Discrete Boost converter output voltage drop

Miracletech

Member

• Oct 30, 2020

• #1

I'm using this circuit design:
"https://www.eevblog.com/forum/proje...onverter/?action=dlattach;attach=294867;image"
But I changed some values:
R1:220 ohms
R3:470 ohms
C1:470PF
M1: TIP41C.
When I turn the converter on and connect a 500MA Load, The output voltage significantly drops from 5.6V to 4.6V, And there is voltage drop across my inductor(???)
My question is; Is there any fault on my side? I used every other thing of same value except the components above.
Is my inductor saturating, Because I'm getting 0.07v drop across it, which is odd.
Any ideas what is going on?
(By the way, I'm stepping up 3.7v to 5V for Led strip. I like discrete.)

 

rjenkinsgb

Well-Known Member

Most Helpful Member

• Oct 30, 2020

• #2

The circuit is designed for a FET.
Your TIP41C is a fairly low gain device typically used for audio, I'd just say it is not very suitable for that type of circuit.

Adding a 10K resistor base-emitter to Q2 may? improve regulation slightly.

What inductor are you using & does it have an adequate DC current rating?

Edit - I'd also connect a 100 Ohm resistor in series with C1 to limit the base current to Q2.

 

Miracletech

Member

• Oct 30, 2020

• #3

2.2UH 2A saturation current.

 

Miracletech

Member

• Oct 30, 2020

• #4

What I had is TIP41C, But I've tested with different others, Like the C3074 And D1207, Same result.

 

Miracletech

Member

• Oct 30, 2020

• #5

Is the inductor saturating? Any idea why there is a voltage drop across it?

 

rjenkinsgb

Well-Known Member

Most Helpful Member

• Oct 30, 2020

• #6

The inductor should have a few volts AC across it.
Anything else is guesswork without seeing actual waveforms...

Can you try a FET rather than a bipolar transistor?

Also, a ceramic cap directly across the circuit input may change the output somewhat, depending on the battery characteristics and the wiring to that.

 

ronsimpson

Well-Known Member

Most Helpful Member

• Oct 30, 2020

• #7

This circuit is very dependent on what transistor for M1. (current gain if TIP41 or on resistance of BSZ0920) (and L1 and Q2, R3)

 

Miracletech

Member

• Oct 30, 2020

• #8

ZD1 is 5.1V Zener diode.
Is the voltage drop normal?

 

Miracletech

Member

• Oct 30, 2020

• #9

I don't have a FET. I'm so messed up without it.
The BJT is okay right? I've seen the TIP41C being used many times, so its okay right?

 

Miracletech

Member

• Oct 30, 2020

• #10

Wait a minute, is it that the on Resistance or VCEsat of the transistor determines the current output of the Boost converter? Lower resistance means more current right?
Or, Am I mistaken somewhere?

 

Nigel Goodwin

Super Moderator

Most Helpful Member

• Oct 30, 2020

• #11

Miracletech said:

I don't have a FET. I'm so messed up without it.
The BJT is okay right? I've seen the TIP41C being used many times, so its okay right?

Click to expand...

No, the circuit is designed for an FET, if you want to use a bi-polar transistor find a circuit designed to use one.

 

Miracletech

Member

• Oct 30, 2020

• #12

"The FET Can Be replaced with a BJT"
The author said. So why can't I?
Its math!

 

Miracletech

Member

• Oct 30, 2020

• #13

My thing is; Is it the transistor that is causing the voltage drop?
If my calculation is correct, The resistance of the transistor determines the output current?

 

S

schmitt trigger

Well-Known Member

Most Helpful Member

• Oct 30, 2020

• #14

There are *very fundamental* differences between bipolar and Mosfet transistors.

If the author said that this circuit would work, as is, with no component changes, and with a VERY SLOW transistor like the TIP41, ask him to show you how.

 

ronsimpson

Well-Known Member

Most Helpful Member

• Oct 30, 2020

• #15

What do you know about L1? Do you have any data? Can you send a picture? If L1 saturates you can not get any more power than that.
--edited--
To get 500mA out, it will take 1A (about) from the 3.7V battery.

 

Last edited: Oct 30, 2020

Miracletech

Member

• Oct 30, 2020

• #16

I know that it will draw high current from the battery, That's why I'm using A 10,000MAH Li-ION Battery.

 

Miracletech

Member

• Oct 30, 2020

• #17

You can call a BJT Old, But Not slow.
BJT's are still wonderful devices that I have used to design circuits.
You can call a BJT Old, But not slow.
I love them, And I still use them.

 

Miracletech

Member

• Oct 30, 2020

• #18

OK, L1 I upgraded to a 4,7Uh inductor from a laptop (I'm good a t salvaging
Edit: The voltage drop across the inductor is still present.
Edit 2: from the wire diameter, It is a 3A inductor, Which is Good enough.

 

Last edited: Oct 30, 2020

rjenkinsgb

Well-Known Member

Most Helpful Member

• Oct 31, 2020

• #19

Did you add the base-emitter resistor I suggested, to Q2?
And a capacitor across the circuit power input?

There is nothing at all wrong with bipolar transistors - IF you use an appropriate one for an application.

A FET gate is purely voltage-driven, with a bit of capacitance.

Bipolar transistors are current driven. The TIP41 could need anything from 30mA up to 100mA to fully turn on with a one amp load, so even ignoring its speed, it's very marginal in a circuit such as that.

 

Miracletech

Member

• Oct 31, 2020

• #20

I did, Same result :-(
I added a 220UF 10V capacitor at the input.

 

Last edited: Oct 31, 2020