Offset remover circuit.

Hi

Im currently doing revision on the circuit attached ,please can someone explain with more detail the circuit operating principle especially how can I know the voltage drop across D1 before signal (Vin) is applied with an example (with numbers)?

Thank you in advanced

SM

Diode D2 acts as a zener stabilizing a voltage of about 0,6 V on its anode.
Through the resistor the voltage is routed to the anode of D1 and on the cathode of D1, if the two diodes are exactly the same, you will see 0 V.
On the positive edge of the sinuosid incoming at the input, it will pass through D1 without any voltage drop on that diode and across the resistor connected from D1 cathode and ground you will see the peaks of the rectified voltage without the 0,6 Drop.
To work properly the two diodes D1 and D2 must be very well matched.
Mandi

R1 is the 1K resistor between +5V and node (a) at D2 anode.
R2 is the 1K resistor between node (a) and node (b) at D1 anode.
R3 is the 100K resistor between node (out or c) at D1 cathode and ground.

By the DC analysis, we get:
From +5V, a current I(R1) will flow to ground via R1 and D2.
Therefore D2 is forward biased hence Va is about 0.7 V.
I(R1) = ( Vcc - Va ) / R1 + I(R2) = ( 5 - 0.7 ) / 1 + I(R2) = 4.3 + I(2) mA

The current I(R2) will flow from node (a) to ground, via R2, D1 and R3.
Also D1 is forward biased, Vbc (or Vb - Vout) is about 0.55 V. It is relatively low because D1 forward current is rather small.
I(R2) = ( Va - Vbc ) / ( R2 + R3 ) = ( 0.7 - 0.55 ) / ( 1 + 100 ) = 0.15 / 101 = 0.0015 mA = 1.5 uA

I(R1) = 4.3 + 0.0015 ≈ 4.3 mA

Vb = Va + I(R2) * R2 = 0.7 + 0.0015 * 1 ≈ 0.7 V.
Vout = Vb - Vbc = 0.7 - 0.55 = 0.15 V

I(R3) = I(D1) = I(R2) = 1.5 uA

Note:
The diode forward voltages 0.7 V @ 4.3 mA and 0.55 V @ 1.5 uA are just estimated values. In a real circuit, they are likely different though the differences won't be big.

KerimF said:

.

By the DC analysis, we get:
From +5V, a current I(R1) will flow to ground via R1 and D2.
Therefore D2 is forward biased hence Va is about 0.7 V.
I(R1) = ( Vcc - Va ) / R1 + I(R2) = ( 5 - 0.7 ) / 1 + I(R2) = 4.3 + I(2) mA

.

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Many thanks Kerim for your reply !
Why you inserted the current flowing in R2 in the equation when (Vcc-Vdiode)/R1 is sufficient

Thanks
sm

Why you inserted the current flowing in R2 in the equation when (Vcc-Vdiode)/R1 is sufficient?

Click to expand...

As you know, the node (a) has 3 branches hence its nodal equation has 3 currents. I included I(R2), though it is neglibible here, just to complete the nodal equation

KerimF said:

R1 is the 1K resistor between +5V and node (a) at D2 anode.
R2 is the 1K resistor between node (a) and node (b) at D1 anode.
R3 is the 100K resistor between node (out or c) at D1 cathode and ground.

By the DC analysis, we get:
From +5V, a current I(R1) will flow to ground via R1 and D2.
Therefore D2 is forward biased hence Va is about 0.7 V.
I(R1) = ( Vcc - Va ) / R1 + I(R2) = ( 5 - 0.7 ) / 1 + I(R2) = 4.3 + I(2) mA

The current I(R2) will flow from node (a) to ground, via R2, D1 and R3.
Also D1 is forward biased, Vbc (or Vb - Vout) is about 0.55 V. It is relatively low because D1 forward current is rather small.
I(R2) = ( Va - Vbc ) / ( R2 + R3 ) = ( 0.7 - 0.55 ) / ( 1 + 100 ) = 0.15 / 101 = 0.0015 mA = 1.5 uA

I(R1) = 4.3 + 0.0015 ≈ 4.3 mA

Vb = Va + I(R2) * R2 = 0.7 + 0.0015 * 1 ≈ 0.7 V.
Vout = Vb - Vbc = 0.7 - 0.55 = 0.15 V

I(R3) = I(D1) = I(R2) = 1.5 uA

Note:
The diode forward voltages 0.7 V @ 4.3 mA and 0.55 V @ 1.5 uA are just estimated values. In a real circuit, they are likely different though the differences won't be big.

