Still not clear to me what your asking. I'm just wanting to "pre-charge" the bootstrap capacitors of two separate half bridges so the circuit will work when called on. This circuit will be totally sequential, no clock signal involve, so the bootstrap caps need to be pre-charged.

What I said was according to the circuit of FAN7390 in page 3, the two outputs HO(High Side) to drive an IGBT and LO(Low Side) to drive another IGBT, and normally two Output Sides will connected together similar as the last output stage of the amplifier, and they also need the input square waves, if you don't want the clock to drive then why don't you just use the MOSFET to be the driver for IGBT?

If you just want the bootstrap function and you want two half bridges working separated then you will need the RBoot, DBoot and CBoot, you can try to pull the inputs HIN and LIN to High, the ic was designed to working with clock, so if you choose to work with your own way then you just try it.



You may refer to the Application Note below --
Design and Application Guide of Bootstrap Circuit for High-Voltage Gate-Drive IC_AN-6076.pdf

 

After posting and going offline, I remembered I didn't specify the logic I'm using. I know to you guy's it is obsolete, but I have a ton of it. I'm using CD4000 series logic, running on 12V.



I looked at the available chips and didn't see any that work for the old Cmos logic.

Most of the time you can get away with a simple C/R charging circuit - but you have to include a diode from the capacitor to Vdd so the collapsing rail at switch off brings down any charge on the capacitor. CMOS inputs have an inherent SCR structure, taking an input out of rail can latch it up - and also kill the internal input protection diode.

If the C/R is too slow - clean it up with a Schmitt inverter or buffer.

 

What I said was according to the circuit of FAN7390 in page 3, the two outputs HO(High Side) to drive an IGBT and LO(Low Side) to drive another IGBT, and normally two Output Sides will connected together similar as the last output stage of the amplifier, and they also need the input square waves, if you don't want the clock to drive then why don't you just use the MOSFET to be the driver for IGBT?

If you just want the bootstrap function and you want two half bridges working separated then you will need the RBoot, DBoot and CBoot, you can try to pull the inputs HIN and LIN to High, the ic was designed to working with clock, so if you choose to work with your own way then you just try it.

Yes I know that. But in reality I'm just using the high side switches. One will be to charge a capacitor bank, the other will discharge that same cap bank. The low side mosfet will just be used to charge up the bootstrap cap so the high side will work. It can't be done with a SCR or other devise , because the cap bank will never be fully discharged to 0V, by design.

 

New question on this topic. Will a "switch debounce" be needed with the "power on reset" or will the power on be long enough to take care of switch bounce? When using one of these type of circuits -

 

Another question. As I am trying to learn this stuff on my own, and have many books on logic "chips" and how they work together, none of them show how to connect them to the "world". By world I mean other things that are needed to make them work with other circuits that the logic will be controlling. Can anyone suggest a good book or website that gives that type of information? Most of you are EE's but you had to learn it somewhere, so can you help?

 

Another question. As I am trying to learn this stuff on my own, and have many books on logic "chips" and how they work together, none of them show how to connect them to the "world". By world I mean other things that are needed to make them work with other circuits that the logic will be controlling. Can anyone suggest a good book or website that gives that type of information? Most of you are EE's but you had to learn it somewhere, so can you help?

I don't know of any useful book on the subject, but http://www.electronics-tutorials.ws/ has a couple of very good pages on interfacing to digital inputs and outputs that covers the basics pretty well. They're worth studying. Also, Googling "how to interface logic chips" brought up a bunch of good stuff.

 

New question on this topic. Will a "switch debounce" be needed with the "power on reset" or will the power on be long enough to take care of switch bounce? When using one of these type of circuits

As long as the RC time constant is larger than any expected bounce time; no. Bounce will increase time reset is asserted.

What is the RC feeding? If it's CMOS as you mentioned, you can use a much smaller capacitor and larger resistor. With a small cap, there's no need to be concerned about contact welding and the 100 ohm resistor can be omitted.


For "long" reset periods, a schmitt trigger would decrease power dissipation.

 

The time constant in 24 is 82 ms, so the time to a CMOS transition level (Vdd/2) is around 57 ms. That should clean up most switches.

ak

 

Thank you all for the help! It's hard learning all of this by yourself, much of what is needed isn't shown in a book, I guess because it's either too basic or what a teacher gives in a classroom. Wish I had gotten interested in this stuff at an earlier age, seems to get harder to learn when you get older.

 

What is the RC feeding? If it's CMOS as you mentioned, you can use a much smaller capacitor and larger resistor. With a small cap, there's no need to be concerned about contact welding and the 100 ohm resistor can be omitted.

It is the 4000 series of CMOS. If the 100 ohm resistor is eliminated, won't that do away with the "R" of the "RC" part of the circuit? Sorry to be so dumb on this stuff.

 

Wish I had gotten interested in this stuff at an earlier age, seems to get harder to learn when you get older.

Learning electronics in a haphazard manner is going to present you with a lot of pot holes to navigate.

Without a firm foundation to build upon you could face endless frustrations. It can be done, but I expect that success stories will be vastly outnumbered by failures...

