CMOS Switch

[mrg1999]

Can anyone recomend a chip or circuit to switch on about 100mA load when the output of a CD4049 goes high. I would also need a complimentary switch (switch on when the 4049 goes LO).
Thanks for your input!

 

[mrg1999]

Sorry, the control voltage would be in the range +3.5 to +6 volts.
I thought about using a MOSFET N type enhancement transistor but I believe the gate voltage needs to be about 10V to fully turn on.
Thanks again!

 

[mrg1999]

The 4016 or 4066 would be perfect for my application except they can only switch 10mA

 

[crutschow]

**broken link removed**, for example has some low resistance switches that should work for you.

 

[Mikebits]

They do make logic level mosfets both in P and N channel. Check out Digikey
**broken link removed**

 

[mrg1999]

I may have misunderstood the saturation region for a MOSFET.

From Wikedpedia: Saturation Mode occurs when Vgs > Vth and Vds > (Vgs - Vth)

If this is true, then a BS170 may work for me.
BS170 Gate Threshold Voltage (Vth) = 2.1V typical

Vgs is going to be around +3-5V
Vds will be about +5-6V

So if I'm reading this right: Vgs(3V) > Vth(2.1V) and Vds(5-6V) > Vgs(3V) - Vth(2.1V) or 0.9V

Continuous Drain Current is rated at 500mA so the BS170 should be ok?

Sorry, I haven't worked with MOSFETs before.
Thanks for the help.

 

[mneary]

MOSFET saturation isn't the same as BJT saturation. When a MOSFET is turned fully "on" as a switch you want it in its "linear" mode where it operates as a low-value resistor. "Saturated" is where the MOSFET acts as a current source (sink). You'll have to find other sources of information as I don't feel qualified to describe the physics.

Vgs needs to be significantly greater than Vth in order to be turned on as a switch. Looking at the BS170 datasheet, (https://www.electro-tech-online.com/custompdfs/2010/01/BS170.pdf) you'll see in figure 2 that with Vgs>4 it acts as a pretty good low current switch. You need Vgs>5V to handle higher currents.

From Figure 1, with Vgs = 3V, the BS170 functions pretty much as a constant-current of <100mA. (This is called "saturation" in a MOSFET).

Finally, the figures show 'typical' operation with Vth=2.1V. But your transistor probably won't be exactly 'typical'. These charts don't cover the full range of 0.8V<Vth<3.0V. The BS170 will probably work as a switch, with your 150mA load, if Vgs is >4.5V.

 

[Hero999]

What is the voltage of the load you're controlling?

Does it need to be isolated from the source?

A CMOS switch is not a solid state relay, the voltage being controlled needs to be within the supply voltage range. A single MOSFET can be used but it can only switch either the low or high side and level shifting will be required for high side operation, if the load voltage is higher than the control voltage.

 

[mrg1999]

My supply voltage is 5-6V; I'm trying switch on a 4017 counter when the output of a 555 timer goes high (it is also powered by the same 5-6V supply).
The output of the 555 drops to about 3.5V (it's also powering other circuits) and the 4017 behaves erratically at this voltage.

I'm looking at a BS108 n-channel and a ZVP4424A p-channel logic level mosfets to do the job (I have other circuits that also need to turn on or off at 4049 outputs as well) - same 5-6 volt supply.

Thanks for any help.

 

[Mikebits]

My supply voltage is 5-6V; I'm trying switch on a 4017 counter when the output of a 555 timer goes high (it is also powered by the same 5-6V supply).
The output of the 555 drops to about 3.5V (it's also powering other circuits) and the 4017 behaves erratically at this voltage.

I'm looking at a BS108 n-channel and a ZVP4424A p-channel logic level mosfets to do the job (I have other circuits that also need to turn on or off at 4049 outputs as well) - same 5-6 volt supply.

Thanks for any help.

Well, you got something wrong there bud. Your 555 should be able to pull way below 3.5 volts. You should forget about a fet and determine your output problem. How about posting your schematic.

 

[Hero999]

Why not use a NPN and a PNP transistor?


Of couse you need to change the power supply to the voltage you're using and replace the bulb with the CD4017 and whatever circuits you're using.

I would also consider replacing R3 with a lower value (470R should be fine) because the base current will be lower at the lower voltage you're using.

