Adding push button to 2 way motor circuit

[rfeyer]

Hello all
First time poster here - and, definately green in electronics!
With the help of others, the attached circuit was made, which is a 2 directional control of a 12V motor - using a MicroController (WebControl) sending a 5v signal to the board.
BTW, the circuit and motor are used to control as chicken coop hen door opener from inside the house or over the web

Anyways, it works great - so then I thought I will need to be able to operate the door locally, not having to use a laptop or tablet - so, need two puch buttons.
I figured it would be as easy as bridging the 1k transistors (see image - lines in heavy blue), but, after a few uses it fries the transistors.
Any thoughts? Any help would be appreciated,

TY in advance

Rainer

[piratepaul]

I can see it would fry the transistors, I should think open circuit the collector.
A two way rocker switch, or something along those lines.

One switch for each transistor.

You don't sound that green.

[piratepaul]

A diagram of the switch would be the same as switch 1, the collector is on and manual is off, or collector is off and manual is on.

[rfeyer]

Ha - here is were my green-ness will show :
So, simply 'jumping' the transistor the way Ihave wil not work, correct?

What I am not understanding is, how I should be changing it

TY for responding already!!

[piratepaul]

You could switch the transistors with a 5v signal from a new chip. Touch button control might be better.

[rfeyer]

piratepaul,
TY for not giving up on me already

But, I may not have explained myself correctly: I wish to have a local push button at the site of the motor which will activate the motor, one button for each direction.

The diagram as is of course reverses polarity of the 12v circuit to change dierction of the motor.
I thought I could simply'jump' the transistor and get the same response as activating the transistor with 5v's. And, I already fried 2 transistors

1-st why is my thought wrong in jumping the transistor - what makes it fry it?
2- where else do I palce the pushbotton / or rocker switch?

Rainer

[piratepaul]

Where it says raise and lower 5V try putting +5V to each of those two points (not at the same time) a phone charger will give a 5V supply...earth to chassis, I will need to think about what switches it off ...carefully. It might be programmed in some way I have not thought of.
Your device to fry at your own risk.

Some sort of latch or flipflop with a momentary switch.

[piratepaul]

I just read the questions at the bottom, you are reverse biasing the transistor ( I think, I am not a transistor expert, I have fried enough to know they fry easy).
Where would you put the switch? Well revised plan is to switch the transistors with a new board, it you look up flipflop/latch there are different types and configurations ... Basically 2 outputs ...when Q (output 1) is 5V ..../Q output 2 is 0 V you can configure a button to switch between out puts ....the problem is stopping it. This can be done but I will need time to study the diagram.

[piratepaul]

Hi, I think I can't see to the right, it is this tablet, I can see as far as parts list, it says motor on off ... If there is any thing to the right of that I can't see it. There are lower and upper limit switches, I don't know what they are but we need to use them to switch a circuit, is to stop when it is open or closed.

It is difficult on this tablet, I need to post this then I can see the left of the diagram.

Any chance of putting any thing to the right on the left and sending it back.

[piratepaul]

I can get my head around it now, I don't need to know the details. You have two sets of inputs and outputs. When the door reaches a limit it sends a signal to what ever is plugged into (raise logic high, upper limit out, lower and lower limit out)
. When the limit is reached a signal is sent to ( its not on the diagram, not on the left any way) a control box, that signal switches circuits which switch the transistor off.
5V to raise logic high or raise logic low will open or close the door, when it reaches its limit,, a 4V signal is sent through upper limit out or lower limit out to the control box.
You can replicate that, I will have a think about it ...tea break.