 

It is the 4000 series of CMOS. If the 100 ohm resistor is eliminated, won't that do away with the "R" of the "RC" part of the circuit?

Assuming that power has been on long enough for the capacitor to have charged fully, the purpose of the switch is to discharge the capacitor so it can charge again.

Sorry to be so dumb on this stuff.

If you never studied this in school, no one could expect you to know it.

It wasn't anything that I recall learning in my first two years of college. I learned about it by seeing it on a schematic in my first job as a technician. But, it wasn't difficult to figure out what it was doing because capacitor charging/discharging is taught early.

 

Learning electronics in a haphazard manner is going to present you with a lot of pot holes to navigate.

Without a firm foundation to build upon you could face endless frustrations. It can be done, but I expect that success stories will be vastly outnumbered by failures...

Tell me about it! I'm slowly getting the hang of things but what a journey it's been. If it wasn't for this site I'd still be lost, or lost worse than I still am. I built a few Heath kits when younger, but this is my first foray into my own circuit. Still not there yet, but with help I will be.

If you never studied this in school, no one could expect you to know it.

It wasn't anything that I recall learning in my first two years of college. I learned about it by seeing it on a schematic in my first job as a technician. But, it wasn't difficult to figure out what it was doing because capacitor charging/discharging is taught early.

This is the stuff I'm lacking in. I'm getting better at seeing what is happening in a schematic, but still get lost sometimes.

 

It is the 4000 series of CMOS. If the 100 ohm resistor is eliminated, won't that do away with the "R" of the "RC" part of the circuit?

No. The 100 ohm resistor is there to prevent a current spike out of the capacitor from damaging the switch contacts. Another way is to make the capacitor smaller so it doesn't store up enough energy to do damage.

Also, for a CMOS input you can scale the reset timing components to a higher impedance network, and get rid of the electrolytic. A 1.0 uf, 0.33 uF, or even a 0.1 uF ceramic works well as a POR component, with lower cost, no possibility of an assembly error, and way better long term stability and reliability. Adjust the resistor for the period you want, but keep it to 1 M or lower.

t = 0.7 x R x C

ak

 

Another related question to this, and shows my lack of understanding. Is there a way to power up the chips involved in a circuit BUT NOT actually start the logic until you want it too? Or stop the actual logic running without powering down?

The circuit I'm trying to do will need to run until the movement of an actuator gets to a predetermined point. A micro switch will determine this point. There is no internal clock in this logic as usual logic uses. it is all dependent on sequencing a step by step process, controlled by voltage levels in different parts the total circuit. In a previous try at this I called it "asynchronous" logic, but was told that was the wrong term but wasn't give a term that was determined "correct".

I'm sorry if I'm not using the "correct" way of explaining myself. But I just don't know all of the "correct" terms.

 

Is there a way to power up the chips involved in a circuit BUT NOT actually start the logic until you want it too? Or stop the actual logic running without powering down?

Post an example.

Devices that aren't clocked evaluate their inputs continuously. Running and stopping only make sense when there is a clock.

 

Post an example.

Devices that aren't clocked evaluate their inputs continuously. Running and stopping only make sense when there is a clock.

That's what I was afraid of. There are two 4013 flip flops that pretty much control the whole thing, and I am using other logic to make them change by using the clock pins on them.

I'm also trying to learn how to do the schematic in a CAD program, DesignSpark. When I get something closer to looking good I'll post it in a new thread. And hopefully I can get you and others that have helped so far to look at it for me. If it is OK with you, I'll tag you when I put it up on the forum.

 

There are two 4013 flip flops that pretty much control the whole thing, and I am using other logic to make them change by using the clock pins on them.

When you use gates to control the inputs of the flip flops, you need to follow the setup and hold times to make sure you get the expected behavior.

Clocked logic is triggered on the edge of the clock. If you look at the circuit for a D FF in CD4013, you'll see that they consist of two level sensitive D latches (the inputs follow the outputs as long as "clock" is HIGH; not just what's present when the clock goes HIGH). By combining two, you get something that behaves like it's edge triggered.

The second half of this presentation shows the differences between D latches and edge triggered D flip flops.

 

Again thank you for working with me on this! I'll be reading the PDF and trying to digest it. There were two "reading" references in the beginning of the PDF, do you know the complete names of the books by chance? One was 'Brown and Vranesic', the other was 'Roth'. I'm more of a book reader and would like to find copies if they are good. Thanks again.

 

Post an example.

@AnalogKid @dl324 I found an older version of what I'm trying/wanting to do. There are some things I'm changing, one is the diode "or" will be changed to a standard logic chip. And there is a timer in the lower right corner that I don't think I will be using. But before I keep doing this, would you please look at what I did with the 4013 part of the circuit? I know it's not a conventional way to use the 4013, but from my view it doesn't seem to break any laws.

Since the switching of the mosfets have to respond to voltage levels at different times in different places in the circuit, and are coming from comparators, I came up with this to give a rising edge for clocking the flip flop. Like I said conventional but think it will work for the "step by step" control I need.