Interesting Transistor arrangements

[BartSimpson]

After my last mess-up , I took extra care to verify circuit 1. It is correct.

The inverters are "low saturation sink drivers".
They usually drive Lamps, LED's, relays, etc.

Div1 and Div2 are driven by the CPU.
When Div1 is high (from the CPU,) the top 2k2 resistor is grounded by the driver.
When Div2 is high, the bottom 2k2 is grounded.

Some other info which might help.
The two 2k2 resistors look to be 1/8 or 1/4 watt.
The 22k and 470R resistor look to be 2 watt packages.
I might breadboard this one in a few days.

Does anyone know why the emitter is the current source ?

 

[Nigel Goodwin]

After my last mess-up , I took extra care to verify circuit 1. It is correct.

The inverters are "low saturation sink drivers".
They usually drive Lamps, LED's, relays, etc.

Div1 and Div2 are driven by the CPU.
When Div1 is high (from the CPU,) the top 2k2 resistor is grounded by the driver.
When Div2 is high, the bottom 2k2 is grounded.

Some other info which might help.
The two 2k2 resistors look to be 1/8 or 1/4 watt.
The 22k and 470R resistor look to be 2 watt packages.
I might breadboard this one in a few days.

Does anyone know why the emitter is the current source ?

Because it's a high side switch, you would normally use a PNP transistor in this type of circuit - due to it's lower voltage drop in this mode.

The circuit is fairly standard, and very simple:

DIV1 does absolutely nothing!.

DIV2 turns the transistor ON, effectively connecting the 470 ohm to 12V.

PWM is presumably a PWM signal to chop the 12V turned on by DIV2 into a PWM signal for feeding some type of analogue device?.

The input part simply includes clipping diodes to prevent it going too high or two low, PIC's actually include these internally for the same reason.

 

[BartSimpson]

DIV1 does absolutely nothing!

Thanks for that. I thought I was going nuts. I will still breadboard it in a few days though, just in case as I can't see why they would waste the resistor and transistor. Unless it is a trap for someone who tries to copy the device - which is not what I am trying to do. Although it did get my attention when I drew it.

 

[Nigel Goodwin]

DIV1 does absolutely nothing!

Thanks for that. I thought I was going nuts. I will still breadboard it in a few days though, just in case as I can't see why they would waste the resistor and transistor. Unless it is a trap for someone who tries to copy the device - which is not what I am trying to do. Although it did get my attention when I drew it.

Presumably the IC opencollector buffer is an unused gate in an existing chip?. It might also be used in alternative designs, are there empty spaces on the PCB?.