Trying to solve exercise in art of electronics involving transistors and diodes

[Robert Hill]

Hi all,

I'm trying to solve the exercise found on page 107 of the 2nd edition of the art of electronics. The question asks:

Design a transistor switch circuit that allows you to swith two loads to ground via saturated NPN transistors. Closing switch A should cause both loads to be powered, whereas clsoing switch B should power only one load. Hint: Use diodes.

Here's a picture of the sketch of my schematic which sort of works:


I've used LEDs for the load. The circuit works in the following way. When button A is pushed sufficient of the voltage gets to the transistor bases that both LEDs light up despite one transistor having a diode in series with its base. When button B is pushed however the circuit voltage is reduced by the variable resistor such that only the LED sinking into the transistor with the diode on its base lights. So the circuit sort of works.

However, it has several issues.
1. The non diode base LED had to be put on the emmitter side of its transistor otherwise it was constantly lit (even when the breadboard power supply was connected to the mains but switched off!?)
2. The diode base LED lights much more dimly when button B is pushed than when button A is pushed, so the loads are not equally powered by the different button for obvious reasons.
3. I would expect the non diode base LED to be the one to light when button B is pushed because that is the easier path when there is limited voltage supply.

If anyone is able to point me back to parts of chapter 2 in the book to go back and re-read to understand the question better then that would be helpful, alternatively if anyone can explain why 1 and 3 above were happening that would be good too.

I expect the first question people ask will be what are the component values. I used 330 resistors, standard LEDs and IN1448 diodes. Not sure what exactly the transistors are. The voltage is 5V and I've no idea of the value of the potentiometer as I salvaged it from some old electrical item.

Cheers,

 

[Colin Mitchell]

Use common-emitter stages and not emitter followers.
Connect the first via a switch and resistor and take a diode to the second transistor.
Connect a sw to the second via a resistor.

 

[Laplace]

Review section "2.02 Transistor switch" on page 63 in the Art of Electronics where it explains a transistor switch in saturation.

 

[CDRIVE]

Keep in mind that you cannot saturate a transistor when emitter resistors are used.
EDIT: The above statement is grossly in error. See subsequent replies for further explanation.
Also, please post your next schematic vertically oriented. My old eyes and neck will appreciate it greatly.

Chris

 

[Colin Mitchell]

"Keep in mind that you cannot saturate a transistor when emitter resistors are used."
This is incorrect.

There are a lot of slight inaccuracies, badly described statements and omissions in the article on P63 of The Art of Electronics. It is no wonder the student does not understand what to do.
The biggest point to pick is this:
A transistor can have a gain of 250 and if the load is a high value of resistance, the collector-emitter voltage will be very small.
Do you call this SATURATION or FULLY TURNED ON ?
The writer is saying the transistor does not have a high gain when the transistor has a small collector to emitter voltage.
Secondly, I have used a staircase of 3 transistors to get a gain of one million and the collector current of the final transistor is not sufficient to drive 1mA through a LED. So, why does he say to put a resistor on the base to prevent collector-to-emitter leakage.
Thirdly, he does not point out that the globe will require about 600mA to start to turn it on.
So a 1amp transistor is needed for a 100mA lamp.
The gain of a transistor drops from 100 to about 20 when it is required to pass a high current so you really cannot pull values of base resistance "out of the air" until you actually buy a transistor and globe and try the circuit.
That's because the globe characteristics change by 500% and the transistor characteristics change by 500%.

 

[Arouse1973]

Keep in mind that you cannot saturate a transistor when emitter resistors are used. Also, please post your next schematic vertically oriented. My old eyes and neck will appreciate it greatly.

Chris

LOL Yeah I had to turn my head 90 degrees, if anyone had seen me they would have said "look at that poor chap with his head on his side, I suspect he was born like it".
Adam

 

[Laplace]

...you cannot saturate a transistor when emitter resistors are used.

Here is a different opinion from http://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/common-collector-amplifier/

 

[CDRIVE]

"Keep in mind that you cannot saturate a transistor when emitter resistors are used."
This is incorrect.

Yes Collin, that was a brain fart!

Chris

 

[CDRIVE]

Here is a different opinion from http://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/common-collector-amplifier/
View attachment 21053

Yes, I've placed a mea culpa in that post.

Chris

 

[Le Sy Hau]

Just follow the circuit I've been uploaded. This is a simple exercise, and the answer also simple. Use a transistor with 2 resistors to make a TRANSISTOR-SWITCH, and calculate these 2 resistor to make the transistor work at 2 mode: off, and saturation, it is simple to find value of these resistor.

 

[CDRIVE]

Just follow the circuit I've been uploaded.

Again! You guys are making me feel like I've got a serious hangover!!
What's with the horizontally oriented schematics?
Are you guys lying down on your side when you post them?

Chris

 

[Le Sy Hau]

Again! You guys are making me feel like I've got a serious hangover!!
What's with the horizontally oriented schematics?
Are you guys lying down on your side when you post them?

Chris

Actually I didn't know why the photo has been rotated, might be the problem came from this website.

 

[Le Sy Hau]

By the way, can you guys help me design the Voltage Amplifier Stage in the picture I've attached. In this case, we know Vcc (dual supply), current source IB, and the value of V1 when no input signal.

 

[Robert Hill]

Ahha I've solved it (just realised I've omitted the power supply to the two LEDs in the schematic.) :
And it should be the right way up. The diode allows voltage from switch A to both bases but the diode stops voltage from switch B getting to one of the bases.

Thanks for all the comments, helpful clarifications on the reduction in gain when there is Low collector emitter voltage.

 

[CDRIVE]

Yes, your schematic doesn't have the LED anodes connected to Vcc as they should be.

Chris

 

[Colin Mitchell]

Your circuit may or may not turn ON.
It is not an engineered circuit.
The LED in the emitters will work and then you can throw away the transistors, and you can throw away the top resistors.