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Help calculating resistance values

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Hello I want to calculate all the resistances shown in the fig. I have used +15V and -9V supply. Current should be 2-3 A through the driving transistor. Please help.
 

MikeMl

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I dont understand what the circuit is suppose to do. Is it supposed to be a linear power amplifier (like a class B audio amp)?
 
It's not an amplifier. All the transistors are used in saturation or cutoff. Circuit is used as a gate driver. You can say isolated gate driver.
 

MikeMl

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Ok, so like a totem-pole output on TTL, except that the driver can source ~2A when high, and sink ~2A when low?

I will LTSpice it later today, but have to be away for a while...
 

Colin

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You are going to have minus 9v on the base of the output transistor. It won't like that.
 

MikeMl

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I modified your circuit, at least so that it works for slowly varying input. I show the current into two separated loads RLt and Rlb as the Opto-isolator current I1 sweeps slowly from zero to 2mA. Normally, I1=0 would be the off state, while (arbitrarily) I1=2mA is the on state. The first question is what happens to the currents through Q3 and Q4 as the input slowly sweeps through its range?

46.png

Ideally, when I1 =0, Q4 is capable of sinking 4A, and Q3 is fully off. As I1 increases, Q4 turns off almost completely, but Q3 begins turning on, with just a bit of overlap, and then as I1 increases a bit more, then Q4 is fully off, and Q3 is fully on. I use a trim pot to establish the amount of overlap, and I show that with the pot wiper 78% up from the bottom, there is just a bit of overlap. The goal would be to reduce the shoot-through current where both PNP and NPN are turned on at the same time.

However, when I tested the driver in the time domain using it to drive a capacitive load at a high slew rate (fast changing I1), the performance is disappointing.

For the high to low transition of I(I1) and V(out), both the PNP and NPN are on at the same time, so the shoot through current is so large as to be unuseable. The delays through Q1 and Q2 are asymmetrical, so the design will have to be modified to solve this problem.

46t.png

I am out of time to work on this, so I include the .asc file in case you want to play with it more...
 

Attachments

Thanks MikeMI. It will be switching at about 16 Khz frequency. And also if we reduce the values of the resistances, will the shoot through be lowered? Anyone please let me know.
Thanks again.
 

ronsimpson

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There are about 100 different versions of IC that are built to do this job. This one FL3100 works up to 18 volts and 2A. There are much bigger available. Higher voltage if you want.
upload_2017-7-9_8-41-39.png
There is a simple way to get a negative output voltage if you really need it. (drive +10V and -4V)
 
Thanks. I already have made a dual power supply with +15V and -9V.

I know there are a lot of ICs available for the job. But availability here makes it difficult. Also I want to make a discrete ( cheaper ) version of the driver.
 

MikeMl

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ronsimpson

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Mike's circuit. It will not pull down (hard) below 1 volt. (1k only below about 1V)
Just saying here is a working circuit.
I usually don't have D1 so Q1 can pull down to 0.3V.
upload_2017-7-9_10-10-43.png
 

ronsimpson

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Will not work at 100%.
Output is AC coupled. Two diodes and pull down resistor sets the off voltage to -1.3 volts.
AC capacitor will need to be a good high current part. Value probably wrong. lol
upload_2017-7-9_10-27-38.png
 

MikeMl

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No reason you couldn't run the circuit in post #17 from split supplies, and then use the opto to isolate it.
 
Thanks for your help. I will definitely try this circuit out. I need to recalculate all the resistors according to my supply.

Isn't there a problem of shoot through? Q2 and Q1 may have overlapping conduction.
 
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