This circuit has me baffled

Thread Starter


Joined Jul 21, 2009
Yes, pin 4 on SKT1 is HI (5V) and the earths are present. In fact, both transistors in both pairs aare passing base to emitter current, but pins 9,10 of the 4053 are both LO and the two resistors R4 & R6 don't appear to be passing any current.


Joined Jul 1, 2009
I pulled the datasheets on your components and walked through the circuit. In looking at it, as I understand it the circuit, you essentially want the option of two keyboards connected to the same computer. I can see the advantage of this in certain circumstances.

Essentially, you have a Triple, Single Pole, Double Throw analog to digital switch. In the circuit shown, X, X0, and X1 are not used, so that switch is unused, leaving you with only two switches to use (Y and Z) Since Switch 'X' is not used, pin 14 needs to be pulled up to Vcc/Vdd or down to Vss/GND. Otherwise the chip won't handle current correctly-- because that pin floats. In other words, you need to fix this on pin 14.

I can see no reason for C1 & C2. C1 & C2 bypass the base on each transistor when Vcc drops, until C1 or C2 discharged. Unless these are 'bypass' capacitors to help regulate over minor spikes/drops.

I can see no reason for 1N4148 and it's section should be removed. SKT1 pin 1 should only hit SKT2 pin 1, via Q2.

R4 and R6 act as pull-down resistors on the used channel selectors (B or C, since A is not used - it is tied to GND). They are pulling pins 9 & 10 LOW.

Essentially, when data comes in from either SKT2 or SKT3 on pin 1, we see this functionality acrosss the HC4053:

0 0 0 | Z0, Y0 | Z, Y = 0 on CLOCK
0 1 0 | Z0, Y1 | Z = 0, Y = 1 on CLOCK
0 0 1 | Z1, Y0 | Z = 1, Y = 0 on CLOCK
0 1 1 | Z1, Y1 | Z = 1, Y = 1 on CLOCK*

If you notice, Vdd/Vcc is +5, and Vss/GND and Vee/Vref are both GROUNDED. That means your acceptable voltage range is between 0 and +5.

The problem I have with this circuit, is that there is no path for Data to travel back and forth between either SKT1 pin 1 and SKT2 pin 1 or SKT3 pin 1. Your circuit is wired thusly:

SKT1 pin 1 - HI operates GATES for SKT2 and SKT3 simultaneously.
SKT2 pin 1 - HI selects Channel B
SKT3 pin 1 - HI selects Channel C

If SKT1 pin 1 is LO, neither SKT2 or SKT3 can choose a channel on the HC4053.

Hope this was useful. I think I would wire it differently, and my observations might be wrong. Perhaps a theory of operation would be useful from you as to why you chose to use the diode, and how you expect the circuit to work.

Thread Starter


Joined Jul 21, 2009
I'm impressed. Your analysis of the circuit is flawless and I have to agree with you on all counts. As I mentioned, the circuit is derived from one used to provide a CNC keyboard to a PC , in addition to the normal keyboard. I had adapted it to use with a radar surveying system, where the master input is from the radar, the main keyboard is the radar controller on SKT2 and a keyboard emulator is to be used on SKT3 to simulate the F7 key. This is used to provide soft markers on the radar data.
I, too, could not see the purpose of the 1N4148 diode, but assumed it was for protection, or to provide a data path from SKT2/1 to SKT1/1.
The two capacitors from collector to emitter of Q1 and Q3 were for protection and are not really necessary. I will remove them.
Your excellent suggestion of a pullup or pull down on pin 14 of the 4053 will be done. I should have seen that myself!
As for the operation analysis, you are quite correct and the assumption was that SKT1/1 would always be high and that SKT2 would be the main keyboard input. SKT3 was to be the F7 emulator input, but a redesign is necessary, because, as you point out, there is no data path from SKT3/1 to SKT1/1, or, without the diode, from SKT2/1 either.
I have to confess that this cct was thrown together in a hurry, as the survey guys have been pressing me for the F7 facility., but it looks like a rethink is necessary.
As you seem to have a good handle on this, any ideas on a rewire solution? I'd sure appreciate it.

Also, to "BeenThere", I hope the above answers your question.