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LM1949 with boost voltage

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Matthias

New Member
Hello all,

I am stuck with the following problem: I want to drive a low impedance fuel injector using the LM1949 (https://www.electro-tech-online.com/custompdfs/2009/10/LM1949.pdf) IC. My application is very similar to that shown on the first page of the data sheet. However I want to apply a boost voltage of 60V for the first 0.5 ms to decrease injector opening time by quickly ramping the coil current in the injector to 10A. I have a simple circuit using a timer and dflip flop to give me a 5V output for the first 0.5ms of the injection event and every thing looks pretty good so far. The only problem I am left with is the darlington. Can you guys recommend a darlington that can be drivven by the LM1949 that can handle 10A and 60V?

Thanks Matthias
 

k7elp60

Active Member
A NTE251 should do the job. It is available from Mouser electronics.
 

Matthias

New Member
k7elp60,

thanks for the info. One more question though, I have added a circuit around the LM1949 to apply the 60V boost voltage for the first 0.5 ms of the injection event. Would you mind taking a look at it and give me some feedback.

The idea is basically to have the flip flop as a kind of rising edge detector. The clk input is connected to the PWM signal of the ECU and the d input is connected to +5 V. Hence every time the PWM has a rising edge (indicating the start of an injection event) the Q output of the flip flop will be set to +5 V. The LM339 will then pull output #1 to GND since Q is greater than the voltage on pin 7 of the LM339 and Q2 should become active. (on a side note: Q2 has to be a pnp transistor correct? - Also, do i need a resistor connecting the base of Q2 and the output of the op-amp to control the max current out of the base?).

Now, as the voltage on the condenser (pin #8 on the lm1949) rises the second op-amp will reset the flip flop to deactivate the boost voltage. Using the potentiometer R3 I can set the time for the boost voltage to be applied. The LM1949 will control the current in the electronic fuel injector (EFI) to 10 A peak and 2.5 A hold. D1 will be conducting, if there is no boost voltage applied and injector operation will be on +12 V.

I am not an electrical engineer and I am not super confident that i picked all components correctly and that his circuit will work, therefore I greatly appreciate any help that will keep me from burning through transistors and other ICs.

Thank you very much.
 

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Roff

Well-Known Member
I haven't analyzed your circuit, but LM339 will only tolerate 36V max. You can use an NPN instead of the LM339 to drive the PNP Darlington. You can also use LM393 instead of LM339. LM393 is the dual version of LM339.

EDIT: Your PNP Darlington was drawn incorrectly.
I replaced Z1 with a diode to +60V, since the zener would have to be greater than 60V anyway.
 

Attachments

  • Injector driver.png
    Injector driver.png
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Matthias

New Member
Thank you all,

I have gone through my circuit again and came up with revision 3 which is enclosed.

A question for Ron: The pnp transistor I picked specifies a max. base-emitter voltage of 5V. Your schematic shows two 4.7kOhm transistors, where i belive one is used as a pull up resistor, while the second is controlling current once the transistor which is connected to the flip flop is active. This however would result in a base voltage of approx. only 30V (not accounting for the voltage drop in the transistor). Wouldn't that cause trouble for the pnp darlington?

I have taken the same approach in my latest revision but sized the resistors differently to obtain a voltage drop of 3.5V across the first resistor only, which should still be sufficient to drive the pnp darlington into saturation.

Any ideas or comments are creatly appreciated!

Thanks.
 

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  • Rev_3.pdf
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Matthias

New Member
Oh i kept the zener, as national semiconductor suggests a zener (breakdown at 75V) for this application of the LM1949 - I guess with the darlington i picked i do not really need it though since that has one built in i believe.

Thanks.
 

Roff

Well-Known Member
Thank you all,

I have gone through my circuit again and came up with revision 3 which is enclosed.

A question for Ron: The pnp transistor I picked specifies a max. base-emitter voltage of 5V. Your schematic shows two 4.7kOhm transistors, where i belive one is used as a pull up resistor, while the second is controlling current once the transistor which is connected to the flip flop is active. This however would result in a base voltage of approx. only 30V (not accounting for the voltage drop in the transistor). Wouldn't that cause trouble for the pnp darlington?

I have taken the same approach in my latest revision but sized the resistors differently to obtain a voltage drop of 3.5V across the first resistor only, which should still be sufficient to drive the pnp darlington into saturation.

