[SOLVED] shoot through prevention

[darryl_co]

How do I prevent a shoot through in this circuit. I am not driving the inputs through a microcontroller, using ne555

 

[barry]

You need to add some delays. You can either use an off-the-shelf driver, or add a couple of one-shots to the circuit. You might be able to replace the whole thing with a single IC.

 

[Vbase]

TIP122 is on before TIP127 is completely off, the current through them can cause a lot of heat.
Put diode 1N4148 across R6 with cathode to the base of the transistor. Add 100pF from the base of the transistor to ground.
Do the same for R3.
For R1 and R9 the same but with the diode reversed (anode to base of transistor).
The idea is to delay the switch on of the power transistors, and not to delay the switch off. You may have to increase the 100pF if the problem is only reduced.

BTW, to my opinion the circuit isn't a clever design, if single transistors instead of darlingtons were used, and also PNP and NPN swapped the efficiency would have been the same but without the problem of both transistors conducting together.

 

[darryl_co]

TIP122 is on before TIP127 is completely off, the current through them can cause a lot of heat.
Put diode 1N4148 across R6 with cathode to the base of the transistor. Add 100pF from the base of the transistor to ground.
Do the same for R3.
For R1 and R9 the same but with the diode reversed (anode to base of transistor).
The idea is to delay the switch on of the power transistors, and not to delay the switch off. You may have to increase the 100pF if the problem is only reduced.

BTW, to my opinion the circuit isn't a clever design, if single transistors instead of darlingtons were used, and also PNP and NPN swapped the efficiency would have been the same but without the problem of both transistors conducting together.

Putting a diode 1N4148 across R6 with cathode to the base of the transistor, will it not bypass R6 in forward conduction. It would be that the ic is driving the transistor without any resistor. Placing a diode across R9 and R1, can the ic handle the discharge current of the capacitor?

 

[D.A.(Tony)Stewart]

For < $0.50 you can get a complete solution in smaller area and guaranteed performance.

0.7A average 3A peak 2~5.5V

Bigger chips cost more, like $2 @45A

and some high power uses external MOSFETs

Why reinvent the wheel?

 

[Vbase]

Putting a diode 1N4148 across R6 with cathode to the base of the transistor, will it not bypass R6 in forward conduction. It would be that the ic is driving the transistor without any resistor. Placing a diode across R9 and R1, can the ic handle the discharge current of the capacitor?

You are right. I missed out important detail. You need also to change Q1-Q4 to 2N7000 (mosfet).

 

[darryl_co]

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You are right. I missed out important detail. You need also to change Q1-Q4 to 2N7000 (mosfet).

will this modifications provide delay and avoid shoot through

 

[Vbase]

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will this modifications provide delay and avoid shoot through

Yes, it will avoid shoot through.

In your circuit Q1,Q2 don't need to be changed to 2N7000.

 

[D.A.(Tony)Stewart]

Yes, it will avoid shoot through.

In your circuit Q1,Q2 don't need to be changed to 2N7000.

Vbase might need a simulator.

Think careful about the asymmetric R values that control slew rate. Assume the transistor is 10 Ohms and Darlington is 1 Ohm.

For well controlled Dead Time you want all drivers to lead to each output switch to turn Off fast and turn On slow in a symmetrical way so it is independent of switching frequency.

Q5 the upper PNP Darl. is driven at base by 10V/1k = -10mA low and only (1+3.9)K pullup
Q6 the lower NPN Darl. is driven at base by 11.6V/1k~12mA pullup and 1 Ohm pulldown

Similarily Q9 emitter presents <1 ohm pullup while Q8 pulls down at 100 Ohm.

So it is poorly balanced.

The topology can be improved greatly and the values slightly.
If you really want to find out why this design was aborted in IC's keep working on it.

 

[Vbase]

Vbase might need a simulator.

I had a simulator once, it kept on finding faults in everything I do, so I took Obama's advice and said: "bucket".

 

[darryl_co]

The basic idea was to solve shoot through problem. Whereas the Darlington biasing is concerned the 1k can be reduced to provide more bias current. Pull up 3k9 could be from base to emitter of tip127. Q9 also present 100 ohms pull up

 

[D.A.(Tony)Stewart]

I had a simulator once, it kept on finding faults

The usual faults are because Caps and Voltage sources in libraries have no ESR so infinite surge currents occur. I simply add where necessary the realistic values.

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Q9 also present 100 ohms pull up

No Rc has no effect on emitter Zout , as Collector is a Current sink ( high Z )
Only Rb, hFE affects Zout on emitter.

 

[darryl_co]

Should I use a fet instead of a transistor for Q9

 

[D.A.(Tony)Stewart]

Of course a FET is far better to drive a Quasi-Complementary push pull BJT output. This is why IGBT's out perform MOSFET in high voltage, high current switching.

The use of MOSFET's exclusively has limitations as Ciss rises with any drop in ESR or RdsON, whereas BJT's do not but are limited by Vce(sat) current gain from 10 to 50 which directly affects cost.

Objectively the Zout or ESR or RdsON or Rce and Voltage * Amps* tPW50 drop with PWM rates must be reduced with each stage, while optimizing power driver efficient and cost and the output final result as a ratio of the load ESR, directly affects ripple voltage in Caps and ripple current in inductive or motor loads and directly controls load regulation % in step R loads.

So rather than myopically suggest 1 part solution, where an integrated solution might be more cost effective, I would prefer if you defined your goals in terms of input, output range and load range with a target for inefficiency, cost and max temperature.

Otherwise no solution can be evaluated against your goal. This procces to a spec.list is not easy to learn, but is the most important part of successful professional design (motus operandi)

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Simple TIP darlingtons use an Ic/Ib = 10 for each stage thus 100 total and when using the same Vcc for bias, the ESR reduction is 100:1 but with Darlington's the Vout drop starts at Vsat =Vbe+Vce and rises with Rce of the output stage.