Sorry I've had this thought rattling around in that empty space between my ears. I haven't built it yet but I have all the parts to:

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when I finish this will be a completed project and I will start a new thread, for now I am opening the subject for ideas and iterations. There will be a 3D printed box to go with it, and if I can get it to work it a frequency counter.

 

One of your MOSFETs is upside down. I don’t know which one, as I don’t know if you intended it is an inverting output, or a complementary source-follower.
(a low-side MOSFET gate driver would make a nice output buffer)

 

I think Q1 is the right way up (judging by the body diode), but the D and S markings are the wrong way round. There will be some shoot-through during the on/off state transitions.

 

At times I have put a small resistor in series with each drain to limit the short circuit current during switching. That can probably be also done by the selection of the MOSFETs so their RdON is high enough.

 

Did something quite similar. Only instead of a rotary switch I used dip switches. It's possible to select one of four caps or a combination of each, all the way up to all of them. Off hand I don't have the schematic handy. It's somewhere. Also featured on this board is the IC socket where you can put much larger caps into the circuit. With all the dip switches open only the applied capacitor will affect the circuit. If the desired frequency is a little too high it's possible to switch into the circuit some of the other caps and slow down the clock rate. If I locate the schematic quickly I'll post it.

 

I have the same kind of breadboard stock, and I have updated my schematic mistake. I will probably use a protoboard for this next design, since soldering is such a pain for me.

 

Latest brain fart:

R4 will be part of the front panel of this piece of equipment.​

 

You have me stumped. Why all those diodes on the adjustable output?

Edit: Is there a chance of hurting Q3 by having it try to drive all those diodes with more than 2.5 volts?

 

What voltage is Vcc?

 

Perhaps I should have drawn this schematic differently, if you look very close the diodes are not connected to the output variable output. The diode chain only crosses the variable output without actually making contact, the combined drop of all those diodes should be around 3.6 volts. Which is what a TTL output should be. Vcc is defined on the protoboard drawing, which is 12 volts. I plan on playing with the actual voltage on Vcc, I may use an external power supply instead of batteries.

 

Hello Wendy,

There are no interconnections between the diodes on the breadboard

Bertus

 

OK, thanks. It shall be fixed.

 

Just a thought on the logic level output:
The CD4050 will run off a 5V supply and take an input of up to 18V, and each section will drive two TTL loads.

 

I may use 8 AA batteries, haven't decided yet. When I start designing the 3D printed box, that's when I will make the final decision.

 

So I have 3D printed a case for this project, and now I am working on the protoboard again. I tried this one:

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the mosfets were a total bust, they loaded down the CMOS 555 to the point where it was not usable. Interestingly, I tried a 7555 and a TS555CN, the TS555CN blew the 7555 out of the water. Now I will addition BJT's to see if they work better. I will let you know.

 

Mosfets have a very high gate to source capacitance that is a heavy load at high frequencies.
Usually a pair of transistors (complementary emitter-followers) is used to drive high frequency Mosfets.

 

If I am going to use BJT transistors I will use only BJT transistors. Latest brain part:

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I will live with the shoot through.

 

Why not use the output transistors as emitter-followers? Then no shoot through current and a lower output impedance.

 

I like the true rail to rail performance, I'm just stupid that way.

 

The Cmos 555 has a rail-to-rail output only when the load current is almost zero. The common-emitter transistors you used need a HUGE base current (10 times the collector current) to saturate. The huge current at the output of the Cmos 555 reduces its output voltage swing.
Emitter-followers do not have a huge base current because they do not saturate, instead they amplify with the low base current determined by the high hFE beta.