Latch ckt not working

Thread Starter

deepak.midri

Joined Jun 2, 2014
11
Attached is latch circuit for powering a microcontroller.

"External Switch" is trigger pressed by user which will turn on the device and wake up MCU.
Once its done, MCU will pull ground to right switch and output should turn off.

On simutaltion this ckt is not working, whats wrong in it?
Even without pressing trigger, I can read output voltage.

1590402726491.png
 

ronsimpson

Joined Oct 7, 2019
684
Depending on what MOSFET, 3.6 volts might not be enough voltage to turn on the transistor.
Even without pressing trigger, I can read output voltage.
Is this only on the computer or did you really build it?
U1B in backwards?
U1B is bad?
U1B is wrong kind P-fet not N-fet.
 
Last edited:

AlbertHall

Joined Jun 4, 2014
9,899
I can see two possible reasons for this.
Leakage through U1B produces enogh output voltage to turn on U1A.
Leakage through U1A produces enough gate voltage on U1B to turn it on.

Disconnect R2. This will let you see the gate voltage on U1B. If this is anywhere near turning the FET on then reduce the value of R3 to reduce that voltage to a safe off state.
Once you have that value for R2, make R4 the same value.

Reducing the value of these two resistors will not increase the current in the off state, though they will contribute to the current in the on state.

Finally, when you are trying to control this switch with an MCU you may need a higher value of R2
 

MisterBill2

Joined Jan 23, 2018
6,107
Attached is latch circuit for powering a microcontroller.

"External Switch" is trigger pressed by user which will turn on the device and wake up MCU.
Once its done, MCU will pull ground to right switch and output should turn off.

On simutaltion this ckt is not working, whats wrong in it?
Even without pressing trigger, I can read output voltage.

View attachment 208072
First of all, "not working" is not an adequate description of a possible problem. Of course you will read a voltage with power connected and no load. Add a 1000 ohm resistor across the load terminals and check the simulation again. It should switch on and then switch off, as you describe. Where did this circuit come from, and what voltage is it intended to operate with. Consider that device U1B must go into full saturation with less than 3.6 volts gate bias, that may not be enough for the device in your simulator, and it may not be enough for many FET type devices. If you had an actual physical version of the circuit. And now as I look at the circuit, it appears that the positive supply is applied to the drain terminal, and it is not clear how grounding the gate will make it more positive than the source terminal.

So I suggest that the TS give us a description of just how the circuit is intended to function, because some things look funny to me.
 

KeithWalker

Joined Jul 10, 2017
918
it appears that the positive supply is applied to the drain terminal, and it is not clear how grounding the gate will make it more positive than the source terminal.
U1B is a P channel MOSFET ( there is a tiny little arrow on the gate), so the gate needs to be negative with respect to the drain to turn it on. The problem is that we still don't know which MOSFETs he is using, so we really have no idea whether the circuit should work or not.
Regards,
Keith
 

MisterBill2

Joined Jan 23, 2018
6,107
U1B is a P channel MOSFET ( there is a tiny little arrow on the gate), so the gate needs to be negative with respect to the drain to turn it on. The problem is that we still don't know which MOSFETs he is using, so we really have no idea whether the circuit should work or not.
Regards,
Keith
OK, I had guessed that it was an N-FET and the arrow was a representation of the internal reverse diode. IF that is the case it would never work.
Certainly we do need to know the numbers of the devices to adequately determine how it would function.
And very likely the similar concept could be made to work with bipolar transistors, although a bit less efficiently. But at least for bipolar devices the symbols are much simpler to understand.
And in simulation the detachment from reality can be rather large. (Bob Pease did not like simulators at all)
 

Thread Starter

deepak.midri

Joined Jun 2, 2014
11
I am simulating with 12V input dc at the moment for testing.
Since 3.6V mosfets are not available in simulation software.

I think its leakage current issue. Can it be solved by using Bjt instead of mosfets?
 

ronsimpson

Joined Oct 7, 2019
684
if both are fets, tried adding load as 1K, same effect, still can see voltage at the output.
What exactly is the voltage on the output?
If the voltage is 0.15 volts then …...
BUT
If the voltage is about one volt less than the input supply then you have:
-Bad MOSFET
-Wrong direction D-S backwards
-N MOSFET while the top should be a P MOSFET

What is the part number for the parts you are using? Which number is for the top transistor?
----------edited-----------
I thought from the first post that U1B is the problem. So make the circuit simple. R2 is load. U1B with R3 to force to off. This should read very close to 0V on the output. We must make this work first.
1590499288885.png
 

KeithWalker

Joined Jul 10, 2017
918
I have never used simulators myself. I design with a calculator, pencil and notepad. I breadboard circuits and test them in real life, cautiously, in stages. You can't beat the real thing.
Keith
 

Thread Starter

deepak.midri

Joined Jun 2, 2014
11
This seems to be working. Instead of N mosfet, NPN is added.
Whenever latch is pressed, output will be seen, and MCU is powered up. Now MCU pin is made inout high impedence.
When need to turn off, this MCU pin is made output low. This will eventually cut off the latch.
Will this work as intended?

We need to make as low power as it can be.


1590583303402.png
 

MisterBill2

Joined Jan 23, 2018
6,107
Should work.

Then you could increase R1 and R2 significantly (100k?).
R2 can be increased, R1 only has current flowing during the switch off time and so it does not have much affect on power use. I suggest trying 10K for R2 first. Is the load actually equal to 36 ohms? That seems like a rather small equivalent resistance for a processor.
 
Top