Simple JFET Switch Question

[Frankchie]

I need a switch circuit to control a diy circuit using a 12v supply voltage. Current is modest at a max of 100ma. The control signal is a 0-8v pulse at a frequency of 50-400 hz. I'm thinking a JFET circuit because I don't want to load the source signal and I also want about a one second turn on/off delay. The high impedance of a JFET should accommodate both of those requirements. Also space is limited so small components are desirable, although I prefer to avoid surface mount devices.

I don't know much about JFETS and my basic question is what type of JFET is applicable (P-channel, N-channel, enhancement/depletion, etc.). I suppose a MOSFET may also work, but I only have a couple of power MOSFETs and their size is kinda big. Also I have a variety of JFETS in stock.

I have attached a rough sketch that may help show what I am thinking. I can figure out the component values I need, but I need some guidance on the JFET type.

Thanks
Frank

 

[Alec_t]

Also I have a variety of JFETS in stock.

Can you list them?

 

[Harald Kapp]

You can't use a JFET as simple as it seems. A JFET is a depletion device, meaning it is conductive for V_GS = 0 V and needs V_GS < 0 V (N-channel) or > 0 V (P channel).
As shown in your circuit you'd have to use a P-cahnnel JFET with source to +12 V. That requires a gate voltage > 12 V to tirn the JFET off, but you have only 0 V ... 8 B control voltage. The JFET will be conducting from gate to source. Not what you want.
If you were to use an N-channel JFET, source would have to be towards the load, drain to + 12 V. Now you'd require a negative gate voltage to turn the JFET off, and ~ 0 V V_GS to turn the JFET on. As the load draws current, the voltage at the source of the JFET will rise towards 12 V, but the gate voltage can reach only 8 V. Therefore the JFET will act as a constant current source.

When you put the JFET into the ground leg of the load, you will find that similar considerations make it unsuitable for that application, too.

In my humble opinion a N-channel MOSFET of even a NPN BJT in the ground leg will serve your purpose much better.

 

[Frankchie]

Can you list them?

I thought I bought an assortment of JFETS a couple of years back, but the only assortment I find are bipolar. I guess my memory failed me. Anyway, I do have some loose JFETs in my junk box, I found 2n5459, j201, and a couple others that I can't read even with magnifying glasses. Many years ago I could read these without any glasses.

If you can identify the basic type of JFET and circuit I need I'll figure out the details.

 

[PETERDECO]

I use this circuit frequently. This mosfet is rated 55A and can supply 2A without a heatsink.

 

[Frankchie]

Harald, Peterdeco,

The problem with using the switch in the ground side is that's also the signal path to my diy circuit. And that circuit also reads the 8v pulse that I want to trigger the on/off function. I know that MOSFETs have low on resistance, but it sounds like a bad practice to add any resistance to a signal ground path.

P.S. Harald, Thanks for the detailed explanation, it was helpful to me. I need to study what you said a littler more. Then I might have some thoughts/questions.

Thanks,
Frank

 

[PETERDECO]

Perhaps the quickest way to achieve that is to have the mosfet turn on a relay or a smaller fet to turn on a PNP transistor with the collector feeding B+ to your circuit.

 

[bertus]

Hello,

You want a high side switch. A P-mosfet will do the trick:
https://learn.sparkfun.com/tutorials/constant-innovation-in-quality-control/the-high-side-switch

Bertus

 

[Frankchie]

It dawned on me that my diy circuit employs a dc-to-dc converter that converts the 12v to 5v that actually powers that circuit. I suspect that converter can operate satisfactorily at 8v input. So I'm thinking that an N-chan FET source follower circuit with the drain on 12v and the gate at about 8v would feed a similar 8v into that converter. I've used bipolar NPN emitter followers before and I think that would work here, but the high FET gate impedance should better facilitate the 1 second turn on/off delay I want. The skeleton circuit in my original post with the details that I just provided hopefully clarifies what I'm saying.

Sound workable?

Thanks everyone for the suggestions.
Frank
P.S. Most of those suggestions were great, but I have very limited space for new components, hence the few small components of the above source follower seems like a good solution.

 

[Harald Kapp]

So I'm thinking that an N-chan FET source follower circuit with the drain on 12v and the gate at about 8v would feed a similar 8v into that converter.

It will not. You'll have to subtract V_GSth from the gate voltage, a few volts, depending on the MOSFET. For high side switching you either employ a P-MOSFET or you need a gate drive voltage much higher than the drain voltage of the MOSFET.

 

[Frankchie]

Harald,
Thank you... you are absolutely correct.

I'm getting dizzy trying to understand JFET switch operation. I think there are a fair number of mistakes on some web sites that may be confusing me.

For example, is below link correct? It seems to me that when the P-JFET is conducting the S is near zero so when the switch closes the G to S voltage is positive. Ok that opens the JFET. But when it opens the S goes to 6v and the gate is now negative turning the JFET on again, no?

http://www.learningaboutelectronics.com/Articles/P-channel-JFET-switch-circuit.php

 

[Harald Kapp]

For example, is below link correct?

Imho no.
As it is, with the switch closed th voltage at the gate will be 4 V. The voltage at the source (assuming the transistor is off, will be + 6 V. Thsu V_GS = V_G - V_S = 4 V - 6 V = -2 V. But a negative gate voltage does not turn off a p-channel JFET, it will cause a current flow into the gate as the gate-channel diode is biased in forward direction.
To create a positive V_GS as required to turn off the P-channel JFET the voltage source for the gate should not be connected to gndn but to + 6 V. Thus with the switch closed V_GS will be -4 V (note the reversed sign), thus turning off the JFET.

Actually the circuit you linked to can be viewed as having source and drain swapped, which is possible for some JFET transistors, but not all (some have special structures that require dedicated use of source and drain) and then source connected to gnd. Then the gate voltage as shown indeed creates a positive V_GS and will pinch off the JFET. If that circuit is meant to operate in his way, that is not how one uses a JFET typically.

I really encourage you to use a MOSFET. You can have them in very small cases, too. Not only in high power TOxxx cases as you seem to think (considering your remarks with respect to available space). Time to go shopping

 

[Frankchie]

Thanks again, Harald,
FETS/MOSFETS appear to be a simple topic, but there's more than meets the eye, IMHO.

I'll have to study your explanation more to fully appreciate your answer, particularly where you said
"Thus with the switch closed VGS will be -4 V (note the reversed sign)".

Anyway, last night I discovered that the dc-dc converter in my diy circuit has an IC with an enable pin tied high via a 100k resistor. So after some judicious SMT soldering, I successfully got the filtered 0-8v square wave to activate that enable pin and accomplish my goal to control the power to the diy circuit. I was also able to get the on/off delay I needed.

Thanks to all who contributed.
Frank

P.S. I happened to discover IC's specifically designed for high-side power switching applications like my application. An example is the TPS22810. I didn't look too closely, but they look like a handy future solution.

 

[bertus]

Hello,

Perhaps the following PDF's might give you a better view on the mosfets.

Bertus

 

[Harald Kapp]

they look like a handy future solution.

yes, but you will usually not find these in your parts bin. Plus the take up space, too.

 

[Frankchie]

Hello,

Perhaps the following PDF's might give you a better view on the mosfets.

Bertus

Thank you.
Frank