Click to expand...

Thanks for your help is much appreciated!
If I am right you assumed that the voltage drop across D1 is about 0.55V and found the currents accordingly to it.
I have two questions,
1. You cant calculate the voltage across D1 you have to assume it, Am I right?
2.If a signal is applied in Vin , Vout=Vin I think the signal is not loaded because the current is already passing into D1 due to Vcc that creates 0.55v , or otherwise? please state

Many Thanks in Advanced
sm

You cant calculate the voltage across D1 you have to assume it, Am I right?

Click to expand...

You are right. Please note that a diode is a non-linear element. One of the methods to analyse a non-linear circuit is by trial and error mainly when very accurate results are not necessary.

If a signal is applied in Vin , Vout=Vin I think the signal is not loaded because the current is already passing into D1

Click to expand...

Let us not forget, the internal dynamic resistance of D1 at its biasing current (slope of its V versus I characteristic). It is higher at low currents.
Typically, R_dyn = 26 Ohm / I (in mA).
So @1.5 uA, R_dyn ≈ 17 Kohm

Also let us not forget the equivalent impedance of the input voltage source.

saviourm said:

2.If a signal is applied in Vin , Vout=Vin I think the signal is not loaded because the current is already passing into D1 due to Vcc that creates 0.55v please state

Click to expand...

Just to clarify

The last statement Is it right?
Last question. The majority of the signal(Vin) passes through the Diode D1 (via nodeC) and not through R3 due to its high resistance(100KOhms) .Am I right?

Thanks you a lot
SaviourM

The last statement Is it right?

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No, unless the forward dynamic resistance of D1 is negligible relative to R3 (100K). In our case, R_dyn ≈ 17 Kohm hence it is not negligible.
Please note that if Vin (unloaded) has a high Vpeak, when it is applied at the circuit input, the positive peak of Vout will be close to Vpeak if the AC source impedance is low (< 10 Ohm) since Vin source is also loaded by R2 (1K).

The majority of the signal(Vin) passes through the Diode D1 (via nodeC) and not through R3 due to its high resistance(100KOhms) .Am I right?

Click to expand...

This is not true always. It depends on the forward dynamic resistance of D1. And don't forget that Vin is also loaded by R2 which is of a relatively low value (1K only).

In any where to remove a offset voltage a capacitor series with resistor connected to the ground will do it (like LPF)... It will remove the DC component and will give the AC in output..

If you need to change the offset value connect the resistor to the voltage level, how much you wanted to put the offset.. It will work as it was..

The input signal Voltage will drop about 0.15V seen on Vout for D1 to become full conductive in other words am I correct?
Thanks

The input signal Voltage will drop about 0.15V seen on Vout for D1 to become full conductive in other words, am I correct

Click to expand...

Sorry I didn't get well your point

The analysis of this circuit is not easy as you might think because it is a non-linear one.

For example, by increasing the AC amplitude of Vin, a DC voltage will be developed on the input capacitor making the average potential of node (b), at D1 anode, lower than of the other terminal at (Vin). The reason is that the capacitor current from left to right is higher than of the opposite direction. The former one sees only R2 (after neglecting the small dynamic resistance of D2 in series and the load 100K with D1). The current from right to left has to pass via R1 and R2 (+5V source acts as a short for AC). Therefore it is smaller (about half). In other words, the input capacitor is charged differently in the positive and negative half cycle.

I suggest using a simulator to observe the exact shape of Vout when Vin is increased from a few millivolts to several volts also the average voltage shift of the input capacitor.
If you didn't have the time to install a simulator (like the one I have, free LTspice) and you cannot test it experimentally, you have no choice but to understand it the best you can since it is non-linear hence the level of Vin is crucial in its analysis.

Dear Kerim
Many thanks for your help!
SaviourM