 

[mrg1999]

The purpose of this circuit is to create different sound effects with the press of a button. It is powered (hopefully) by four D cell batteries.
The ISD1790 referred to in the circuit is a multi-message voice record & playback chip manufactured by Winbond, the datasheet can be found here: http://www.datasheetcatalog.org/datasheets2/36/368701_1.pdf

When the circuit is first powered up only the ISD1790, U1 (a 4017 counter) and a 7555 pulse generator for the U1 clock signal are “hot”.
U1 cycles through it’s outputs at a 7Hz rate (as dictated by the pulse generator) waiting for one of the momentary pushbutton switches (at each of it’s outputs) to be pressed.

When a switch is pressed, it triggers another 7555 timer (U4) to send a high signal to the ENABLE Pin of U1 stopping U1’s cycling and holding the pin of U1 associated with the particular pushbutton pressed high until U4 times out (approx 3 secs).

The reason for this is to essentially lock-out all the other pushbuttons until the initially pressed button has performed it’s function (ISD1790 has played it’s associated sound and the timer has timed out. – I set it up this way because little fingers are not going to wait to push other buttons.

Once the button has been pressed, U4 goes high and powers up a second 4017 counter whose clock input is controlled by the ISD1790’s RDY pin. RDY is normally HI but goes LO during a FORWARD or PLAY pulse to it’s respected inputs.

The first step is to send a LO pulse to the ISD1790’s RESET pin which causes this chip to reset to it’s initial state (all counters reset to zero).

Then a HI pulse is sent to the 4017 (U2) clock and a LO pulse to ISD1790’s FORWARD pin causing it to skip to the next recorded sound. Once LO goes to FORWARD it causes RDY to send a LO to U2’s clock signal causing U2 to switch to the next output- this progression will continue until U2’s output matches the HI output of U1 (through the 74HC08 AND gate).

Once U2 matches U1’s HI output, a HI signal is sent to U2’s ENABLE pin through the AND gate. This cause U2 to stop cycling, a LO pulse is sent to ISD1790’s PLAY pin causing it to play the associated sound.

See how the HI to 1790’s PLAY pin also causes the U4 (7555) to time out yet still holds U1 (4017) ENABLE pin HI until it has finished playing it’s sound effect.

Any suggestions / critiques are welcome. Thank you!!

 

[Mikebits]

Well I can see why your input to the 4017 enable pin has the wrong voltage level ie. not going below 3,5v. The triple or-gate thingy needs a pull down resistor on the cathodes. Your better off using a logic or-gate, but try hanging a 10k to ground on your 4017 enable pin.

**broken link removed**

 

[kchriste]

Mrq1999,
You've got some odd circuitry there. I don't think that the P-FET will reset the ISD1790 properly. Use an N-FET but connect it's gate to pin2 of the 4069, ground it's source, and connect the drain to the ISD1790's reset pin. I don't really like the idea of switching the supply voltage to various chips off while leaving others on. A unpowered CMOS chip can be "unintentionally powered up" if one of it's inputs are driven high by another, already powered, chip.

 

[Boncuk]

Here is a logic SPST switch (if still required). Using two of them you can make a SPDT switch of them.

Boncuk

 

[mrg1999]

Mrq1999,
You've got some odd circuitry there. I don't think that the P-FET will reset the ISD1790 properly. Use an N-FET but connect it's gate to pin2 of the 4069, ground it's source, and connect the drain to the ISD1790's reset pin. I don't really like the idea of switching the supply voltage to various chips off while leaving others on. A unpowered CMOS chip can be "unintentionally powered up" if one of it's inputs are driven high by another, already powered, chip.

My reasoning for turning the logic gates off and on as well as the 4017 #2 was to conserve battery power, but I may take your advice and leave them on. I also changed the P-FET. I also made some other changes with problems I saw and have uploaded a new schematic.

Thanks!

 

[Mikebits]

You still have a problem with the enable pin of your 4017. The 3 diode or gate will not work properly. You need a pull down resistor on the enable pin.

Have a look here to see what I mean.
**broken link removed**

 

[kchriste]

My reasoning for turning the logic gates off and on as well as the 4017 #2 was to conserve battery power

When idle, and with no inputs floating, the 4017 only draws a max of 80uA at 25C with 15V supply. The typical draw is much less. The 4049? A max of 16uA at 25C with 15V supply. CMOS gates draw very little current when they aren't changing state.

 

[bwearl]

Does anyone have an idea how to create or but a switch that is activated by a weak magetic field. I want to place a magnet in a stationary location and when the switch passes by it is closed. I would appreciate any help on this.

 

[Hero999]

Use a hall effect sensor or reed switch and create your own thread in future.