# [Help] Analyzing battery charger circuit

#### simpsons310

Joined Jun 3, 2017
6
Hi, I'm stucking with this battery charger and have some questions.
1, The current on R2 ( in circuit is 1mA ) but my result isn't similar. My result is 1.7mA, i have used 39V, when the current flow through 2 diode, the voltage is 37.6V, add with Vbe = 0.62V, i got 38.22. Thus, IR2 = 1.7mA. If I used detail in circuit, Vbe would be 0.93. So which is correct and how to calculate Vbe, I have read in datasheet, it only shows me the maximum of Vbe?
2, The second problem is beta of transistor, you can see Q1, Q2 in circuit, with IB2 = 0.2mA, i have IE2 = 10mA -> beta = 50 and IB1 = 10mA, IE = 1A -> beta = 100. So how to determine beta?
3, I have tried to simulate on Proteus, but result is different (Image below). I replace source circuit by DC source and add R7 resistor in my simulation.
4, I'm not clearly understand of diode D4, Q4 and Q5 function. Anyone can explain those ones?

I have to use google translate to translate my questions , so please ignore English grammar error.
Thank you very much.

#### Jony130

Joined Feb 17, 2009
5,230
The current on R2 ( in circuit is 1mA ) but my result isn't similar. My result is 1.7mA, i have used 39V, when the current flow through 2 diode, the voltage is 37.6V, add with Vbe = 0.62V, i got 38.22. Thus, IR2 = 1.7mA. If I used detail in circuit, Vbe would be 0.93. So which is correct and how to calculate Vbe, I have read in datasheet, it only shows me the maximum of Vbe?
It is impossible to know the exact value for Vbe. Typical we assume Vbe = 0.6V...0.7V.
The voltage across R2 will be around (Vd1 + Vd2 ) - VbeQ3 ≈ 1.2V -0.6V = 0.6V and the current around 1.3mA.
2, The second problem is beta of transistor, you can see Q1, Q2 in circuit, with IB2 = 0.2mA, i have IE2 = 10mA -> beta = 50 and IB1 = 10mA, IE = 1A -> beta = 100. So how to determine beta?
Why you whant to know the beta value ?
I have tried to simulate on Proteus, but result is different (Image below). I replace source circuit by DC source and add R7 resistor in my simulation.
But this is normal this values are "load dependent" and transistor dependemt. Transistor beta is not constant but will change with the collector current. Also the Vbe value is also not constant but Ic current dependent.
4, I'm not clearly understand of diode D4, Q4 and Q5 function
D4 is a reverse battery protection diode.
Q4 is a voltage regulator "error amplifier". This BJT together with D3 and R6, R7 and R7 sets the maximum charging voltage.
Vo_max = (1 +(R6+R7)/R8 ) * Vzd3 + VbeQ4 ≈ 6.8V* (1 + 57/10 ) = 45V

Q5 is a current limiting transistor. This BJT together with R5 set the maximum charging current (I_Lmax ≈ 0.7V/R5 ≈ 1.2A).

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#### ebp

Joined Feb 8, 2018
2,332
D4 will not protect against a reversed battery.

#### MrAl

Joined Jun 17, 2014
8,136
Hello there,

To start, D4 is not a reverse battery protection diode it is a "input power disconnect" diode. That is, if the input power goes away the battery does not try to conduct to the emitters of the other transistors and thus case a reverse bias that is too high for the transistors base emitter diodes, although the circuit may not be complete as is because there is no pulldown resistors. It could have also been put there by mistake thinking it was going to protect against a battery being connected in reverse, but analyze what happens when you connect the negative of a battery to a diode cathode and you can see that it will actually conduct more than if you connected the positive lead to the cathode. Thus, no battery reverse protection The transistor Q5 conducts when the current gets too high and then it would steal base current from Q1, thus limiting the output current level.

The transistor Q4 conducts when the base is higher than the zener voltage Vz plus the base emitter diode drop Vd which is Vz+Vd. When it conducts, it also steals current from the base node of Q1 thus limiting the voltage, which acts as voltage regulation.
The temperature coefficient of the zener could be better than the base emitter diode, so the temperature stability depends on how much the base emitter diode voltage changes with temperature. For a perfect zener of 6.2v and base emitter drop of 0.7v just for reference, the temperature stability would be roughly 300 parts per million, but some 6.2v zeners are made to work specifically with silicon diodes so that their temp co is opposite to the diode, thus achieving a lot better temperature compensation, and that was the old school way of doing things while a more modern approach would use an op amp and voltage reference diode.

The base emitter diode of a transistor is not an exact figure but you can get some idea sometimes from the data sheet for the transistor. What you would look for in this circuit mainly is a constant base voltage over a useful temperature range, and the more constant it is the better the zener and base emitter diode work together. So you dont really need to know the base emitter voltage, just the base voltage, and that's only to make sure it works right. Otherwise you just adjust the pot If i had to guess, i would say the designer thought the diode D4 was for protection against a reverse battery connection and that is because the diode does not function yet as it should because there is a missing part.
One more possibility though is that the diode is needed to get the transistor outputs a tiny bit higher than the output so that the feedback circuit and pass transistor remain biased correctly for all possible operation modes. We'd have to analyze the Q point of the circuit for various input/output/load conditions to find out if this is the case.

To add true reverse battery connection protection, you have to add a diode in anti parallel with the output and a fuse in series with the very output of the circuit.
.

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