A Simple Circuit concept to study biasing transistors

Thread Starter

fredric58

Joined Nov 28, 2014
203
wow that looks much simpler. I will read over this do some studying. I will probably have a couple questions this evening. thanks for the input!
 

MrAl

Joined Jun 17, 2014
7,761
wow that looks much simpler. I will read over this do some studying. I will probably have a couple questions this evening. thanks for the input!
Hi again,

Sure no problem :)

I just want to remind you that a CMOS latch is even simpler, although not as generic because you have to order a specific IC chip while with the two transistor circuit you can use a lot of different types of transistors.
Either way though, i think you will be happy with the results.
 

Thread Starter

fredric58

Joined Nov 28, 2014
203
I'm not understanding. if you have some time can you show me the current flow. in this schematic is the circuit WITHOUT an arduino signal ON?
 

MrAl

Joined Jun 17, 2014
7,761
Hi,

Current flows from one collector lead through the base resistor of the other transistor and into the base. That turns the transistor on. When the transistor turns on, it turns the other transistor off. But to switch stages yuou must connect a pin of the Arduino to each collector, so one pin to Q1 and one pin to Q2 collectors.
When you drive that pin low, it turns the other transistor off and so the collector goes high and that then keeps the first transistor on.

The diagram shows the Arduino pin connections.
 

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hobbyist

Joined Aug 10, 2008
887
I know I took the long way to get there but I learned about other things as well along the way. which gave me more ideas as well as questions. this is great fun. I sincerely appreciate the time you devoted and your instructions. I learned exactly how to accomplish the task. there may be a more efficient to do it which I will find out as I learn more. .
That's what designing circuits is all about, experiment try new ideas and keep learning along the way.

Glad to be of help.
Happy designing.
 

Thread Starter

fredric58

Joined Nov 28, 2014
203
On my post 57 I posted a circuit idea. Change S1 to a transistor with a resistor on the base. A signal from the arduino will turn it on, another signal will turn it off. But I was told, that it could end up being backwards from what I wanted to do. The reference is from the voltage divider, a fixed current. When you push the button, the capacitor sucks current and it's out of balance. So it turns on While it is on, the capacitor charges. When you push the button again the input is higher than the fixed current. How could it possibly get backwards? Thanks
 

hobbyist

Joined Aug 10, 2008
887
Hi
I know nothing about 555's or any Ic's for that matter, my question is, is the switch S1 used to introduce some sort of RC timing into the input of the 555 to make it pulsate?

Thanks.
 

Thread Starter

fredric58

Joined Nov 28, 2014
203
Hi
I know nothing about 555's or any Ic's for that matter, my question is, is the switch S1 used to introduce some sort of RC timing into the input of the 555 to make it pulsate?

Thanks.
Hi
I know nothing about 555's or any Ic's for that matter, my question is, is the switch S1 used to introduce some sort of RC timing into the input of the 555 to make it pulsate?

Thanks.[/QUOT
yes, pins 2 and 7 have equal current. When you push s1 the cap begins to steal current ( and charge ) so now the trigger side is lower ( not equal to the divider) so it powers pin 3. At the same time the cap begins charging. When s1 is pushed again the cap boosts the current and it goes higher than the threshold and it turns off pin 3. I just use it as a switch. Or a manual timer. MrAI circuit requires 2 pins from the arduino. This one only 1
 

MrAl

Joined Jun 17, 2014
7,761
Can you do a ladder schematic and add the + and - of the load? I almost have this……thanks

Hello,

The Arduino pins are connected to the collectors as shown.

When the Arduino starts up you drive one pin to the low state and one pin to the high impedance state. The collectors are the outputs and you may need an additional driver to drive say an LED. That would be a transistor or gate.

When you want to switch states you just drive the opposite pins to the opposite states.

For example, we drive pin 1 to low and pin 2 to high Z. That means pin 2 is high.
When we want to switch states, we drive pin 1 to high Z first, then pin 2 to low state, then pin 2 is low.
We can then drive pin 2 to high Z right way if we wish but pin 2 will will still stay low because of the latch.
When we want to switch states again, we drive pin 1 to low and pin 2 to high Z.

Now at some point we want to turn the power to the Arduino off, so the Arduino will have no knowledge of the last state of the latch. This means we must first read the state, then make the change in which pin we drive low. If pin 2 was high, then we drive pin 2 low, if pin 2 was low, then we drive pin 1 low. We always drive the other pin to high Z when we drive one pin low.

There are only two possibilities for the state of the latch. Either:
1. pin 1 is high, pin 2 is low.
2. pin 1 is low, pin 2 is high.

If we use diodes as shown in the schematic, then we must also use something like a 10k resistor across at least one of the diodes so that we can read at least one output when we power the Arduino up. That's so that we can detect the initial state on power up.

I dont think another diagram will help because these transistors are interconnected.

Can you visualize this better if the transistors were replaced with relays?
 

hobbyist

Joined Aug 10, 2008
887
On my post 57 I posted a circuit idea. Change S1 to a transistor with a resistor on the base. A signal from the arduino will turn it on, another signal will turn it off. ...
Are you still looking for a way to replace S1 with a transistor, if so, then I need some circuit info.

What is the voltage at the top of the capacitor when S1 is OPEN?

Is the 100K ohm resistor, used for positive feedback from the ouput back into the input of the 555, when S1 is CLOSED in order for there to be some sort of timing pulse?

What is the output status of the 555 when S1 is OPEN, and what is the output status of the 555 when S1 is CLOSED.

That way it would give some sort of idea if a unilateral device (bjt) can even be used in place of S1.

Thanks again.
 
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