Any ideas or comments are creatly appreciated!

Thanks.
You don't understand transistors. The 5V spec is for maximum reverse base-emitter voltage. The forward drop of a BJT is ≈0.7V, so the Darlington Vbe is about 1.4V. I assumed the Darlington needs Ib>(Ic/1000). The 4.7k between base and emitter has about (1.4V/4.7k)=300uA. The 4.7k resistor between the PNP base and the 2N5551 collector forces about ((60V-1.4V)/4.7k)-300uA=12.2mA Darlington PNP base current. The base current of the 2N5551 is about (5V-0.7V)/4.7k=900uA.
The 4.7k from base to emitter of the PNP Darlington helps speed up turnoff of the Darlington, and shunts any collector-base leakage to +60V, to ensure that it stays off when it is supposed to be off, even if it gets hot.
 

Matthias

New Member
Ron,

thank you so much - I understand now! I am a mechanical engineer and my knowledge about transistors is truly limited. All my knowledge is based on reading some data sheets one class that covered the basics of electronics about 5 years back and I think we spent about a week on transistors...

Again thank you very much for your help I greatly appreciate it! I will put this circuit together and put it to the test! Once, this project is completed I am sure I will know a little more about transistors.

Matthias
 

Roff

Well-Known Member
Ron,

thank you so much - I understand now! I am a mechanical engineer and my knowledge about transistors is truly limited. All my knowledge is based on reading some data sheets one class that covered the basics of electronics about 5 years back and I think we spent about a week on transistors...

Again thank you very much for your help I greatly appreciate it! I will put this circuit together and put it to the test! Once, this project is completed I am sure I will know a little more about transistors.

Matthias
I hope the RC network on the 1949 gives you the delay you want. I didn't look at that very closely
I think you've done a great job for an ME. Reminds me of a job I had once where my boss was out to get me. He had me design a turret to rotate lenses in front of a camera. I'm an EE. I think I did a pretty good job, but I never found out, as I managed to escape after being there for a few months.
 

Matthias

New Member
Ron,

the RC network with the flip flop works great and I do get the delay I need - I have already built that section of the circuit on a rapid prototyping board and checked the performance with an oscilloscope.

I will let you know about the performance of the entire circuit once i have it built!

Thanks,
Matthias
 

Roff

Well-Known Member
I'm curious how you came up with 60V, 500uSecs, 10A. Do you have an inductance spec or measurement? What is the series resistance?
 

Mike odom

Active Member
You don't understand transistors. The 5V spec is for maximum reverse base-emitter voltage. The forward drop of a BJT is ≈0.7V, so the Darlington Vbe is about 1.4V. I assumed the Darlington needs Ib>(Ic/1000). The 4.7k between base and emitter has about (1.4V/4.7k)=300uA. The 4.7k resistor between the PNP base and the 2N5551 collector forces about ((60V-1.4V)/4.7k)-300uA=12.2mA Darlington PNP base current. The base current of the 2N5551 is about (5V-0.7V)/4.7k=900uA.
The 4.7k from base to emitter of the PNP Darlington helps speed up turnoff of the Darlington, and shunts any collector-base leakage to +60V, to ensure that it stays off when it is supposed to be off, even if it gets hot.

OK guys, but don't forget I squared R... your 4.7K base resistor will dissipate about 3/4W, so you'll need a 2 or 3W resistor there, unless you can gaurantee a duty cycle of less than 8% (1/12th). Remember, automotive applications are at elevated temperatures so we have to derate. Use 1/2watt resistor for 1/4 watt application...etc...

Also, diode D2 does absolutely nothing for you except dissipate power. You already have a pn junction pointed in that direction, one you're turning on and off. Are you expecting the 12V to go higher than 60? Because that's the only thing in the world D2 would be for...

Also, on your darlington base current calculation, you forgot to subtract another .7v, the collector sat. voltage on the NPN you're using to turn the darlington on... it's 60v - 1.4v (darlington) - 0.7v (NPN) = 57.9V divided by 4700 then minus 300uA.

