low power SCR?

[tattee]

Yes, I can turn OFF the circuit by directly connecting the Gate to ground. And when I disconnect the battery and put it back, the SCR remains OFF since that was the last state(OFF state) before the battery was disconnected. But if the last state was ON before removing the battery waht happens is that when I put the battery back the circuit and SCR conducts automatically as if there's some kind of memory that tells it waht the exact state(ON state) it was last in.

If what you meant was, if the circuit would turn OFF if I remove the batteries while the whole circuit and SCR is conducting, Yes, it will.

 

[dknguyen]

Yes, I can turn OFF the circuit by directly connecting the Gate to ground. And when I disconnect the battery and put it back, the SCR remains OFF since that was the last state(OFF state) before the battery was disconnected. But if the last state was ON before removing the battery waht happens is that when I put the battery back the circuit and SCR conducts automatically as if there's some kind of memory that tells it waht the exact state(ON state) it was last in.

If what you meant was, if the circuit would turn OFF if I remove the batteries while the whole circuit and SCR is conducting, Yes, it will.

Hehe, I was asking your first paragraph. If you got something that did the second one, you'd be rich.

I was also staring at the BJT model of an SCR and made a big mistake earlier. SCRs can be turned off it seems by bringing the base low (however, they remember retain their current conduction state, on or off, if the base goes high impedance) and then the base goes high impedance). It is a triac (modelled by two anti-parallel SCRs) that cannot be turned off using the base once turned on due. As a result I deleted my first post completely since almost all the information in there applied to triacs. THe difference being that while a triac and SCR can block current in both directions (unlike MOSFETs which can only block in one direction), a triac is able to conduct in both directions while an SCR can only conduct in one direciton.

You mentioned something about the capacitor in your memory problem.
DOes it occur with or without the capacitor?
Or does it occur whether the capacitor is there or not?
And how long does the circuit hold its memory for?

I can sort of see how the capacitor might act like the battery after the battery is disconnected thus keeping the SCR on causing a memory effect...but the capacitor is too small and would get drained too fast for you to notice this particular memory effect.

 

[tattee]

By base you mean the gate?? SO if the base or gate goes high impedance, it would retain its last state? Is that what you mean?

 

[dknguyen]

By base you mean the gate??

Yeah. SOrry about that. I first started calling it the gate (like on a MOSFET) but this is more BJT-based so it should be the base.

SO if the base or gate goes high impedance, it would retain its last state? Is that what you mean?

Yes, but ONLY if there is still current flowing through the triac somehow. The PNP needs current to keep the NPN on and the NPN Needs current to keep the PNP on. So energy has to be coming from somewhere in order for the two BJTs to keep each other on. Once there is no more energy the cycle, it shuts off. THe only place I can see that coming from with no battery is the capacitor...but the capacitor is too small unless I am missing something.

THe capacitor is 1uF right? With 1uF the memory effect from it would last at most 10ms- too short for you to notice. With 1mF the effect would last for 11 seconds absolute maximum which you might notice. How long does the memory effect last for (how long can you wait before reconnecting the battery and not have it remember?) DOes it happen without the capacitor?

 

[tattee]

yeah, I know the capacitor is too small for that. Thats why im wondering why and how it stores the last state.. Is there any way I can drain that energy from the transistors?

 

[dknguyen]

BJTs have very little capacitance as far as transistors go. How long does the memory effect last for?

If you wait 1 minute, 5 minutes, 10 minutes, or an hour to reconnect the battery, does the circuit still remember?

 

[tattee]

I will try to have some test on that. I will just update you tomorrow on the results coz I have to go now. Thank you very much for your prompt response. I appreciate your help!

 

[tattee]

The PIC will not be able to turn off the circuit as shown, only turn it on. You need to pull point G to ground to shut the circuit off. To do that you also need to add a resistor in series with the collector of Q1.

If you place the PIC output in a high impedance state, then you don't need the diode to the PIC. To turn it off, just put the PIC output in a logic 0 state. Or you could add another transistor from point G to ground contolled by the PIC.

Just want to ask, what is meant when you say place PIC output in high impedance state?

 

[crutschow]

Just want to ask, what is meant when you say place PIC output in high impedance state?

