill)

Many of your cmos logic chips will drive it directly also, or use a line driver.

If you want isolation you can use opto isolators and tie the collector to +12, emitter to your 47k load.

Automotive voltage is only nominal 12v, and may be as high as 15 volts, so you should have lots of latitude. Your 47k suggests you are driving logic circuits and not lamps or motors, so you far more options.

Try your favourite circuit, it will probably work with a little bread board tweaking.

It depends what the 47k load feeds into!

E.g. is it connecting into the base of another transistor? or a CMOS input or what?

You might be fine with an emitter follower as it is a 47k load. In which case 1 transistor will do, as someone else said it won't go right upto 12v...but that's presumably ok?

Try it and see is the simplest answer!

If you are going to build a lot of these...then make sure you do plenty of testing!

Things work on the bench then fail in the field! So try resistor values higher and lower than your final chosen value, test it with 9v 15v and any other variables you can think of!

Good luck

Del

If you like that relay, you could switch to a reed relay. much smaller and still keeps the isolation.

Ah, I think I'm begining to understand the problem!

I don't think a transistor is necessary if you have only a 47k load...

The point of an emitter follower is current gain...but you don't need much current!

And if you have a 12v signal then you don't need votage gain!

The relay is providing isolation and also noise immunity (due to it's magnetic hysteresis and inertia)

It seems to me a simple resistor capacitor network will probably work just as well for noise immunity.

Say for example 12v signal into two 22k resistors in series with a 0.1uF capacitor going from the junction of the two resistors to 0v.

It still depends on what this signal is driving... if say it only needs a 6v signal, you can divide your signal by half with the addition of another resistor to 0v..

If you divide the signal you also divide the noise.

I hope I'm on the right track here...

First you need to know the Ic-saturated(the current of the collector)as

Ic-sat=Vcc-Vcesat/Rc where Rc is your load Vcesat is less than 1V so you can estimate the Ic=Vcc/Rc

then you need the Ibase that drove the transistor to saturate

I bsat=Ic-sat/Bdc

Then you can find what transistor is ok for your application and what resistors you can use. A rule is 1 ohm for collector ten for the base but is not for all situations

Thanks to all, hope you had a nice Memorial weekend!

Del the Cat,

The only thing that the body control module (BCM) of the vehicle requires is 12v 47k input which would now obviously be a logic input as pointed out by others. Originally the input was given direct but for some reason, transient signals are finding its way back and thats why I used a relay because I know that a relay would physically isolate the circuit.

To all, if I can't figure this out after trying a few things, would it be improper to hire help from someone here for piecework jobs like I'm doing now?