The purpose of the zener is to protect the darlington driving the Injector. When that device turns off, it's collector will go to twice the VCC (24V if 12, 120V if 60... see LM1949 data sheet Figure 2 page 7 Q1 collector voltage... they show it going up to VZ... the zener clamping voltage) by the nature of inductors. Therefore, the zener should be selected to be less than the max collector to emitter voltage of the darlington. If the transistor you select has one built in, then you don't need it. If you connect it to 60V, you'd be using it as a flyback diode, and you'd need one to 12 also, but you can't if you want that point at 60v so more properly it would be placed across the injector (on the injector side of the fuse, even). It's purpose is to help the field collapse when the injector current is turned off it can no longer flow from Vcc to ground, so it now still flows out from the injector bottom, through the diode (now forward biased since the collector is at 2 x Vcc), back into the top of the inductor to let the field collapse and ease the voltage seen on the collector of the driving device. Remember, current can't change instantaneously in an inductor, so when you turn off the switch, it still needs to flow somewhere...
By connecting the diode to 60v and with the upper darlington turned off, the diode becomes a clamp that will clamp the collector to 60v (not let it go above 60.7V) but does nothing to help collapse the field, and protects a device with a collector to emitter breakdown greater than 60V.

Mike
 
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Roff

Well-Known Member
This thread is over 4 months old, so all this is probably irrelevant to the OP, but here goes anyhow...
OK guys, but don't forget I squared R... your 4.7K base resistor will dissipate about 3/4W, so you'll need a 2 or 3W resistor there, unless you can gaurantee a duty cycle of less than 8% (1/12th). Remember, automotive applications are at elevated temperatures so we have to derate. Use 1/2watt resistor for 1/4 watt application...etc...
No argument here.

Also, diode D2 does absolutely nothing for you except dissipate power. You already have a pn junction pointed in that direction, one you're turning on and off. Are you expecting the 12V to go higher than 60? Because that's the only thing in the world D2 would be for...
One other thing in the world... If the 60V supply comes up after the 12V (most likely it will), or if it fails, then the base-emitter junction of the PNP will break down, with no current limiting, causing it to fail.

Also, on your darlington base current calculation, you forgot to subtract another .7v, the collector sat. voltage on the NPN you're using to turn the darlington on... it's 60v - 1.4v (darlington) - 0.7v (NPN) = 57.9V divided by 4700 then minus 300uA.
That's wrong. Vce(sat) on 2N5551 is guaranteed to be less than 150mV at Ic=10mA, and less than 200mV at 20mA. Typically it is less than 100mV. These voltages are in the same range for almost any general purpose transistor. Ignoring Vce(sat) in this situation is acceptable.

The purpose of the zener is to protect the darlington driving the Injector. When that device turns off, it's collector will go to twice the VCC (24V if 12, 120V if 60... see LM1949 data sheet Figure 2 page 7 Q1 collector voltage... they show it going up to VZ... the zener clamping voltage) by the nature of inductors. Therefore, the zener should be selected to be less than the max collector to emitter voltage of the darlington. If the transistor you select has one built in, then you don't need it. If you connect it to 60V, you'd be using it as a flyback diode, and you'd need one to 12 also, but you can't if you want that point at 60v so more properly it would be placed across the injector (on the injector side of the fuse, even). It's purpose is to help the field collapse when the injector current is turned off it can no longer flow from Vcc to ground, so it now still flows out from the injector bottom, through the diode (now forward biased since the collector is at 2 x Vcc), back into the top of the inductor to let the field collapse and ease the voltage seen on the collector of the driving device. Remember, current can't change instantaneously in an inductor, so when you turn off the switch, it still needs to flow somewhere...
By connecting the diode to 60v and with the upper darlington turned off, the diode becomes a clamp that will clamp the collector to 60v (not let it go above 60.7V) but does nothing to help collapse the field, and protects a device with a collector to emitter breakdown greater than 60V.

Mike
When an inductor is switched off, the flyback voltage can go far above Vcc. It is not restrained to 2 x Vcc. Even with Vcc=12V, it can reach hundreds of volts or more.
A flyback clamp cannot work if there is not a return path for the inductor current (where would the clamp current come from?), so if you can clamp, you will provide for field collapse.
I think the zener (in series with a diode) across the inductor is a good idea. A zener to ground will also work.
The diode to +60V is a bad idea in retrospect because +60V would have to sink current, which most positive V power supplies can't do, except through their output capacitor. Plus, there's the issue of +60V possibly not coming up immediately, or failing.
 
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