Many microprocessor digital outputs can have three states (tri-state): high, low, or off. The off state is an open-circuit, or high impedance state. It's normally to prevent interaction when there is more than one driver on a bus.

 

[kchriste]

In simpler terms, change the PIC output into an input by setting the appropriate TRISx bit. This makes the pin high impedance.

 

[Sceadwian]

Actually crut, there's 4 states. Hi, low, hi-z and pullup. Pullup means the line is connected internally to VCC via a pullup resistor the pullup resistor value can vary widly but is typically 30-60k ohms.

When the pin is set to output toggling the I/O line will make it go from high to low.
When the pin is set to an input toggling the I/O line will make it go from pullup to hi-z

 

[Ubergeek63]

Actually crut, there's 4 states. Hi, low, hi-z and pullup. Pullup means the line is connected internally to VCC via a pullup resistor the pullup resistor value can vary widly but is typically 30-60k ohms.

When the pin is set to output toggling the I/O line will make it go from high to low.
When the pin is set to an input toggling the I/O line will make it go from pullup to hi-z

Ummmm.....no. The pullup is actually a mosfet current source.

While the effective resistance might be considderd to be 30-60K it is only there when you are pulling it to ground, it is actually nonlinear and if you look at the spec close enough you see it stated as 20µA typical.

Others have speedup sources for the output mode that can cause grief in the intput mode.

Dan

 

[Sceadwian]

There's still four states ubergeek. Though you're probably right about it being a mosfet current source.

 

[Ubergeek63]

There's still four states ubergeek. Though you're probably right about it being a mosfet current source.

Oh I know...I was really commenting on just the 30-60K pull up statement. I don't know how many times I have corrected my boss, the owner of the company, on that detail. It does happen to appear as 30-60K when you are pulling it to near ground, but it is really a current source that peters out as you approach the positive rail.

Dan

 

[Sceadwian]

I guess I just consider it a resistor because if you're trying to pull it all the way to ground it's equivalent resistance is what matters. I don't think it'd be particularly useful to attempt to bias an I/O line at it's mid point for any reason.

 

[tattee]

Thanks for your reply guys. Bt the way i've attached the circuit diagram.

• The switch will trigger the SCR to turn ON
• SCR will continue to run no matter the condition of the switch
• PIC will turn OFF the SCR when a "low" signal is set

I placed a diode on the PIC to protect it from the +voltage. I was worried that the +voltage might damge the PIC port. Are my assumptions right? Also, I can't set the PIC port with the corresponding TRIS* at the beggining of the program since it may affect the operation of the SCR.

 

[Ubergeek63]

I guess I just consider it a resistor because if you're trying to pull it all the way to ground it's equivalent resistance is what matters. I don't think it'd be particularly useful to attempt to bias an I/O line at it's mid point for any reason.

but it makes a big difference if you have a marginal source pulling it down or are hoping to rely on it to pull up something other than a CMOS gate input.

In the former case you might not get it a good low input and in the latter you might not get a good low output. In another situation you might just say you have an internal pull up to a CMOS input and either not have the speed or be very susceptible to noise. The speed up sources cause even more trouble since they add current drive only in the transition zone.

Dan

 

[Ubergeek63]

Thanks for your reply guys. Bt the way i've attached the circuit diagram.

• The switch will trigger the SCR to turn ON
• SCR will continue to run no matter the condition of the switch
• PIC will turn OFF the SCR when a "low" signal is set

I placed a diode on the PIC to protect it from the +voltage. I was worried that the +voltage might damge the PIC port. Are my assumptions right? Also, I can't set the PIC port with the corresponding TRIS* at the beggining of the program since it may affect the operation of the SCR.

nope...

the PIC would have no trouble with the voltage since it is running off a higher one.

The direction of the diode would be to turn on not off the SCR.

Assuming you could turn off the SCR that way the PIC would have no where near enough output current to do so.

Why are you not just using a FET?

 

[tattee]

By the way, the PIC will get its supply voltage from the output of the SCR. So, it will have a supply voltage of 3.3V. I don't know how to use FET. Is it MOSFET I should use? Will it function the same way the SCR works in my circuit?

 

[kchriste]

One the first page of this thread, I showed you a way to do it with transistors:
https://www.electro-tech-online.com/threads/low-power-scr.86544/#post673211